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checked that these do not depend on the choices of representatives for the equivalence classes, and that we obtain in this way an S 1A-module S 1M D f m s j m 2 M; s 2 S g and a homomorphism m 7! m 1 W M iS! S 1M of A-modules whose kernel is fa 2 M j sa D 0 for some s 2 S g: PROPOSITION 1.17. The elements of S act invertibly on S 1M, and every homomorphism from M to an A-module N with this property factors uniquely through iS, iS M S 1M 9Š N: PROOF. Similar to the proof of 1.10. PROPOSITION 1.18. The functor M S 1M is exact. In other words, if the sequence of A-modules ˇ! M 00 is exact, then so also is the sequence of S 1A-modules M 0 ˛! M S 1M 0 S 1˛! S 1M S 1ˇ! S 1M 00: PROOF. Because ˇ ı˛ D 0, we have 0 D S 1.ˇ ı˛/ D S 1ˇ ıS 1˛. Therefore Im.S 1˛/ 2 Ker.S 1ˇ/, where m 2 M and s 2 S. Then Ker.S 1ˇ/. For the reverse inclusion, let m s D 0 and so, for some t 2 S, we have tˇ.m/ D 0. Then ˇ.tm/ D 0, and so tm D ˛.m0/ ˇ.m/ s for some m0 2 M 0. Now m s D tm ts D ˛.m0/ ts 2 Im.S 1˛/: PROPOSITION 1.19. Let A be a ring, and let M be an A-module. The canonical map M! YfMm j m a maximal ideal in Ag is injective. 22 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let m 2 M map to zero in all Mm. The annihilator a D fa 2 A j am D 0g of m is an ideal in A. Because m maps to zero Mm, there exists an s 2 A X m such

that sm D 0. Therefore a is not contained in m. Since this is true for all maximal ideals m, a D A, and so it contains 1. Now m D 1m D 0. COROLLARY 1.20. An A-module M D 0 if Mm D 0 for all maximal ideals m in A. PROOF. Immediate consequence of the lemma. PROPOSITION 1.21. Let A be a ring. A sequence of A-modules is exact if and only if is exact for all maximal ideals m. M 0 ˛! M ˇ! M 00 M 0 m ˛m! Mm ˛m! M 00 m (*) (**) PROOF. The necessity is a special case of Proposition 1.18. For the sufficiency, let N D Ker.ˇ/= Im.˛/. Because the functor M Mm is exact, Nm D Ker.ˇm/= Im.˛m/: If (**) is exact for all m, then Nm D 0 for all m, and so N D 0 (by 1.20). But this means that (*) is exact. COROLLARY 1.22. A homomorphism M! N of A-modules is injective (resp. surjective) if and only if Mm! Nm is injective (resp. surjective) for all maximal ideals m: PROOF. Apply the proposition to 0! M! N (resp. M! N! 0). Direct limits A directed set is a pair.I; / consisting of a set I and a preorder4 on I such that for all i; j 2 I, there exists a k 2 I with i; j k. Let.I; / be a directed set, and let A be a ring. A direct system of A-modules indexed W Mi! Mj /i j of k all i j k.5 An A-module M j for all i j is the direct by.I; / is a family.Mi /i 2I of A-modules together with a family.˛i j D idMi and ˛j A-linear maps such that ˛i i together with A-linear maps ˛i W Mi! M such that ˛i D ˛j ı ˛i limit of the system.Mi

; ˛j D ˛i ı ˛i j k (a) M D S (b) mi 2 Mi maps to zero in M if and only if it maps to zero in Mj for some j i. i / if i 2I ˛i.Mi /, and Direct limits of A-algebras are defined similarly. PROPOSITION 1.23. For every multiplicative subset S of A, S 1A'lim! Ah, where h runs over the elements of S (partially ordered by division). 4A preorder is a reflexive transitive binary relation. 5Regard I as a category with Hom.a; b/ empty unless a b, in which case it contains a single element. Then a direct system is a functor from I to the category of A-modules. c. Unique factorization 23 PROOF. When hjh0, say, h0 D hg, there is a canonical homomorphism a h0 W Ah! h Ah0, and so the rings Ah form a direct system indexed by the set S. When h 2 S, the W Ah! S 1A homomorphism A! S 1A extends uniquely to a homomorphism a h (1.10), and these homomorphisms are compatible with the maps in the direct system. Now it is easy to see that S 1A satisfies the conditions to be the direct limit of the Ah. 7! a h 7! ag c. Unique factorization Let A be an integral domain. An element a of A is irreducible if it is not zero, not a unit, and admits only trivial factorizations, i.e., a D bc H) b or c is a unit. An element a is said to be prime if.a/ is a prime ideal, i.e., ajbc H) ajb or ajc. An integral domain A is called a unique factorization domain (or a factorial domain) if every nonzero nonunit in A can be written as a finite product of irreducible elements in exactly one way up to units and the order of the factors: Principal ideal domains, for example, Z and kŒX, are unique factorization domains, PROPOSITION 1.24. Let A be an integral domain, and let a be an element of A that is neither zero nor a unit. If a is prime, then a

is irreducible, and the converse holds when A is a unique factorization domain. PROOF. Assume that a is prime. If a D bc, then a divides bc and so a divides b or c. Suppose the first, and write b D aq. Now a D bc D aqc, which implies that qc D 1 because A is an integral domain, and so c is a unit. Therefore a is irreducible. For the converse, assume that a is irreducible and that A is a unique factorization domain. If ajbc, then bc D aq, some q 2 A: On writing each of b, c, and q as a product of irreducible elements, and using the uniqueness of factorizations, we see that a differs from one of the irreducible factors of b or c by a unit. Therefore a divides b or c. COROLLARY 1.25. Let A be an integral domain. If A is a unique factorization domain, then every prime ideal of height 1 is principal. PROOF. Let p be a prime ideal of height 1. Then p contains a nonzero element, and hence an irreducible element a. We have p.a/.0/. As.a/ is prime and p has height 1, we must have p D.a/. PROPOSITION 1.26. Let A be an integral domain in which every nonzero nonunit element is a finite product of irreducible elements. If every irreducible element of A is prime, then A is a unique factorization domain. PROOF. Suppose that a1 am D b1 bn (9) 24 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA with the ai and bi irreducible elements in A. As a1 is prime, it divides one of the bi, which we may suppose to be b1. As b1 is irreducible, b1 D ua1 for some unit u. On cancelling a1 from both sides of (9), we obtain the equality a2 am D.ub2/b3 bn: Continuing in this fashion, we find that the two factorizations are the same up to units and the order of the factors. PROPOSITION 1.27 (GAUSS’S LEMMA). Let A be a unique factorization domain

with field of fractions F. If f.X/ 2 AŒX factors into the product of two nonconstant polynomials in F ŒX, then it factors into the product of two nonconstant polynomials in AŒX. PROOF. Let f D gh in F ŒX. For suitable c; d 2 A, the polynomials g1 D cg and h1 D dh have coefficients in A, and so we have a factorization cdf D g1h1 in AŒX. If an irreducible element p of A divides cd, then, looking modulo.p/, we see that 0 D g1 h1 in.A=.p// ŒX. According to Proposition 1.24,.p/ is prime, and so.A=.p// ŒX is an integral domain. Therefore, p divides all the coefficients of at least one of the polynomials g1; h1, say g1, so that g1 D pg2 for some g2 2 AŒX. Thus, we have a factorization.cd=p/f D g2h1 in AŒX. Continuing in this fashion, we can remove all the irreducible factors of cd, and so obtain a factorization of f in AŒX. Let A be a unique factorization domain. A nonzero polynomial f D a0 C a1X C C amX m in AŒX is said to be primitive if the coefficients ai have no common factor (other than units). Every polynomial f in F ŒX can be written f D c.f / f1 with c.f / 2 F and f1 primitive. The element c.f /, which is well-defined up to multiplication by a unit, is called the content of f. Note that f 2 AŒX if and only if c.f / 2 A. LEMMA 1.28. The product of two primitive polynomials is primitive. PROOF. Let f D a0 C a1X C C amX m g D b0 C b1X C C bnX n; be primitive polynomials, and let p be an irreducible element of A. Let ai0, i0 m, be the first coefficient of f not divisible by p,

and let bj0, j0 n, the first coefficient of g not divisible by p. Then all the terms in the sum P i Cj Di0Cj0 ai bj are divisible by p, except ai0bj0, which is not divisible by p. Therefore, p doesn’t divide the.i0 C j0/th-coefficient of fg. We have shown that no irreducible element of A divides all the coefficients of fg, which must therefore be primitive. c. Unique factorization 25 PROPOSITION 1.29. Let A be a unique factorization domain with field of fractions F. For polynomials f; g 2 F ŒX, c.fg/ D c.f / c.g/I hence every factor in AŒX of a primitive polynomial is primitive. PROOF. Let f D c.f / f1 and g D c.g/ g1 with f1 and g1 primitive. Then fg D c.f / c.g/ f1g1 with f1g1 primitive, and so c.fg/ D c.f /c.g/. COROLLARY 1.30. An element f 2 AŒX is irreducible if and only if either (a) f is constant, say f D a, with a an irreducible element of A, or (b) f is a nonconstant primitive polynomial that is irreducible in F ŒX. PROOF. (: If f is as in (a) and f D gh in AŒX, then g and h both lie in A and one must be a unit in A, and hence a unit in AŒX. If f is as in (b) and f D gh, then one of g or h must be constant because otherwise f would be reducible in F ŒX. If it is g that is constant, then, because f is primitive, g must be a unit in A, hence in AŒX. ): Let f 2 AŒX be irreducible. If f is a constant polynomial, say f D a, then a is obviously irreducible in A. If f nonconstant, then it must be primitive because otherwise f D c.f / f1 would be a nontrivial factor

ization in AŒX. It must also be irreducible in F ŒX, because otherwise it would have a nontrivial factorization in AŒX (by 1.27). PROPOSITION 1.31. If A is a unique factorization domain, then so also is AŒX. PROOF. We shall check that A satisfies the conditions of Proposition 1.26. Let f 2 AŒX, and write f D c.f /f1. Then c.f / is a product of irreducible elements in A, and f1 is a product of irreducible primitive polynomials. This shows that f is a product of irreducible elements in AŒX. Let a be an irreducible element of A. If a divides fg, then it divides c.fg/ D c.f /c.g/. As a is prime (1.24), it divides c.f / or c.g/, and hence also f or g. Let f be an irreducible primitive polynomial in AŒX. Then f is irreducible in F ŒX, and so if f divides the product gh of g; h 2 AŒX, then it divides g or h in F ŒX. Suppose the first, and write f q D g with q 2 F ŒX. Then c.q/ D c.f /c.q/ D c.f q/ D c.g/ 2 A, and so q 2 AŒX. Therefore f divides g in AŒX. We have shown that every element of AŒX is a product of irreducible elements and that every irreducible element of AŒX is prime, and so AŒX is a unique factorization domain (1.26). Polynomial rings Let k be a field. The elements of the polynomial ring kŒX1; : : : ; Xn are finite sums X ca1anX a1 1 X an n ; ca1an 2 k; aj 2 N; with the obvious notions of equality, addition, and multiplication. In particular, the monomials form a basis for kŒX1; : : : ; Xn as a k-vector space. The degree, deg.f /, of a nonzero po

lynomial f is the largest total degree of a monomial occurring in f with nonzero coefficient. Since deg.fg/ D deg.f / C deg.g/, kŒX1; : : : ; Xn is an integral domain and kŒX1; : : : ; Xn D k. An element f of kŒX1; : : : ; Xn is irreducible if it is nonconstant and f D gh H) g or h is constant. 26 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA THEOREM 1.32. The ring kŒX1; : : : ; Xn is a unique factorization domain. PROOF. Note that kŒX1; : : : ; Xn1ŒXn D kŒX1; : : : ; Xn: This simply says that every polynomial f in n symbols X1; : : : ; Xn can be expressed uniquely as a polynomial in Xn with coefficients in kŒX1; : : : ; Xn1, f.X1; : : : ; Xn/ D a0.X1; : : : ; Xn1/X r n C C ar.X1; : : : ; Xn1/: Since, as we noted, kŒX is a unique factorization domain, the theorem follows by induction from Proposition 1.31. COROLLARY 1.33. A nonzero proper principal ideal.f / in kŒX1; : : : ; Xn is prime if and only if f is irreducible. PROOF. Special case of Proposition 1.24. d. Integral dependence Let A be a subring of a ring B. An element ˛ of B is said to be6 integral over A if it is a root of a monic7 polynomial with coefficients in A, i.e., if it satisfies an equation ˛n C a1˛n1 C C an D 0; ai 2 A: If every element of B is integral over A, then B is said to be integral over A. In the next proof, we shall need to apply a variant of Cramer’s rule: if x1; : : : ; xm is a solution to the system of linear equations m X j D1

cij xj D 0; i D 1; : : : ; m; with coefficients in a ring A, then det.C / xj D 0; j D 1; : : : ; m; (10) where C is the matrix of coefficients. To prove this, expand out the left hand side of 0 B @ det c11 ::: cm1 : : : : : : c1 j 1 ::: cm j 1 P P i c1i xi ::: i cmi xi c1 j C1 ::: cm j C1 : : : : : : c1m ::: cmm 1 C A D 0 using standard properties of determinants. An A-module M is faithful if aM D 0, a 2 A, implies that a D 0. PROPOSITION 1.34. Let A be a subring of a ring B. An element ˛ of B is integral over A if and only if there exists a faithful AŒ˛-submodule M of B that is finitely generated as an A-module. 6More generally, if f W A! B is an A-algebra, an element ˛ of B is integral over A if it satisfies an equation ˛n C f.a1/˛n1 C C f.an/ D 0; ai 2 A: Thus, ˛ is integral over A if and only if it is integral over the subring f.A/ of B. 7A polynomial is monic if its leading coefficient is 1, i.e., f.X/ D X nC terms of degree less than n. d. Integral dependence 27 PROOF. )W Suppose that ˛n C a1˛n1 C C an D 0; ai 2 A: Then the A-submodule M of B generated by 1, ˛,..., ˛n1 has the property that ˛M M, and it is faithful because it contains 1. (W Let M be a faithful AŒ˛-submodule of B admitting a finite set fe1; : : : ; eng of generators as an A-module. Then, for each i, ˛ei D P aij ej, some aij 2 A: We can rewrite this system of equations as.˛ a11/e1 a12e2 a13e3 D

0 a21e1 C.˛ a22/e2 a23e3 D 0 D 0: Let C be the matrix of coefficients on the left-hand side. Then Cramer’s formula tells us that det.C / ei D 0 for all i. As M is faithful and the ei generate M, this implies that det.C / D 0. On expanding out the determinant, we obtain an equation ˛n C c1˛n1 C c2˛n2 C C cn D 0; ci 2 A: PROPOSITION 1.35. An A-algebra B is finite if it is generated as an A-algebra by a finite set of elements each of which is integral over A. PROOF. Suppose that B D AŒ˛1; : : : ; ˛m and that C ai1˛ni 1 C C ai ni D 0; ˛ni i i aij 2 A; i D 1; : : : ; m. Any monomial in the ˛i divisible by some ˛ni is equal (in B) to a linear combination of i monomials of lower degree. Therefore, B is generated as an A-module by the finite set of monomials ˛r1 1 m, 1 ri < ni. ˛rm COROLLARY 1.36. An A-algebra B is finite if and only if it is finitely generated and integral over A. PROOF. (: Immediate consequence of 1.35. ): We may replace A with its image in B. Then B is a faithful AŒ˛-module for all ˛ 2 B (because 1B 2 B), and so 1.34 shows that every element of B is integral over A. As B is finitely generated as an A-module, it is certainly finitely generated as an A-algebra. PROPOSITION 1.37. Consider rings A B C. If B is integral over A and C is integral over B, then C is integral over A. PROOF. Let 2 C. Then n C b1 n1 C C bn D 0 for some bi 2 B. Now AŒb1; : : : ; bn is finite over A (see 1.35), and AŒb1; : : :

; bnŒ is finite over AŒb1; : : : ; bn, and so it is finite over A. Therefore is integral over A by 1.34. THEOREM 1.38. Let A be a subring of a ring B. The elements of B integral over A form an A-subalgebra of B. 28 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let ˛ and ˇ be two elements of B integral over A. Then AŒ˛; ˇ is finitely generated as an A-module (1.35). It is stable under multiplication by ˛ ˙ ˇ and ˛ˇ and it is faithful as an AŒ˛ ˙ ˇ-module and as an AŒ˛ˇ-module (because it contains 1A). Therefore 1.34 shows that ˛ ˙ ˇ and ˛ˇ are integral over A. DEFINITION 1.39. Let A be a subring of the ring B. The integral closure of A in B is the subring of B consisting of the elements integral over A. PROPOSITION 1.40. Let A be an integral domain with field of fractions F, and let ˛ be an element of some field containing F. If ˛ is algebraic over F, then there exists a d 2 A such that d˛ is integral over A. PROOF. By assumption, ˛ satisfies an equation ˛m C a1˛m1 C C am D 0; ai 2 F: Let d be a common denominator for the ai, so that dai 2 A for all i, and multiply through the equation by d m:.d˛/m C a1d.d˛/m1 C C amd m D 0: As a1d; : : : ; amd m 2 A, this shows that d˛ is integral over A. COROLLARY 1.41. Let A be an integral domain and let E be an algebraic extension of the field of fractions of A. Then E is the field of fractions of the integral closure of A in E. PROOF. In fact, the proposition shows that every element of E is a quotient ˇ=d with ˇ integral

over A and d 2 A. DEFINITION 1.42. An integral domain A is said to be integrally closed if it is equal to its integral closure in its field of fractions F, i.e., if ˛ 2 F; ˛ integral over A H) ˛ 2 A: An integrally closed integral domain is called an integrally closed domain or normal domain. PROPOSITION 1.43. Unique factorization domains are integrally closed. PROOF. Let A be a unique factorization domain, and let a=b be an element of its field of fractions. If a=b … A, then b divisible by some prime element p not dividing a. If a=b is integral over A, then it satisfies an equation.a=b/n C a1.a=b/n1 C C an D 0; ai 2 A: On multiplying through by bn, we obtain the equation an C a1an1b C C anbn D 0: The element p then divides every term on the left except an, and hence divides an. Since it doesn’t divide a, this is a contradiction. d. Integral dependence 29 Let F E be fields, and let ˛ 2 E be algebraic over F. The minimal polynomial of ˛ over F is the monic polynomial of smallest degree in F ŒX having ˛ as a root. If f is the minimal polynomial of ˛, then the homomorphism X 7! ˛W F ŒX! F Œ˛ defines an isomorphism F ŒX =.f /! F Œ˛, i.e., F Œx'F Œ˛, x $ ˛. PROPOSITION 1.44. Let A be an integrally closed domain, and let E be a finite extension of the field of fractions F of A. An element of E is integral over A if and only if its minimal polynomial over F has coefficients in A. PROOF. Let ˛ 2 E be integral over A, so that ˛m C a1˛m1 C C am D 0; some ai 2 A; m > 0. Let f.X / be the minimal polynomial of ˛ over F, and let ˛0 be a

conjugate of ˛, i.e., a root of f in some splitting field of f. Then f is also the minimal polynomial of ˛0 over F, and so (see above), there is an F -isomorphism W F Œ˛! F Œ˛0;.˛/ D ˛0: On applying to the above equation we obtain the equation ˛0m C a1˛0m1 C C am D 0; which shows that ˛0 is integral over A. As the coefficients of f are polynomials in the conjugates of ˛, it follows from (1.38) that the coefficients of f.X/ are integral over A. They lie in F, and A is integrally closed, and so they lie in A. This proves the “only if” part of the statement, and the “if” part is obvious. COROLLARY 1.45. Let A F E be as in the proposition, and let ˛ be an element of E integral over A. Then NmE=F.˛/ 2 A, and ˛ divides NmE=F.˛/ in AŒ˛. PROOF. Let f.X/ D X m C a1X m1 C C am be the minimal polynomial of ˛ over F. Then Nm.˛/ D.1/mnan (FT 5.45), and so Nm.˛/ 2 A. Because f.˛/ D 0, m, where n D ŒEW F Œ˛ 0 D an1 m.˛m C a1˛m1 C C am/ m ˛m1 C C an1 D ˛.an1 m am1/ C.1/mn Nm.˛/; and so ˛ divides NmE=F.˛/ in AŒ˛. COROLLARY 1.46. Let A be an integrally closed domain with field of fractions F, and let f.X/ be a monic polynomial in AŒX. Then every monic factor of f.X/ in F ŒX has coefficients in A. PROOF. It suffices to prove this for an irreducible mon

ic factor g of f in F ŒX. Let ˛ be a root of g in some extension field of F. Then g is the minimal polynomial of ˛. As ˛ is a root of f, it is integral over A, and so g has coefficients in A. PROPOSITION 1.47. Let A B be rings, and let A0 be the integral closure of A in B. For any multiplicative subset S of A, S 1A0 is the integral closure of S 1A in S 1B. 30 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let b s 2 S 1A0 with b 2 A0 and s 2 S. Then bn C a1bn1 C C an D 0 for some ai 2 A, and so n b s C a1 s b s n1 C C an sn D 0: Therefore b=s is integral over S 1A. This shows that S 1A0 is contained in the integral closure of S 1B. For the converse, let b=s (b 2 B, s 2 S ) be integral over S 1A. Then n b s C a1 s1 b s n1 C C an sn D 0: for some ai 2 A and si 2 S. On multiplying this equation by sns1 sn, we find that s1 snb 2 A0, and therefore that b s 2 S 1A0. D s1snb ss1sn COROLLARY 1.48. Let A B be rings, and let S be a multiplicative subset of A. If A is integrally closed in B, then S 1A is integrally closed in S 1B. PROOF. Special case of the proposition in which A0 D A. PROPOSITION 1.49. The following conditions on an integral domain A are equivalent: (a) A is integrally closed; (b) Ap is integrally closed for all prime ideals p; (c) Am is integrally closed for all maximal ideals m. PROOF. The implication (a))(b) follows from 1.48, and (b))(c) is obvious. It remains to prove (c))(a). If c is integral over A, then it is integral over each Am, and hence lies in each Am. It follows that the

ideal consisting of the a 2 A such that ac 2 A is not contained in any maximal ideal m, and therefore equals A. Hence 1 c 2 A. Let E=F be a finite extension of fields. Then.˛; ˇ/ 7! TrE=F.˛ˇ/W E E! F (11) is a symmetric bilinear form on E regarded as a vector space over F. LEMMA 1.50. If E=F is separable, then the trace pairing (11) is nondegenerate. PROOF. Let ˇ1; :::; ˇm be a basis for E as an F -vector space. We have to show that the discriminant det.Tr.ˇi ˇj // of the trace pairing is nonzero. Let 1; :::; m be the distinct F -homomorphisms of E into some large Galois extension ˝ of F. Recall (FT 5.45) that TrL=K.ˇ/ D 1ˇ C C mˇ (12) By direct calculation, we have det.Tr.ˇi ˇj // D det.P k k.ˇi ˇj // D det.P k k.ˇi / k.ˇj // D det.k.ˇi // det.k.ˇj // D det.k.ˇi //2: (by 12) d. Integral dependence 31 Suppose that det.i ˇj / D 0. Then there exist c1; :::; cm 2 ˝ such that ci i.ˇj / D 0 all j: X i By linearity, it follows that P theorem on the independence of characters (FT 5.14). i ci i.ˇ/ D 0 for all ˇ 2 E, but this contradicts Dedekind’s PROPOSITION 1.51. Let A be an integrally closed domain with field of fractions F, and let B be the integral closure of A in a separable extension E of F of degree m. There exist free A-submodules M and M 0 of E such that M B M 0. (13) If A is noetherian, then B is a finite A-algebra. PROOF. Let fˇ1; :::; ˇ

mg be a basis for E over F. According to Proposition 1.40, there exists a d 2 A such that d ˇi 2 B for all i. Clearly fd ˇ1; : : : ; d ˇmg is still a basis for E as a vector space over F, and so we may assume to begin with that each ˇi 2 B. Because the trace pairing is nondegenerate, there is a dual basis fˇ0 g of E over F with the property that Tr.ˇi ˇ0 j / D ıij for all i; j. We shall show that 1; :::; ˇ0 m Aˇ1 C Aˇ2 C C Aˇm B Aˇ0 1 C Aˇ0 2 C C Aˇ0 m: Only the second inclusion requires proof. Let ˇ 2 B. Then ˇ can be written uniquely as a linear combination ˇ D P bj ˇ0 j with coefficients bj 2 F, and we have to show that each bj 2 A. As ˇi and ˇ are in B, so also is ˇ ˇi, and so Tr.ˇ ˇi / 2 A (1.44). But j of the ˇ0 Tr.ˇ ˇi / D Tr. X bj ˇ0 j ˇi / D X bj Tr.ˇ0 j ˇi / D X bj ıij D bi : j j j Hence bi 2 A. If A is Noetherian, then M 0 is a Noetherian A-module, and so B is finitely generated as an A-module. LEMMA 1.52. Let A be a subring of a field K. If K is integral over A, then A is also a field. PROOF. Let a be a nonzero element of A. Then a1 2 K, and it is integral over A:.a1/n C a1.a1/n1 C C an D 0; ai 2 A: On multiplying through by an1, we find that a1 C a1 C C anan1 D 0; from which it follows that a1 2 A. THEOREM 1.53 (GOING-UP THEOREM). Let A

B be rings with B integral over A. (a) For every prime ideal p of A, there is a prime ideal q of B such that q \ A D p. (b) Let p D q \ A; then p is maximal if and only if q is maximal. 32 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. (a) If S is a multiplicative subset of a ring A, then the prime ideals of S 1A are in one-to-one correspondence with the prime ideals of A not meeting S (see 1.14). It therefore suffices to prove (a) after A and B have been replaced by S 1A and S 1B, where S D A p. Thus we may assume that A is local, and that p is its unique maximal ideal. In this case, for all proper ideals b of B, b \ A p (otherwise b A 3 1/. To complete the proof of (a), we shall show that for all maximal ideals n of B, n \ A D p. Consider B=n A=.n \ A/. Here B=n is a field, which is integral over its subring A=.n \ A/, and n \ A will be equal to p if and only if A=.n \ A/ is a field. This follows from Lemma 1.52. (b) The ring B=q contains A=p, and it is integral over A=p. If q is maximal, then Lemma 1.52 shows that p is also. For the converse, note that any integral domain integral over a field is a field because it is a union of integral domains finite over the field, which are automatically fields (left multiplication by an element is injective, and hence surjective, being a linear map of a finite-dimensional vector space). COROLLARY 1.54. Let A B be rings with B integral over A. Let p p0 be prime ideals of A, and let q be a prime ideal of B such that q \ A D p. Then there exists a prime ideal q0 of B containing q and such that q0 \ A D p0, B A q q0 p p0: PROOF. We have A=p B=q, and B=q is integral over A=p. According to the (1.53), there exists a prime ideal q00 in B=

q such that q00 \.A=p/ D p0=p. The inverse image q0 of q00 in B has the required properties. ASIDE 1.55. Let A be a noetherian integral domain, and let B be the integral closure of A in a finite extension E of the field of fractions F of A. Is B always a finite A-algebra? When A is integrally closed and E is separable over F, or A is a finitely generated k-algebra, then the answer is yes (1.51, 8.3). However, in 1935, Akizuki found an example of a noetherian integral domain whose integral closure in its field of fractions is not finite (according to Matsumura 1986, finding the example cost him a year’s hard struggle). F.K. Schmidt found another example at about the same time.8 e. Tensor Products Tensor products of modules Let A be a ring, and let M, N, and P be A-modules. A map W M N! P of A-modules is said to be A-bilinear if.x C x0; y/ D.x; y/ C.x0; y/;.x; y C y0/ D.x; y/ C.x; y0/;.ax; y/ D a.x; y/;.x; ay/ D a.x; y/; x; x0 2 M; y 2 N x 2 M; y; y0 2 N a 2 A; a 2 A; x 2 M; y 2 N x 2 M; y 2 N; i.e., if is A-linear in each variable. 8For a discussion of the examples Akizuki and Schmidt and generalizations, see Olberding, Bruce, One- dimensional bad Noetherian domains. Trans. Amer. Math. Soc. 366 (2014), no.8, 4067–4095. e. Tensor Products 33 An A-module T together with an A-bilinear map W M N! T is called the tensor product of M and N over A if it has the following universal property: every A-bilinear map 0W M N! T 0 factors uniquely through. M N T 0 9Š linear T 0: As usual, the universal property determines the tensor product uniquely up to a unique isomorphism.

We write it M ˝A N. Note that HomA-bilinear.M N; T /'HomA-linear.M ˝A N; T /: CONSTRUCTION Let M and N be A-modules, and let A.M N / be the free A-module with basis M N. Thus each element A.M N / can be expressed uniquely as a finite sum X ai.xi ; yi /; ai 2 A; xi 2 M; yi 2 N: Let P be the submodule of A.M N / generated by the following elements.x C x0; y/.x; y/.x0; y/; x; x0 2 M; y 2 N.x; y C y0/.x; y/.x; y0/; x 2 M; y; y0 2 N.ax; y/ a.x; y/; a 2 A;.x; ay/ a.x; y/; a 2 A; x 2 M; y 2 N x 2 M; y 2 N; and define M ˝A N D A.M N /=P: Write x ˝ y for the class of.x; y/ in M ˝A N. Then.x; y/ 7! x ˝ yW M N! M ˝A N is A-bilinear — we have imposed the fewest relations necessary to ensure this. Every element of M ˝A N can be written as a finite sum9 X ai.xi ˝ yi /; ai 2 A; xi 2 M; yi 2 N; and all relations among these symbols are generated by the following relations.x C x0/ ˝ y D x ˝ y C x0 ˝ y x ˝.y C y0/ D x ˝ y C x ˝ y0 a.x ˝ y/ D.ax/ ˝ y D x ˝ ay: The pair.M ˝A N;.x; y/ 7! x ˝ y/ has the correct universal property because any bilinear map 0W M N! T 0 defines an A-linear map A.M N /! T 0, which factors through A.M N /=K, and gives

a commutative triangle. 9“An element of the tensor product of two vector spaces is not necessarily a tensor product of two vectors, but sometimes a sum of such. This might be considered a mathematical shenanigan but if you start with the state vectors of two quantum systems it exactly corresponds to the notorious notion of entanglement which so displeased Einstein.” Georges Elencwajg on mathoverflow.net. 34 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA Tensor products of algebras Let A and B be k-algebras. A k-algebra C together with homomorphisms i W A! C and j W B! C is called the tensor product of A and B if it has the following universal property: for every pair of homomorphisms (of k-algebras) ˛W A! R and ˇW B! R, there is a unique homomorphism W C! R such that ı i D ˛ and ı j D ˇ: A i C j B ˛ 9Š ˇ R: If it exists, the tensor product, is uniquely determined up to a unique isomorphism by this property. We write it A ˝k B. Note that Homk.A ˝k B; R/'Homk.A; R/ Homk.B; R/ (homomorphisms of k-algebras). CONSTRUCTION Form the tensor product A ˝k B of A and B regarded as k-vector spaces. There is a multiplication map A ˝k B A ˝k B! A ˝k B for which.a ˝ b/.a0 ˝ b0/ D aa0 ˝ bb0. This makes A ˝k B into a ring, and the homomorphism c 7! c.1 ˝ 1/ D c ˝ 1 D 1 ˝ c makes it into a k-algebra. The maps a 7! a ˝ 1W A! C and b 7! 1 ˝ bW B! C are homomorphisms, and they make A ˝k B into the tensor product of A and B in the above sense. EXAMPLE 1.56. The algebra B,

equipped with the given map k! B and the identity map B! B, has the universal property characterizing k ˝k B, so k ˝k B'B. In terms of the constructive definition of tensor products, the isomorphism is c ˝ b 7! cbW k ˝k B! B. EXAMPLE 1.57. The ring kŒX1; : : : ; Xm; XmC1; : : : ; XmCn, equipped with the obvious inclusions kŒX1; : : : ; Xm,! kŒX1; : : : ; XmCn - kŒXmC1; : : : ; XmCn is the tensor product of kŒX1; : : : ; Xm and kŒXmC1; : : : ; XmCn. To verify this we only have to check that, for every k-algebra R, the map Homk-alg.kŒX1; : : : ; XmCn; R/! Homk-alg.kŒX1; : : :; R/ Homk-alg.kŒXmC1; : : :; R/ induced by the inclusions is a bijection. But this map can be identified with the obvious bijection RmCn! Rm Rn: In terms of the constructive definition of tensor products, the isomorphism is f ˝ g 7! fgW kŒX1; : : : ; Xm ˝k kŒXmC1; : : : ; XmCn! kŒX1; : : : ; XmCn. f. Transcendence bases 35 REMARK 1.58. (a) If.b˛/ is a family of generators (resp. basis) for B as a k-vector space, then.1 ˝ b˛/ is a family of generators (resp. basis) for A ˝k B as an A-module. (b) Let k,! ˝ be fields. Then ˝ ˝k kŒX1; : : : ; Xn'˝Œ1 ˝ X1; : : : ; 1 ˝ X

n'˝ŒX1; : : : ; Xn: If A D kŒX1; : : : ; Xn=.g1; : : : ; gm/, then ˝ ˝k A'˝ŒX1; : : : ; Xn=.g1; : : : ; gm/: (c) If A and B are algebras of k-valued functions on sets S and T respectively, then.f ˝ g/.x; y/ D f.x/g.y/ realizes A ˝k B as an algebra of k-valued functions on S T. f. Transcendence bases We review the theory of transcendence bases. For the proofs, see Chapter 9 of FT. 1.59. Elements ˛1; :::; ˛n of a k-algebra A are said to be algebraically dependent over k there exists a nonzero polynomial f.X1; :::; Xn/ 2 kŒX1; :::; Xn such that f.˛1; :::; ˛n/ D 0. Otherwise, the ˛i are said to be algebraically independent over k. Now let ˝ be a field containing k. 1.60. For a subset A of ˝, we let k.A/ denote the smallest subfield of ˝ containing k and A. For example, if A D fx1; : : : ; xmg, then k.A/ consists of the quotients f.x1;:::;xmg g.x1;:::;xmg with f; g 2 kŒX1; : : : ; Xm. A subset B of ˝ is algebraically dependent on A if each element of B is algebraic over k.A/. 1.61 (FUNDAMENTAL THEOREM). Let A D f˛1; :::; ˛mg and B D fˇ1; :::; ˇng be two subsets of ˝. Assume that (a) A is algebraically independent (over k), and (b) A is algebraically dependent on B (over k). Then m n. The reader should note the similarity of this to the statement in linear algebra with “algebraically

” replaced by “linearly”. 1.62. A transcendence basis for ˝ over k is an algebraically independent set A such that ˝ is algebraic over k.A/: 1.63. Assume that there is a finite subset A ˝ such that ˝ is algebraic over k.A/. Then (a) every maximal algebraically independent subset of ˝ is a transcendence basis; (b) every subset A minimal among those such that ˝ is algebraic over k.A/ is a transcen- dence basis; (c) all transcendence bases for ˝ over k have the same finite number of elements (called the transcendence degree, tr degk˝, of ˝ over k). 1.64. Let k L ˝ be fields. Then tr degk˝ D tr degkL C tr degL˝. More precisely, if A is a transcendence basis for L=k and B is a transcendence basis for ˝=L, then A [ B is a transcendence basis for ˝=k. 36 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA Exercises 1-1. Let k be an infinite field (not necessarily algebraically closed). Show that an f 2 kŒX1; : : : ; Xn that is identically zero on kn is the zero polynomial (i.e., has all its coefficients zero). 1-2. Find a minimal set of generators for the ideal.X C 2Y; 3X C 6Y C 3Z; 2X C 4Y C 3Z/ in kŒX; Y; Z. What standard algorithm in linear algebra will allow you to answer this question for any ideal generated by homogeneous linear polynomials? Find a minimal set of generators for the ideal.X C 2Y C 1; 3X C 6Y C 3X C 2; 2X C 4Y C 3Z C 3/: 1-3. A ring A is said to be normal if Ap is a normal integral domain for all prime ideals p in A. Show that a noetherian ring is normal if and only if it is a finite product of normal integral domains. 1-4. Prove the statement in 1.64. CHAPTER 2 Algebraic Sets a. Definition of an algebraic set An

algebraic subset V.S/ of kn is the set of common zeros of some collection S of polynomials in kŒX1; : : : ; Xn, V.S/ D f.a1; : : : ; an/ 2 kn j f.a1; : : : ; an/ D 0 all f 2 Sg: We refer to V.S/ as the zero set of S. Note that S S 0 H) V.S/ V.S 0/I — more equations means fewer solutions. Recall that the ideal a generated by a set S consists of the finite sums X fi gi ; fi 2 kŒX1; : : : ; Xn; gi 2 S: Such a sum P fi gi is zero at every point at which the gi are all zero, and so V.S/ V.a/, but the reverse conclusion is also true because S a. Thus V.S/ D V.a/ — the zero set of S is the same as the zero set of the ideal generated by S. Therefore the algebraic subsets of kn can also be described as the zero sets of ideals in kŒX1; : : : ; Xn. An empty set of polynomials imposes no conditions, and so V.;/ D kn. Therefore kn is n for kn regarded an algebraic subset. It is also the zero set of the zero ideal.0/. We write A as an algebraic set. Examples 2.1. If S is a set of homogeneous linear equations, ai1X1 C C ai nXn D 0; i D 1; : : : ; m; then V.S/ is a subspace of kn. If S is a set of nonhomogeneous linear equations, ai1X1 C C ai nXn D di ; i D 1; : : : ; m; then V.S/ is either empty or is the translate of a subspace of kn. 37 38 2. ALGEBRAIC SETS 2.2. If S consists of the single equation Y 2 D X 3 C aX C b; 4a3 C 27b2 ¤ 0; then V.S/ is an elliptic curve. For example.X 2 1/ We generally visualize algebraic sets as though the field k were R, i.e., we draw the real locus of the

curve. However, this can be misleading — see the examples 4.11 and 4.17 below. 2.3. If S consists of the single equation Z2 D X 2 C Y 2; then V.S/ is a cone. 2.4. A nonzero constant polynomial has no zeros, and so the empty set is algebraic. 2.5. The proper algebraic subsets of k are the finite subsets, because a polynomial f.X/ in one variable X has only finitely many roots. 2.6. Some generating sets for an ideal will be more useful than others for determining what the algebraic set is. For example, the ideal a D.X 2 C Y 2 C Z2 1; X 2 C Y 2 Y; X Z/ can be generated by1 X Z; Y 2 2Y C 1; Z2 1 C Y: The middle polynomial has (double) root 1, from which it follows that V.a/ consists of the single point.0; 1; 0/. b. The Hilbert basis theorem In our definition of an algebraic set, we didn’t require the set S of polynomials to be finite, but the Hilbert basis theorem shows that, in fact, every algebraic set is the zero set of a finite set of polynomials. More precisely, the theorem states that every ideal in kŒX1; : : : ; Xn can be generated by a finite set of elements, and we have already observed that a set of generators of an ideal has the same zero set as the ideal. 1This is, in fact, a Gr¨obner basis for the ideal. c. The Zariski topology 39 THEOREM 2.7 (HILBERT BASIS THEOREM). The ring kŒX1; : : : ; Xn is noetherian. As we noted in the proof of 1.32, kŒX1; : : : ; Xn D kŒX1; : : : ; Xn1ŒXn: Thus an induction argument shows that the theorem follows from the next statement. THEOREM 2.8. If A is noetherian, then so also is AŒX. PROOF. We shall show that every ideal in AŒX is finitely generated. Recall that for a polynomial f.X/ D a0

X r C a1X r1 C C ar ; ai 2 A; a0 ¤ 0; a0 is called the leading coefficient of f. Let a be a proper ideal in AŒX, and let a.i / denote the set of elements of A that occur as the leading coefficient of a polynomial in a of degree i (we also include 0). Clearly, a.i/ is an ideal in A, and a.i/ a.i C 1/ because, if cX i C 2 a, then X.cX i C / 2 a. Let b be an ideal of AŒX contained in a. Then b.i/ a.i/, and if equality holds for all i, then b D a. To see this, let f be a polynomial in a. Because b.deg f / D a.deg f /, there exists a g 2 b such that deg.f g/ < deg.f /. In other words, f D g C f1 with g 2 b and deg.f1/ < deg.f /. Similarly, f1 D g1 C f2 with g1 2 b and deg.f2/ < deg.f1/. Continuing in this fashion, we find that f D g C g1 C g2 C 2 b.. As A is noetherian, the sequence a.1/ a.2/ a.i/ eventually becomes constant, say a.d / D a.d C 1/ D : : : (and then a.d / contains the leading coefficient of every polynomial in a). For each i d, there exists a finite generating set fai1; ai2; : : : ; ai ni g of a.i/, and for each.i; j /, there exists an fij 2 a with leading coefficient aij. The ideal b of AŒX generated by the (finitely many) fij is contained in a and has the property that b.i/ D a.i / for all i. Therefore b D a, and a is finitely generated. ASIDE 2.9. One may ask how many elements are needed to generate a given ideal a in kŒX1; : : : ; Xn, or, what is not quite the same thing, how many equations are needed to define a given algebraic set V. For n D 1, the ring

kŒX is a principal ideal domain, which means that every ideal is generated by a single element. Also, if V is a linear subspace of kn, then linear algebra shows that it is the zero set of n dim.V / polynomials. All one can say in general, is that at least n dim.V / polynomials are needed to define V (see 3.45), but often more are required. Determining exactly how many is an area of active research — see 3.55. c. The Zariski topology Recall that, for ideals a and b in kŒX1; : : : ; Xn, a b H) V.a/ V.b/. PROPOSITION 2.10. There are the following relations: (a) V.0/ D kn; V.kŒX1; : : : ; Xn/ D ;I (b) V.ab/ D V.a \ b/ D V.a/ [ V.b/I (c) V.P i 2I ai / D T i 2I V.ai / for every family of ideals.ai /i 2I. 40 2. ALGEBRAIC SETS PROOF. (a) This is obvious. (b) Note that ab a \ b a; b H) V.ab/ V.a \ b/ V.a/ [ V.b/: For the reverse inclusions, observe that if a … V.a/ [ V.b/, then there exist f 2 a, g 2 b such that f.a/ ¤ 0, g.a/ ¤ 0; but then.fg/.a/ ¤ 0, and so a … V.ab/. (c) Recall that, by definition, P ai consists of all finite sums of the form P fi, fi 2 ai. Thus (c) is obvious. Proposition 2.10 shows that the algebraic subsets of A n satisfy the axioms to be the n: both the whole space and the empty set are algebraic; closed subsets for a topology on A a finite union of algebraic sets is algebraic; an arbitrary intersection of algebraic sets is n for which the closed subsets are exactly the algebraic. Thus, there is a topology on A n. The induced topology on a algebraic subsets — this is called

the Zariski topology on A subset V of A n is called the Zariski topology on V. The Zariski topology has many strange properties, but it is nevertheless of great importance. For the Zariski topology on k, the closed subsets are just the finite sets and the whole space, and so the topology is not Hausdorff (in fact, there are no disjoint nonempty open subsets at all). We shall see in 2.68 below that the proper closed subsets of k2 are finite 2 are much coarser unions of points and curves. Note that the Zariski topologies on C and C (have fewer open sets) than the complex topologies. d. The Hilbert Nullstellensatz n and the ideals of We wish to examine the relation between the algebraic subsets of A kŒX1; : : : ; Xn more closely, but first we must answer the question of when a collection S of polynomials has a common zero, i.e., when the system of equations g.X1; : : : ; Xn/ D 0; g 2 S; is “consistent”. Obviously, equations gi.X1; : : : ; Xn/ D 0; i D 1; : : : ; m are inconsistent if there exist fi 2 kŒX1; : : : ; Xn such that P fi gi D 1, i.e., if 1 2.g1; : : : ; gm/ or, equivalently,.g1; : : : ; gm/ D kŒX1; : : : ; Xn. The next theorem provides a converse to this. THEOREM 2.11 (HILBERT NULLSTELLENSATZ). 2 Every proper ideal a in kŒX1; : : : ; Xn has a zero in kn. A point P D.a1; : : : ; an/ in kn defines a homomorphism “evaluate at P ” kŒX1; : : : ; Xn! k; f.X1; : : : ; Xn/ 7! f.a1; : : : ; an/; whose kernel contains a if P 2 V.a/. Conversely, from a homomorphism 'W kŒX1; : : : ; Xn!

k of k-algebras whose kernel contains a, we obtain a point P in V.a/, namely, 2Nullstellensatz = zero-points-theorem. P D.'.X1/; : : : ; '.Xn//: e. The correspondence between algebraic sets and radical ideals 41 Thus, to prove the theorem, we have to show that there exists a k-algebra homomorphism kŒX1; : : : ; Xn=a! k. Since every proper ideal is contained in a maximal ideal (see p. 16), it suffices to prove this for a maximal ideal m. Then K defD kŒX1; : : : ; Xn=m is a field, and it is finitely generated as an algebra over k (with generators X1 C m; : : : ; Xn C m/. To complete the proof, we must show that K D k. The next lemma accomplishes this. In the next lemma, we need to allow k to be arbitrary in order to make the induction in the proof work. We shall also need to use that kŒX has infinitely many distinct monic irreducible polynomials. When k is infinite, the polynomials X a, a 2 k, are distinct and irreducible. When k is finite, we can adapt Euclid’s argument: if p1; : : : ; pr are monic irreducible polynomials in kŒX, then p1 pr C 1 is divisible by a monic irreducible polynomial distinct from p1; : : : ; pr. LEMMA 2.12 (ZARISKI’S LEMMA). Let k K be fields, not necessarily algebraically closed. If K is finitely generated as an algebra over k, then K is algebraic over k. (Hence K D k if k is algebraically closed.) In other words, if K is finitely generated as a ring over k, then it is finitely generated as a module. PROOF. We shall prove this by induction on r, the minimum number of elements required to generate K as a k-algebra. The case r D 0 being trivial, we may suppose that K D kŒx1; : : : ; xr ; r 1: If K is not algebraic over k,

then at least one xi, say x1, is not algebraic over k. Then, kŒx1 is a polynomial ring in one symbol over k, and its field of fractions k.x1/ is a subfield of K. Clearly K is generated as a k.x1/-algebra by x2; : : : ; xr, and so the induction hypothesis implies that x2; : : : ; xr are algebraic over k.x1/. From 1.40, we see that there exists a c 2 kŒx1 such that cx2; : : : ; cxr are integral over kŒx1. Let f 2 k.x1/. Then f 2 K D kŒx1; : : : ; xr and so, for a sufficiently large N, cN f 2 kŒx1; cx2; : : : ; cxr. Therefore cN f is integral over kŒx1 by 1.38, which implies that cN f 2 kŒx1 because kŒx1 is integrally closed in k.x1/ (1.43). But this contradicts the fact that that kŒx1 has infinitely many distinct monic irreducible polynomials that can occur as the denominator of an f in k.x1/. e. The correspondence between algebraic sets and radical ideals The ideal attached to a subset of kn For a subset W of kn, we write I.W / for the set of polynomials that are zero on W : I.W / D ff 2 kŒX1; : : : ; Xn j f.P / D 0 all P 2 W g: Clearly, it is an ideal in kŒX1; : : : ; Xn. There are the following relations: (a) V W H) I.V / I.W /I (b) I.;/ D kŒX1; : : : ; Xn; I.kn/ D 0I (c) I.S Wi / D T I.Wi /. Only the statement I.kn/ D 0 is (perhaps) not obvious. It says that every nonzero polynomial in kŒX1; : : : ; Xn is nonzero at some point of kn. This is true for any infinite field k (see Exercise 1-1

). Alternatively, it follows from the strong Hilbert Nullstellensatz (2.19 below). 42 2. ALGEBRAIC SETS EXAMPLE 2.13. Let P be the point.a1; : : : ; an/, and let mP D.X1 a1; : : : ; Xn an/. Clearly I.P / mP, but mP is a maximal ideal, because “evaluation at.a1; : : : ; an/” defines an isomorphism kŒX1; : : : ; Xn=.X1 a1; : : : ; Xn an/! k: As I.P / is a proper ideal, it must equal mP : PROPOSITION 2.14. Let W be a subset of kn. Then V I.W / is the smallest algebraic subset of kn containing W. In particular, V I.W / D W if W is an algebraic set. PROOF. Certainly V I.W / is an algebraic set containing W. Let V D V.a/ be another algebraic set containing W. Then a I.W /, and so V.a/ V I.W /. Radicals of ideals The radical of an ideal a in a ring A is rad.a/ defD ff j f r 2 a, some r 2 Ng: PROPOSITION 2.15. Let a be an ideal in a ring A. (a) The radical of a is an ideal. (b) rad.rad.a// D rad.a/. PROOF. (a) If a 2 rad.a/, then clearly f a 2 rad.a/ for all f 2 A. Suppose that a; b 2 rad.a/, with say ar 2 a and bs 2 a. When we expand.a C b/rCs using the binomial theorem, we find that every term has a factor ar or bs, and so lies in a. (b) If ar 2 rad.a/, then ars D.ar /s 2 a for some s. An ideal is said to be radical if it equals its radical. Thus a is radical if and only if the ring A=a is reduced, i.e., without nonzero nilpotent elements. Since integral domains are reduced, prime ideals (a fortiori, maximal ideals) are radical. Note that rad.a/

is radical (2.15b), and hence is the smallest radical ideal containing a. If a and b are radical, then a \ b is radical, but a C b need not be: consider, for example, a D.X 2 Y / and b D.X 2 C Y /; they are both prime ideals in kŒX; Y, but X 2 2 a C b, X … a C b. (See 2.22 below.) The strong Nullstellensatz For a polynomial f and point P 2 kn, f r.P / D f.P /r. Therefore f r is zero on the same set as f, and it follows that the ideal I.W / is radical for every subset W kn. In particular, I V.a/ rad.a/. The next theorem states that these two ideals are equal. THEOREM 2.16 (STRONG NULLSTELLENSATZ). For every ideal a in kŒX1; : : : ; Xn, in particular, I V.a/ D a if a is a radical ideal. I V.a/ D rad.a/I e. The correspondence between algebraic sets and radical ideals 43 PROOF. We have already noted that I V.a/ rad.a/. For the reverse inclusion, we have to show that if a polynomial h vanishes on V.a/, then hN 2 a for some N > 0. We may assume h ¤ 0. Let g1; : : : ; gm generate a, and consider the system of m C 1 equations in n C 1 symbols, gi.X1; : : : ; Xn/ D 0; 1 Y h.X1; : : : ; Xn/ D 0: i D 1; : : : ; m; If.a1; : : : ; an; b/ satisfies the first m equations, then.a1; : : : ; an/ 2 V.a/; consequently, h.a1; : : : ; an/ D 0, and.a1; : : : ; an; b/ doesn’t satisfy the last equation. Therefore, the equations are inconsistent, and so, according to the original Nullstellensatz, there exist fi 2 kŒX1; : : : ; Xn; Y such that 1 D m X i D1 fi gi C f

mC1.1 Y h/ (in the ring kŒX1; : : : ; Xn; Y ). On applying the homomorphism Xi 7! Xi Y 7! h1 W kŒX1; : : : ; Xn; Y! k.X1; : : : ; Xn/ to the above equality, we obtain the identity 1 D m X i D1 fi.X1; : : : ; Xn; h1/ gi.X1; : : : ; Xn/ (*) in k.X1; : : : ; Xn/. Clearly fi.X1; : : : ; Xn; h1/ D polynomial in X1; : : : ; Xn hNi for some Ni. Let N be the largest of the Ni. On multiplying (*) by hN we obtain an equation hN D m X i D1 (polynomial in X1; : : : ; Xn/ gi.X1; : : : ; Xn/; which shows that hN 2 a. COROLLARY 2.17. The map a 7! V.a/ defines a one-to-one correspondence between the set of radical ideals in kŒX1; : : : ; Xn and the set of algebraic subsets of kn; its inverse is I. PROOF. We know that I V.a/ D a if a is a radical ideal (2.16), and that V I.W / D W if W is an algebraic set (2.14). Therefore, I and V are inverse bijections. COROLLARY 2.18. The radical of an ideal in kŒX1; : : : ; Xn is equal to the intersection of the maximal ideals containing it. PROOF. Let a be an ideal in kŒX1; : : : ; Xn. Because maximal ideals are radical, every maximal ideal containing a also contains rad.a/, and so rad.a/ \ ma m. 44 2. ALGEBRAIC SETS For each P D.a1; : : : ; an/ 2 kn, the ideal mP D.X1 a1; : : : ; Xn an/ is maximal in kŒX1; : : : ; Xn, and (see 2.13). Thus f 2 mP ”

f.P / D 0 mP a ” P 2 V.a/. If f 2 mP for all P 2 V.a/, then f is zero on V.a/, and so f 2 I V.a/ D rad.a/. We have shown that rad.a/ \ P 2V.a/ mP \ ma m. Remarks 2.19. Because V.0/ D kn, I.kn/ D I V.0/ D rad.0/ D 0I in other words, only the zero polynomial is zero on the whole of kn. In fact, this holds whenever k is infinite (Exercise 1-1). 2.20. The one-to-one correspondence in Corollary 2.17 is order reversing. Therefore the maximal proper radical ideals correspond to the minimal nonempty algebraic sets. But the maximal proper radical ideals are simply the maximal ideals in kŒX1; : : : ; Xn, and the minimal nonempty algebraic sets are the one-point sets. As I..a1; : : : ; an// D.X1 a1; : : : ; Xn an/ (see 2.13), this shows that the maximal ideals of kŒX1; : : : ; Xn are exactly the ideals.X1 a1; : : : ; Xn an/ with.a1; : : : ; an/ 2 kn. 2.21. The algebraic set V.a/ is empty if and only if a D kŒX1; : : : ; Xn (Nullstellensatz, 2.11). 2.22. Let W and W 0 be algebraic sets. As W \ W 0 is the largest algebraic subset contained in both W and W 0, I.W \ W 0/ must be the smallest radical ideal containing both I.W / and I.W 0/: I.W \ W 0/ D rad.I.W / C I.W 0//: For example, let W D V.X 2 Y / and W 0 D V.X 2 C Y /; then I.W \ W 0/ D rad.X 2; Y / D.X; Y / (assuming characteristic ¤ 2/. Note that W \ W 0 D f.0; 0/g, but when realized as the intersection of Y D X 2 and Y D X

2, it has “multiplicity 2”. P Q! P! and V W be the set of subsets of kn and let 2.23. Let Then I W and (see FT 7.19). It follows that I and V define a one-to-one correspondence between I. and V. ideals, and (by definition) V. 2.17. Q / P / consists exactly of the radical / consists of the algebraic subsets. Thus we recover Corollary define a simple Galois correspondence between /. But the strong Nullstellensatz shows that I. be the set of subsets of kŒX1; : : : ; Xn. Q Q Q Q P P P •V(X2−Y)V(X2+Y) f. Finding the radical of an ideal 45 n capture only part of the ideal theory of kŒX1; : : : ; Xn ASIDE 2.24. The algebraic subsets of A because two ideals with the same radical correspond to the same algebraic subset. There is a finer n are in notion of an algebraic scheme over k for which the closed algebraic subschemes of A one-to-one correspondence with the ideals in kŒX1; : : : ; Xn (see Chapter 11 on my website). f. Finding the radical of an ideal Typically, an algebraic set V is defined by a finite set of polynomials fg1; : : : ; gsg, and we need to find I.V / D rad.g1; : : : ; gs/. PROPOSITION 2.25. A polynomial h 2 rad.a/ if and only if 1 2.a; 1 Y h/ (the ideal in kŒX1; : : : ; Xn; Y generated by the elements of a and 1 Y h). PROOF. We saw that 1 2.a; 1 Y h/ implies h 2 rad.a/ in the course of proving 2.16. Conversely, from the identities 1 D Y N hN C.1 Y N hN / D Y N hN C.1 Y h/.1 C Y h C C Y N 1hN 1/ we see that, if hN 2 a, then 1 2 a C.1 Y h/. Given a set of generators of an ideal, there is an algorithm for deciding whether

or not a polynomial belongs to the ideal, and hence an algorithm for deciding whether or not a polynomial belongs to the radical of the ideal. There are even algorithms for finding a set of generators for the radical. These algorithms have been implemented in the computer algebra systems CoCoA and Macaulay 2. g. Properties of the Zariski topology n. n and on an algebraic subset of A We now examine more closely the Zariski topology on A n, V I.W / is the closure of W, and 2.17 says Proposition 2.14 says that, for a subset W of A n and the radical that there is a one-to-one correspondence between the closed subsets of A ideals of kŒX1; : : : ; Xn. Under this correspondence, the closed subsets of an algebraic set V correspond to the radical ideals of kŒX1; : : : ; Xn containing I.V /. PROPOSITION 2.26. Let V be an algebraic subset of A n. (a) The points of V are closed for the Zariski topology. (b) Every ascending chain of open subsets U1 U2 of V eventually becomes constant. Equivalently, every descending chain of closed subsets of V eventually becomes constant. (c) Every open covering of V has a finite subcovering. PROOF. (a) We have seen that f.a1; : : : ; an/g is the algebraic set defined by the ideal.X1 a1; : : : ; Xn an/. (b) We prove the second statement. A sequence V1 V2 of closed subsets of V gives rise to a sequence of radical ideals I.V1/ I.V2/ : : :, which eventually becomes constant because kŒX1; : : : ; Xn is noetherian. (c) Given an open covering of V, let can be expressed as a finite union of sets in the covering. If every element of ascending chain of sets in be the collection of open subsets of V that does not contain V, then is properly contained in another element, and so there exists an infinite (axiom of dependent choice), contradicting (b). U U U U 46 2. ALGEBRAIC SETS A topological space whose points are closed is said to be T1; the condition means that,

for any pair of distinct points, each has an open neighbourhood not containing the other. A topological space having the property (b) is said to be noetherian. The condition is equivalent to the following: every nonempty set of closed subsets of V has a minimal element. A topological space having property (c) is said to be quasicompact (by Bourbaki at least; others call it compact, but Bourbaki requires a compact space to be Hausdorff). The proof of (c) shows that every noetherian space is quasicompact. Since an open subset of a noetherian space is again noetherian, it is also quasicompact. h. Decomposition of an algebraic set into irreducible algebraic sets A topological space is said to be irreducible if it is not the union of two proper closed subsets. Equivalent conditions: every pair of nonempty open subsets has nonempty intersection; every nonempty open subset is dense. By convention, the empty space is not irreducible. Obviously, every nonempty open subset of an irreducible space is irreducible. In a Hausdorff topological space, any two points have disjoint open neighbourhoods. Therefore, the only irreducible Hausdorff spaces are those consisting of a single point. PROPOSITION 2.27. An algebraic set W is irreducible if and only if I.W / is prime. PROOF. Let W be an irreducible algebraic set, and let fg 2 I.W / — we have to show that either f or g is in I.W /. At each point of W, either f is zero or g is zero, and so W V.f / [ V.g/. Hence W D.W \ V.f // [.W \ V.g//: As W is irreducible, one of these sets, say W \ V.f /, must equal W. But then f 2 I.W /. Let W be an algebraic set such that I.W / is prime, and let W D V.a/ [ V.b/ with a and b radical ideals — we have to show that W equals V.a/ or V.b/. The ideal a \ b is radical, and V.a \ b/ D V

.a/ [ V.b/ (2.10); hence I.W / D a \ b. If W ¤ V.a/, then there exists an f 2 a X I.W /. Let g 2 b. Then fg 2 a \ b D I.W /, and so g 2 I.W / (because I.W / is prime). We conclude that b I.W /, and so V.b/ V.I.W // D W. SUMMARY 2.28. There are one-to-one correspondences, radical ideals in kŒX1; : : : ; Xn $ algebraic subsets of A prime ideals in kŒX1; : : : ; Xn $ irreducible algebraic subsets of A n n maximal ideals in kŒX1; : : : ; Xn $ one-point sets of A n: EXAMPLE 2.29. Let f 2 kŒX1; : : : ; Xn. We saw (1.32) that kŒX1; : : : ; Xn is a unique factorization domain, and so.f / is a prime ideal if and only if f is irreducible (1.33). Thus f is irreducible H) V.f / is irreducible. On the other hand, suppose f factors as f D Y f mi i ; fi distinct irreducible polynomials. h. Decomposition of an algebraic set into irreducible algebraic sets 47 Then.f / D T f mi i rad.f / D T.fi / V.f / D S V.fi / f mi i distinct ideals.fi / distinct prime ideals V.fi / distinct irreducible algebraic sets. LEMMA 2.30. Let W be an irreducible topological space. If W D W1 [ : : : [ Wr with each Wi closed, then W is equal to one of the Wi. PROOF. When r D 2, the statement is the definition of “irreducible”. Suppose that r > 2. Then W D W1 [.W2 [ : : : [ Wr /, and so W D W1 or W D.W2 [ : : : [ Wr /; if the latter, then W D W2 or W3 [ :

: : [ Wr, etc. PROPOSITION 2.31. Let V be a noetherian topological space. Then V is a finite union of irreducible closed subsets, V D V1 [ : : : [ Vm. If the decomposition is irredundant in the sense that there are no inclusions among the Vi, then the Vi are uniquely determined up to order. PROOF. Suppose that V cannot be written as a finite union of irreducible closed subsets. Then, because V is noetherian, there will be a nonempty closed subset W of V that is minimal among those that cannot be written in this way. But W itself cannot be irreducible, and so W D W1 [ W2, with W1 and W2 proper closed subsets of W. Because W was minimal, each Wi is a finite union of irreducible closed subsets. Hence W is also, which is a contradiction. Suppose that V D V1 [ : : : [ Vm D W1 [ : : : [ Wn are two irredundant decompositions of V. Then Vi D S j.Vi \ Wj /, and so, because Vi is irreducible, Vi D Vi \ Wj for some j. Consequently, there is a function f W f1; : : : ; mg! f1; : : : ; ng such that Vi Wf.i/ for each i. Similarly, there is a function gW f1; : : : ; ng! f1; : : : ; mg such that Wj Vg.j / for each j. Since Vi Wf.i / Vgf.i /, we must have gf.i/ D i and Vi D Wf.i/; similarly fg D id. Thus f and g are bijections, and the decompositions differ only in the numbering of the sets. The Vi given uniquely by the proposition are called the irreducible components of V. They are exactly the maximal irreducible closed subsets of V.3 In Example 2.29, the V.fi / are the irreducible components of V.f /. An algebraic set with two irreducible components. 3In fact, they are exactly the maximal irreducible subsets of V because the closure of an irreducible subset is also irreduc

ible. 48 2. ALGEBRAIC SETS COROLLARY 2.32. The radical of an ideal a in kŒX1; : : : ; Xn is a finite intersection of prime ideals, a D p1 \ : : : \ pn. If there are no inclusions among the pi, then the pi are uniquely determined up to order (and they are exactly the minimal prime ideals containing a). PROOF. Write V.a/ as a union of its irreducible components, V.a/ D Sn pi D I.Vi /. Then rad.a/ D p1 \ : : : \ pn because they are both radical ideals and i D1 Vi, and let V.rad.a// D V.a/ D [ V.pi / 2.10bD V. \ p/: i The uniqueness similarly follows from the proposition. Remarks 2.33. An irreducible topological space is connected, but a connected topological space 2, need not be irreducible. For example, V.X1X2/ is the union of the coordinate axes in A n is disconnected if and which is connected but not irreducible. An algebraic subset V of A only if there exist radical ideals a and b such that V is the disjoint union of V.a/ and V.b/, that is, V D V.a/ [ V.b/ D V.a \ b/ ” a \ b D I.V / ; D V.a/ \ V.b/ D V.a C b/ ” a C b D kŒX1; : : : ; Xn: Note that then kŒV'kŒX1; : : : ; Xn a kŒX1; : : : ; Xn b (Chinese remainder theorem, 1.1). 2.34. A Hausdorff space is noetherian if and only if it is finite, in which case its irreducible components are the one-point sets. 2.35. In kŒX1; : : : ; Xn, a principal ideal.f / is radical if and only if f is square-free, in which case f is a product of distinct irreducible polynomials, f D f1 : : : fr, and.f / D.f1

/ \ : : : \.fr /. 2.36. In a noetherian ring, every proper ideal a has a decomposition into primary ideals: a D T qi (see CA 19). For radical ideals, this becomes a simpler decomposition into prime ideals, as in the corollary. For an ideal.f / with f D Q f mi, the primary decomposition is the decomposition.f / D T.f mi / in Example 2.29. i i i. Regular functions; the coordinate ring of an algebraic set Let V be an algebraic subset of A n, and let I.V / D a. The coordinate ring of V is kŒV defD kŒX1; : : : ; Xn=a. This is a finitely generated k-algebra. It is reduced (because a is radical), but not necessarily an integral domain. An f 2 kŒX1; : : : ; Xn defines a function P 7! f.P /W V! k: i. Regular functions; the coordinate ring of an algebraic set 49 Functions of this form are said to be regular. Two polynomials f; g 2 kŒX1; : : : ; Xn define the same regular function on V if and only if they define the same element of kŒV, and so kŒV is the ring of regular functions on V. The coordinate function xi W V! k;.a1; : : : ; an/ 7! ai is regular, and kŒV D kŒx1; : : : ; xn. In other words, the coordinate ring of an algebraic set V is the k-algebra generated by the coordinate functions on V. For an ideal b in kŒV, set V.b/ D fP 2 V j f.P / D 0, all f 2 bg — it is a closed subset of V. Let W D V.b/. The quotient maps kŒX1; : : : ; Xn kŒV D kŒX1; : : : ; Xn a kŒW D kŒV b send a regular function on kn to its restriction to V and then to its restriction to W. Write for the quotient map kŒX1; : : : ; Xn kŒV.

Then b 7! 1.b/ is a bijection from the set of ideals of kŒV to the set of ideals of kŒX1; : : : ; Xn containing a, under which radical, prime, and maximal ideals correspond to radical, prime, and maximal ideals (because each of these conditions can be checked on the quotient ring, and kŒX1; : : : ; Xn= 1.b/'kŒV =b). Clearly V. 1.b// D V.b/; and so b 7! V.b/ is a bijection from the set of radical ideals in kŒV to the set of algebraic sets contained in V. Now 2.28 holds for ideals in kŒV and algebraic subsets of V, radical ideals in kŒV $ algebraic subsets of V prime ideals in kŒV $ irreducible algebraic subsets of V maximal ideals in kŒV $ one-point sets of V: Moreover (see 2.33), the decompositions of a closed subset W of V into a disjoint union of closed subsets correspond to pairs of radical ideals a; b 2 kŒV such that kŒW D kŒV =a \ b'kŒV =a kŒV =b: For h 2 kŒV, set D.h/ D fa 2 V j h.a/ ¤ 0g: It is an open subset of V, because its complement is the closed set V..h//. It is empty if and only if h is zero (2.19). PROPOSITION 2.37. The sets D.h/, h 2 kŒV, are a base for the topology on V, i.e., each D.h/ is open, and every open set is a (finite) union of this form. PROOF. We have already observed that D.h/ is open. Every open subset U V is the complement of a set of the form V.b/, with b an ideal in kŒV. If f1; : : : ; fm generate b, then U D S D.fi /. 50 2. ALGEBRAIC SETS The D.h/ are called the basic (or principal) open subsets of V. We sometimes write Vh for D.

h/. Note that D.h/ D.h0/ ” V.h/ V.h0/ ” rad..h// rad..h0// ” hr 2.h0/ some r ” hr D h0g, some g: Some of this should look familiar: if V is a topological space, then the zero set of a family of continuous functions f W V! R is closed, and the set where a continuous function is nonzero is open. Let V be an irreducible algebraic set. Then I.V / is a prime ideal, and so kŒV is an integral domain. Let k.V / be its field of fractions — k.V / is called the function field of V or the field of rational functions on V. j. Regular maps Let W km and V kn be algebraic sets. Let xi denote the ith coordinate function.b1; : : : ; bn/ 7! bi W V! k on V. The i th component of a map 'W W! V is 'i D xi ı '. Thus,'is the map P 7! '.P / D.'1.P /; : : : ; 'n.P //W W! V kn: DEFINITION 2.38. A continuous map 'W W! V of algebraic sets is regular if each of its components 'i is a regular function. As the coordinate functions generate kŒV, a continuous map'is regular if and only if f ı'is a regular function on W for every regular function f on V. Thus a regular map 'W W! V of algebraic sets defines a homomorphism f 7! f ı 'W kŒV! kŒW of k-algebras, which we sometimes denote by '. k. Hypersurfaces; finite and quasi-finite maps A hypersurface in A polynomial, nC1 is the algebraic set H defined by a single nonzero nonconstant We examine the regular map H! A n defined by the projection H W f.T1; : : : ; Tn; X/ D 0..t1; : : : ; tn; x/ 7!.t1; : : : ; tn/: We can write f in the form f D a0X m C a1X m1 C C am;

ai 2 kŒX1; : : : ; Xm; a0 ¤ 0; m 2 N: k. Hypersurfaces; finite and quasi-finite maps 51 We assume that m ¤ 0, i.e., that X occurs in f (otherwise, H is a cylinder over a hypersurface n over.t1; : : : ; tn/ 2 kn is the set of points.t1; : : : ; tn; c/ n). The fibre of the map H! A in A such that c is a root of the polynomial a0.t /X m C a1.t/X m1 C C am.t/; ai.t/ defD ai.t1; : : : ; tn/ 2 k: Suppose first that a0 2 k, so that a0.t/ is a nonzero constant independent of t. Then the fibre over t consists of the roots of the polynomial a0X m C a1.t/X m1 C C am.t/; (14) in kŒX. Counting multiplicities, there are exactly m of these. More precisely, let D be the discriminant of the polynomial a0X m C a1X m1 C C am: Then D 2 kŒX1; : : : ; Xm, and the fibre has exactly m points over the open subset D ¤ 0, and fewer then m points over the closed subset D D 0.4 We can picture it schematically as follows (m D 3): Now drop the condition that a0 is constant. For certain t, the degree of (14) may drop, which means that some roots have “disappeared off to infinity”. Consider, for example, f.T; X/ D TX 1; for each t ¤ 0, there is one point.t; 1=t/, but there is no point with t D 0 (see the figure p. 71). Worse, for certain t all coefficients may be zero, in which case n such that the number the fibre is a line. There is a nested collection of closed subsets of A of points in the fibre (counting multiplicities) drops as you pass to a smaller subset, except that over the smallest subset the fibre may be a full line. DEFINITION 2.39. Let 'W

W! V be a regular map of algebraic subsets, and let 'W kŒV! kŒW be the map f 7! f ı '. (a) The map'is dominant if '.W / is dense in V. (b) The map'is quasi-finite if '1.P / is finite for all P 2 V. (c) The map'is finite if kŒW is a finite kŒV -algebra. As we shall see (8.28), finite maps are indeed quasi-finite. As kŒW is finitely generated as a k-algebra, a fortiori as a kŒV -algebra, to say that kŒW is a finite kŒV -algebra means that it is integral over kŒV (1.36). The map H! A n considered above is finite if and only if a0 is constant, and quasi-finite if and only if the polynomials a0; : : : ; am have no common zero in kn. 4I’m ignoring the possibility that D is identically zero. Then the open set where D ¤ 0 is empty. This case occurs when the characteristic is p ¤ 0, and f is a polynomial in T1; : : : ; Tn; and X p. HAn 52 2. ALGEBRAIC SETS PROPOSITION 2.40. A regular map 'W W! V is dominant if and only if 'W kŒV! kŒW is injective. PROOF. Let f 2 kŒV. If the image of'is dense, then f ı'D 0 H) f D 0: On the other hand, if the image of'is not dense, then its closure Z is a proper closed subset of V, and so there exists a nonzero regular function f zero on Z. Then f ı'D 0. PROPOSITION 2.41. A dominant finite map is surjective. PROOF. Let 'W W! V be dominant and finite. Then 'W kŒV! kŒW is injective, and kŒW is integral over the image of kŒV. According to the going-up theorem (1.53), for every maximal ideal m of kŒV there exists a maximal

ideal n of kŒW such that m D n \ kŒV. Because of the correspondence between points and maximal ideals, this implies that'is surjective. l. Noether normalization theorem Let H be a hypersurface in A projection map.x1; : : : ; xnC1/ 7!.x1; : : : ; xn/W A nC1. We show that, after a linear change of coordinates, the n. n defines a finite map H! A nC1! A PROPOSITION 2.42. Let H W f.X1; : : : ; XnC1/ D 0 be a hypersurface in A nC1. There exist c1; : : : ; cn 2 k such that the map H! A n defined by.x1; : : : ; xnC1/ 7!.x1 c1xnC1; : : : ; xn cnxnC1/ is finite. PROOF. Let c1; : : : ; cn 2 k. In terms of the coordinates x0 i H is the zero set of D xi ci xnC1, the hyperplane f.X1 C c1XnC1; : : : ; Xn C cnXnC1; XnC1/ D a0X m nC1 C a1X m1 nC1 C : The next lemma shows that the ci can be chosen so that a0 is a nonzero constant. This n defined by.x1; : : : ; xnC1/ 7!.x0 implies that the map H! A n/ is finite. 1; : : : ; x0 LEMMA 2.43. Let k be an infinite field (not necessarily algebraically closed), and let f 2 kŒX1; : : : ; Xn; T. There exist c1; : : : ; cn 2 k such that f.X1 C c1T; : : : ; Xn C cnT; T / D a0T m C a1T m1 C C am with a0 2 k and all ai 2 kŒX1; : : : ; Xn. PROOF. Let F be the homogeneous part of highest degree of f and let r D deg.F /

. Then F.X1 C c1T; : : : ; Xn C cnT; T / D F.c1; : : : ; cn; 1/T r C terms of degree < r in T, because the polynomial F.X1 C c1T; : : : ; Xn C cnT; T / is still homogeneous of degree r in X1; : : : ; Xn; T, and so the coefficient of the monomial T r can be obtained by setting each Xi equal to zero in F and T to 1. As F.X1; : : : ; Xn; T / is a nonzero homogeneous l. Noether normalization theorem 53 polynomial, F.X1; : : : ; Xn; 1/ is a nonzero polynomial, and so we can choose the ci so that F.c1; : : : ; cn; 1/ ¤ 0 (Exercise 1-1). Now f.X1 C c1T; : : : ; Xn C cnT; T / D F.c1; : : : ; cn; 1/T r C terms of degree < r in T; with F.c1; : : : ; cn; 1/ 2 k, as required. In fact, every algebraic set V admits a finite surjective map to A d for some d. THEOREM 2.44. Let V be an algebraic set. For some natural number d, there exists a finite surjective map 'W V! A d. This follows from the next statement applied to A D kŒV : the regular functions d, which is finite and surjective because kŒx1; : : : ; xd! A x1; : : : ; xd define a map V! A is finite and injective. THEOREM 2.45 (NOETHER NORMALIZATION THEOREM). Let A be a finitely generated k-algebra. There exist elements x1; : : : ; xd 2 A that are algebraically independent over k, and such that A is finite over kŒx1; : : : ; xd. It is not necessary to assume that A is reduced in Theorem 2.45, nor that k is algebraically closed, although the proof we give requires

it to be infinite (for the general proof, see CA 8.1). Let A D kŒx1; : : : ; xn. We prove the theorem by induction on n. If the xi are algebraically independent, there is nothing to prove. Otherwise, the next lemma shows that A is finite over a subring B D kŒy1; : : : ; yn1. By induction, B is finite over a subring C D kŒz1; : : : ; zd with z1; : : : ; zd algebraically independent, and A is finite over C. LEMMA 2.46. Let A D kŒx1; : : : ; xn be a finitely generated k-algebra, and let fx1; : : : ; xd g be a maximal algebraically independent subset of fx1; : : : ; xng. If n > d, then there exist c1; : : : ; cd 2 k such that A is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1. PROOF. By assumption, the set fx1; : : : ; xd ; xng is algebraically dependent, and so there exists a nonzero f 2 kŒX1; : : : ; Xd ; T such that f.x1; : : : ; xd ; xn/ D 0: (15) Because fx1; : : : ; xd g is algebraically independent, T occurs in f, and so f.X1; : : : ; Xd ; T / D a0T m C a1T m1 C C am with ai 2 kŒX1; : : : ; Xd, a0 ¤ 0, and m > 0. If a0 2 k, then (15) shows that xn is integral over kŒx1; : : : ; xd. Hence x1; : : : ; xn are integral over kŒx1; : : : ; xn1, and so A is finite over kŒx1; : : : ; xn1. If a0 … k, then, for a suitable choice of.c1; : : : ; cd / 2 k, the po

lynomial g.X1; : : : ; Xd ; T / defD f.X1 C c1T; : : : ; Xd C cd T; T / takes the form g.X1; : : : ; Xd ; T / D bT r C b1T C C br with b 2 k (see 2.43). As g.x1 c1xn; : : : ; xd cd xn; xn/ D 0 this shows that xn is integral over kŒx1 c1xn; : : : ; xd cd xn, and so A is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1 as before. (16) 54 Remarks 2. ALGEBRAIC SETS 2.47. For an irreducible algebraic subset V of A prove the following more precise statement: n, the above argument can be modified to Let x1; : : : ; xn be the coordinate functions on V ; after possibly renumbering the coordinates, we may suppose that fx1; : : : ; xd g is a maximal algebraically independent subset of fx1; : : : ; xng; then there exist cij 2 k such that the map.x1; : : : ; xn/ 7! x1 n X j Dd C1 c1j xj ; : : : ; xd n X W A n! A d cdj xj j Dd C1 induces a finite surjective map V! A d : Indeed, Lemma 2.46 shows that there exist c1; : : : ; cn 2 k such that kŒV is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : : : ; xn1. Now fx1; : : : ; xd g is algebraically dependent on fx1 c1xn; : : : ; xd cd xng. If the second set were not algebraically independent, we could drop one of its elements, but this would contradict 1.61. Therefore fx1 c1xn; : : : ; xd cd xng is a maximal algebraically independent subset of fx1 c1xn; :

: : ; xd cd xn; xd C1; : : : ; xn1g and we can repeat the argument. m. Dimension The dimension of a topological space Let V be a noetherian topological space whose points are closed. DEFINITION 2.48. The dimension of V is the supremum of the lengths of the chains V0 V1 Vd of distinct irreducible closed subsets (the length of the displayed chain is d ). 2.49. Let V1; : : : ; Vm be the irreducible components of V. Then (obviously) dim.V / D max i.dim.Vi //: 2.50. Assume that V is irreducible, and let W be a proper closed subspace of V. Then every chain W0 W1 in W extends to a chain V W0, and so dim.W / C 1 dim.V /. If dim.V / < 1, then dim.W / < dim.V /. Thus an irreducible topological space V has dimension 0 if and only if it is a point; it has dimension 1 if and only if every proper closed subset is a point; and, inductively, V has dimension n if and only if every proper closed subset has dimension n 1. The dimension of an algebraic set DEFINITION 2.51. The dimension of an algebraic set is its dimension as a topological space. Because of the correspondence between the prime ideals in kŒV and irreducible closed subsets of V, dim.V / D Krull dimension of kŒV : m. Dimension 55 Note that, if V1; : : : ; Vm are the irreducible components of V, then dim V D max i dim.Vi /: When the Vi all have the same dimension d, we say that V has pure dimension d. A one-dimensional algebraic set is called a curve; a two-dimensional algebraic set is called a surface; and an n-dimensional algebraic set is called an n-fold. Let V be an irreducible algebraic set and W an algebraic subset of V. If W is irreducible, then its codimension in V is codimV W D dim V dim W: Dimension and transcendent degree THEOREM 2.52. Let V be an irreducible algebraic set.

Then dim.V / D tr degkk.V /: The proof will occupy the rest of this subsection. Let A be an arbitrary commutative ring. Let x 2 A, and let Sfxg denote the multiplicative subset of A consisting of the elements of the form xn.1 ax/; n 2 N; a 2 A: The boundary Afxg of A at x is defined to be the ring of fractions S 1 fxgA. We write dim.A/ for the Krull dimension of A. PROPOSITION 2.53. Let A be a ring and let n 2 N. Then dim.A/ n ” for all x 2 A, dim.Afxg/ n 1: PROOF. We shall use (1.14) that there is a one-to-one correspondence between the prime ideals of S 1A and the prime ideals of A disjoint from S. We begin with two observations. (a) For every x 2 A and maximal ideal m A, m \ Sfxg ¤ ;. Indeed, if x 2 m, then certainly x 2 m \ Sfxg. On the other hand, if x … m, then it is invertible modulo m, and so there exists an a 2 A such that 1 ax 2 m (hence also m \ Sfxg). (b) Let m be a maximal ideal, and let p be a prime ideal contained in m; for every x 2 m X p, we have p \ Sfxg D ;. Indeed, if xn.1 ax/ 2 p, then 1 ax 2 p (as x … p/; hence 1 ax 2 m, and so 1 2 m, which is a contradiction. Statement (a) shows that every chain of prime ideals beginning with a maximal ideal is shortened when passing from A to Afxg, while statement (b) shows that a maximal chain of length n is shortened only to n 1 when x is chosen appropriately. From this, the proposition is follows. PROPOSITION 2.54. Let A be an integral domain, and let k be a subfield of A. Then dim.A/ tr degkF.A/; where F.A/ is the field of fractions of A. 56 2. ALGEBRAIC SETS PROOF. If tr degkF.A/ D 1, there is nothing to prove, and so we suppose that

tr degkF.A/ D n 2 N. We argue by induction on n. We can replace k with its algebraic closure in A without changing tr degkF.A/. Let x 2 A. If x … k, then it is transcendental over k, and so tr degk.x/F.A/ D n 1 by 1.64; since k.x/ Afxg, this implies (by induction) that dim.Afxg/ n 1. If x 2 k, then 0 D 1 x1x 2 Sfxg, and so Afxg D 0; again dim.Afxg/ n 1. We deduce from 2.53 that dim.A/ n. COROLLARY 2.55. The polynomial ring kŒX1; : : : ; Xn has Krull dimension n. PROOF. The existence of the sequence of prime ideals.X1; : : : ; Xn/.X1; : : : ; Xn1/.X1/.0/ shows that kŒX1; : : : ; Xn has Krull dimension at least n. Now 2.54 completes the proof. COROLLARY 2.56. Let A be an integral domain and let k be a subfield of A. If A is finitely generated as a k-algebra, then tr degkF.A/ D dim.A/: PROOF. According to the Noether normalization theorem (2.45), A is integral over a polynomial subring kŒx1; : : : ; xn. Clearly n D tr degkF.A/. The going up theorem (1.54), implies that a chain of prime ideals in kŒx1; : : : ; xn lifts to a chain in A, and so dim.A/ dim.kŒx1; : : : ; xn/ D n. Now 2.54 shows that dim.A/ D n. COROLLARY 2.57. Let V be an irreducible algebraic set. Then V has dimension n if and only if there exists a finite surjective map V! A n. PROOF. The d in Theorem 2.44 is the dimension of V. ASIDE 2.58. In linear algebra, we justify saying that a vector space V has dimension n by proving

that its elements are parametrized by n-tuples. It is not true in general that the points of an algebraic set of dimension n are parametrized by n-tuples. All we can say is Corollary 2.57. ASIDE 2.59. The inequality in Proposition 2.54 may be strict; for example, A D k.x/ has dimension 0 but its field of fractions k.x/ has transcendence degree 1 over k. It is possible to deduce 2.54 from 2.56 — see mo79959. NOTES. The above proof of 2.55 is based on that in Coquand and Lombardi, Amer. Math. Monthly 112 (2005), no. 9, 826–829. Examples EXAMPLE 2.60. Let V D A X1; : : : ; Xn over k, and so dim.V / D n. n. Then k.V / D k.X1; : : : ; Xn/, which has transcendence basis m. Dimension 57 EXAMPLE 2.61. If V is a linear subspace of kn (or a translate of a linear subspace), then the dimension of V as an algebraic set is the same as its dimension in the sense of linear algebra — in fact, kŒV is canonically isomorphic to kŒXi1; : : : ; Xid, where the Xij are the “free” variables in the system of linear equations defining V. More specifically, let c be an ideal in kŒX1; : : : ; Xn generated by linear forms `1; : : : ; `r, which we may assume to be linearly independent. Let Xi1; : : : ; Xinr be such that f`1; : : : ; `r ; Xi1; : : : ; Xinr g is a basis for the linear forms in X1; : : : ; Xn. Then kŒX1; : : : ; Xn=c'kŒXi1; : : : ; Xinr : This is obvious if the forms are X1; : : : ; Xr. In the general case, because fX1; : : : ; Xng g are both bases for the linear forms, each element of one set and f`1; : : : ; `r ; Xi1; : : : ; Xin

r can be expressed as a linear combination of the elements of the other. Therefore, kŒX1; : : : ; Xn D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr ; and so kŒX1; : : : ; Xn=c D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr =c'kŒXi1; : : : ; Xinr : EXAMPLE 2.62. If W is a proper algebraic subset of an irreducible algebraic set V, then dim.W / < dim.V / (see 2.50). EXAMPLE 2.63. Every nonempty algebraic set contains a point, which is a closed irreducible subset. Therefore an irreducible algebraic set has dimension 0 if and only if it consists of a single point. n has dimension n 1. It suffices to prove this for an EXAMPLE 2.64. A hypersurface in A irreducible hypersurface H. Such an H is the zero set of an irreducible polynomial f (see 2.29). Let kŒx1; : : : ; xn D kŒX1; : : : ; Xn=.f /; xi D Xi C.f /; and let k.x1; : : : ; xn/ be the field of fractions of kŒx1; : : : ; xn. As f is not the zero polynomial, some Xi, say, Xn, occurs in it. Then Xn occurs in every nonzero multiple of f, and so no nonzero polynomial in X1; : : : ; Xn1 belongs to.f /. This means that x1; : : : ; xn1 are algebraically independent. On the other hand, xn is algebraic over k.x1; : : : ; xn1/, and so fx1; : : : ; xn1g is a transcendence basis for k.x1; : : : ; xn/ over k. (Alternatively, use 2.57.) EXAMPLE 2.65. Let F.X; Y / and G.X; Y / be nonconstant polynomials with no common factor.

Then V.F.X; Y // has dimension 1 by 2.64, and so V.F.X; Y // \ V.G.X; Y // must have dimension zero; it is therefore a finite set. PROPOSITION 2.66. Let W be a closed set of codimension 1 in an algebraic set V ; if kŒV is a unique factorization domain, then I.W / D.f / for some f 2 kŒV. PROOF. Let W1; : : : ; Ws be the irreducible components of W ; then I.W / D T I.Wi /, and so if we can prove I.Wi / D.fi /, then I.W / D.f1 fr /. Thus we may suppose that W is irreducible. Let p D I.W /; it is a prime ideal, and it is not zero because otherwise 58 2. ALGEBRAIC SETS dim.W / D dim.V /. Therefore it contains an irreducible polynomial f. From (1.33) we know.f / is prime. If.f / ¤ p, then we have p.f /.0/ (distinct prime ideals) and hence W D V.p/ V.f / V (distinct irreducible closed subsets). But then (2.62) dim.W / < dim.V.f // < dim V, which contradicts the hypothesis. COROLLARY 2.67. The closed sets of codimension 1 in A n are exactly the hypersurfaces. PROOF. Combine 2.64 and 2.66. 2. If V has dimension EXAMPLE 2.68. We classify the irreducible algebraic sets V of A 2. If V has dimension 1, then 2, then (by 2.62) it can’t be a proper subset of A V D V.f /, where f is any irreducible polynomial in I.V / (see 2.66 and its proof). Finally, if V has dimension zero, then it is a point. Correspondingly, the following is a complete list of the prime ideals in kŒX; Y : 2, so it is A.0/;.f / with f irreducible,.X a; Y b/ with

a; b 2 k: Exercises 2-1. Find I.W /, where W D.X 2; XY 2/. Check that it is the radical of.X 2; X Y 2/. 2-2. Identify kmn with the set of m n matrices, and let r 2 N. Show that the set of matrices with rank r is an algebraic subset of kmn. 2-3. Let V D f.t; t 2; : : : ; t n/ j t 2 kg. Show that V is an algebraic subset of kn, and that kŒV kŒX (polynomial ring in one variable). (Assume k has characteristic zero.) 2-4. Let f1; : : : ; fm 2 QŒX1; : : : ; Xn. If the fi have no common zero in C, prove that there exist g1; : : : ; gm 2 QŒX1; : : : ; Xn such that f1g1 C C fmgm D 1. (Hint: linear algebra). 2-5. Let k K be algebraically closed fields, and let a be an ideal in kŒX1; : : : ; Xn. Show that if f 2 KŒX1; : : : ; Xn vanishes on V.a/, then it vanishes on VK.a/. Deduce that the zero set V.a/ of a in kn is dense in the zero set VK.a/ of a in Kn. [Hint: Choose a basis.ei /i2I for K as a k-vector space, and write f D P ei fi (finite sum) with fi 2 kŒX1; : : : ; Xn.] 2-6. Let A and B be (not necessarily commutative) Q-algebras of finite dimension over Q, and let Qal be the algebraic closure of Q in C. Show that if there exists a C-algebra homomorphism C ˝ Q B, then there exists a Qal-algebra homomorphism Qal ˝ Q A! Qal ˝ Q B. (Hint: The proof takes only a few lines.) Q A! C ˝ 2-7. Let A be finite dimensional k-algebra, where k is an infinite field,

and let M and N be A-modules. Show that if kal ˝k M and kal ˝k N are isomorphic kal ˝k A-modules, then M and N are isomorphic A-modules. 2-8. Show that the subset f.z; ez/ j z 2 Cg is not an algebraic subset of C 2. CHAPTER 3 Affine Algebraic Varieties In this chapter, we define the structure of a ringed space on an algebraic set. In this way, we are led to the notion of an affine algebraic variety — roughly speaking, this is an algebraic set n. This is in preparation for Chapter 5, where we define with no preferred embedding into A an algebraic variety to be a ringed space that is a finite union of affine algebraic varieties satisfying a natural separation axiom. a. Sheaves Let k be a field (in this section 3a, k need not be algebraically closed). DEFINITION 3.1. Let V be a topological space, and suppose that, for every open subset U V.U / is a sheaf of k-algebras of V we have a set if the following statements hold for every open U in V : V.U / of functions U! k. Then U O O (a) V.U / is a k-subalgebra of the algebra of all k-valued functions on U, i.e., V.U / V.U /, then so also do f C g and O O contains the constant functions and, if f; g lie in fg; O (b) the restriction of an f in O (c) a function f W U! k lies in V.U / to an open subset U 0 of U is in V.U 0/I V.U / if there exists an open covering U D S O i 2I Ui of U such that f jUi 2 O V.Ui / for all i 2 I. O V is a sheaf if, for all U, In other words, lies in lies in V.U / is a k-subalgebra and a function f W U! k V.U / if and only if every point P of U has a neighbourhood UP such that f jUP V.UP / (so the condition for f to lie in V.U / is local). O O

Note that, for disjoint open subsets Ui of V, condition (c) says that O O O V.U /'Q O V.Ui /. i O Examples 3.2. Let V be a topological space, and for each open subset U of V let of all continuous real-valued functions on U. Then V is a sheaf of R-algebras. O V.U / be the set O 3.3. Recall that a function f W U! R on an open subset U of R n is said to be smooth (or infinitely differentiable) if its partial derivatives of all orders exist and are continuous. Let n, and for each open subset U of V, let V be an open subset of R V.U / be the set of all smooth functions on U. Then V is a sheaf of R-algebras. O O 59 60 3. AFFINE ALGEBRAIC VARIETIES 3.4. Recall that a function f W U! C on an open subset U of C n, is said to be analytic (or holomorphic) if it is described by a convergent power series in a neighbourhood of each V.U / be point of U. Let V be an open subset of C the set of all analytic functions on U. Then n, and for each open subset U of V, let V is a sheaf of C-algebras. O O 3.5. Let V be a topological space, and, for each open subset U of V, let V.U / be the set of all constant functions U! k. If V is not connected, then V is not a sheaf: let U1 and U2 be disjoint open subsets of V, and let f be the function on U1 [ U2 that takes the V.U1 [ U2/, and constant value 0 on U1 and the constant value 1 on U2; then f is not in so condition (3.1c) fails. When “constant” is replaced with “locally constant”, V becomes a sheaf of k-algebras (in fact, the smallest such sheaf). O O O O 3.6. Let V be a topological space, and, for each open subset U of V, let of all functions U! k. The k-algebras are subshe

aves of this one. V.U / be the set V is a sheaf of k-algebras. By definition, all our sheaves of O O b. Ringed spaces O A pair.V; V / consisting of a topological space V and a sheaf of k-algebras on V will be called a k-ringed space (or just a ringed space when the k is understood). For historical reasons, we sometimes write.U; over U. V.U / and call its elements the sections of V / for O O O V O Let.V; V / be a k-ringed space. For each open subset U of V, the restriction the collection of open subsets of U is a sheaf of k-algebras on U. O Let.V; V / be a k-ringed space, and let P 2 V. A germ of a function at P is an equivalence class of pairs.U; f / with U an open neighbourhood of P and f 2 V.U /; two pairs.U; f / and.U 0; f 0/ are equivalent if the functions f and f 0 agree on some open neighbourhood of P in U \ U 0. The germs of functions at P form a k-algebra V;P, called the stalk of O O V to O V at P. In other words, O V;P D lim! O In the interesting cases, are zero at P. We often write P for O O O V;P. O V.U / (direct limit over open neighbourhoods U of P ). V;P is a local ring with maximal ideal mP the set of germs that O V be the sheaf of holomorphic functions on V D C, and let c 2 C. EXAMPLE 3.7. Let A power series P n0 an.z c/n, an 2 C, is said to be convergent if it converges on some open neighbourhood of c. The set of such power series is a C-algebra, and I claim that it is canonically isomorphic to the stalk O To prove this, let f be a holomorphic function on a neighbourhood U of c. Then f has a unique power series expansion f D P an.z c/n in some (possibly smaller) open neighbourhood of c (Cartan 19631, II 2.6). Moreover, another holomorphic function f 0

on a neighbourhood U 0 of c defines the same power series if and only if f and f 0 agree on some neighbourhood of c contained in U \ U 0 (ibid. I, 4.3). Thus we have a well-defined injective map from the ring of germs of holomorphic functions at c to the ring of convergent power series, which is obviously surjective. V at c. V;c of O 1Cartan, Henri. Elementary theory of analytic functions of one or several complex variables. Hermann, Paris; Addison-Wesley; 1963. c. The ringed space structure on an algebraic set 61 c. The ringed space structure on an algebraic set Let V be an algebraic subset of kn. Recall that the basic open subsets of V are those of the form D.h/ D fQ j h.Q/ ¤ 0g; h 2 kŒV : A pair g; h 2 kŒV with h ¤ 0 defines a function Q 7! g.Q/ h.Q/ W D.h/! k. A function on an open subset of V is regular if it is locally of this form. More formally: DEFINITION 3.8. Let U be an open subset of V. A function f W U! k is regular at P 2 U if there exist g; h 2 kŒV with h.P / ¤ 0 such that f.Q/ D g.Q/= h.Q/ for all Q in some neighbourhood of P. A function f W U! k is regular if it is regular at every P 2 U. Let O V.U / denote the set of regular functions on an open subset U of V. PROPOSITION 3.9. The map U V.U / is a sheaf of k-algebras on V. O PROOF. We have to check the conditions of (3.1). (a) Clearly, a constant function is regular. Suppose f and f 0 are regular on U, and let P 2 U. By assumption, there exist g; g0; h; h0 2 kŒV, with h.P / ¤ 0 ¤ h0.P / such that f h0 respectively on a neighbourhood U 0 of P. Then f C f 0 agrees and f 0 agree with g with gh0Cg 0h hh0 on U 0

, and so is regular at P. on U 0, and so f C f 0 is regular at P. Similarly, ff 0 agrees with gg 0 h and g 0 hh0 (b,c) The definition is local. We next determine O V.U / when U is a basic open subset of V. LEMMA 3.10. Let g; h 2 kŒV with h ¤ 0. The function P 7! g.P /= h.P /mW D.h/! k is zero if and only if and only if gh D 0 in kŒV. PROOF. If g= hm is zero on D.h/, then gh is zero on V because h is zero on the complement of D.h/. Therefore gh is zero in kŒV. Conversely, if gh D 0, then g.P /h.P / D 0 for all P 2 V, and so g.P / D 0 for all P 2 D.h/. Let kŒV h denote the ring kŒV with h inverted (see 1.11). The lemma shows that V.D.h// sending g= hm to the regular function P 7! g.P /= h.P /m is the map kŒV h! well-defined and injective. O PROPOSITION 3.11. The above map kŒV h! V.D.h// is an isomorphism of k-algebras. O PROOF. It remains to show that every regular function f on D.h/ arises from an element of kŒV h. By definition, we know that there is an open covering D.h/ D S Vi and elements gi, hi 2 kŒV with hi nowhere zero on Vi such that f jVi D gi. We may assume that each hi set Vi is basic, say, Vi D D.ai / for some ai 2 kŒV. By assumption D.ai / D.hi /, and so aN i 2 kŒV (see p. 50). On D.ai /, i for some N 2 N and g0 D hi g0 i f D gi hi D gi g0 i hi g0 i D gi g0 i aN i : 62 3. AFFINE ALGEBRAIC VARIETIES

Note that D.aN assume that Vi D D.hi /. i / D D.ai /. Therefore, after replacing gi with gi g0 i and hi with aN i, we can We now have that D.h/ D S D.hi / and that f jD.hi / D gi. Because D.h/ is quasicomhi on D.hi / \ D.hj / D D.hi hj /, pact, we can assume that the covering is finite. As gi hi D gj hj hi hj.gi hj gj hi / D 0, i.e., hi h2 (*) — this follows from Lemma 3.10 if hi hj ¤ 0 and is obvious otherwise. Because D.h/ D S D.hi / D S D.h2 i /, j gi D h2 i hj gj V..h// D V..h2 1; : : : ; h2 m//; and so h lies in rad.h2 1; : : : ; h2 m/: there exist ai 2 kŒV such that hN D m X i D1 ai h2 i : for some N. I claim that f is the function on D.h/ defined by (**) : P ai gi hi hN Let P be a point of D.h/. Then P will be in one of the D.hi /, say D.hj /. We have the following equalities in kŒV : h2 j m X i D1 ai gi hi./D m X i D1 ai gj h2 i hj./D gj hj hN. But f jD.hj / D gj hj following equality of functions on D.hj /:, i.e., f hj and gj agree as functions on D.hj /. Therefore we have the h2 j m X i D1 ai gi hi D f h2 j hN : Since h2 on D.hj / equals that defined by P ai gi hi. j is never zero on D.hj /, we can cancel it, to find that, as claimed, the function f hN On taking h D 1 in the proposition, we see that the definition of a regular function on V in this section

agrees with that in Section 2i. COROLLARY 3.12. For every P 2 V, there is a canonical isomorphism where mP is the maximal ideal I.P /. P! kŒV mP, O PROOF. In the definition of the germs of a sheaf at P, it suffices to consider pairs.f; U / with U lying in a some basis for the neighbourhoods of P, for example, the basis provided by the basic open subsets. Therefore, P D lim! O h.P /¤0.D.h/; V / O (3.11)' lim! h…mP kŒV h (1.23)' kŒV mP : c. The ringed space structure on an algebraic set 63 Remarks 3.13. Let V be an algebraic set and let P be a point on V. Proposition 1.14 shows that there is a one-to-one correspondence between the prime ideals of kŒV contained in mP and the P. In geometric terms, this says that there is a one-to-one correspondence prime ideals of between the irreducible closed subsets of V passing through P and the prime ideals in P. The irreducible components of V passing through P correspond to the minimal prime ideals P. The ideal p corresponding to an irreducible closed subset Z consists of the elements in P represented by a pair.U; f / with f jZ\U D 0. of O O O O O V;P depends only on.U; V;P is an integral domain if P lies on a single irreducible component 3.14. The local ring V jU / for U an open neighbourhood of P, we may of V. As suppose that V itself is irreducible, in which case the statement follows from 3.12. On the other hand, if P lies on more than one irreducible component of V, then P contains more than one minimal prime ideal 3.13, and so the ideal.0/ can’t be prime. O O O 3.15. Let V be an algebraic subset of kn, and let A D kŒV. Propositions 2.37 and 3.11 allow us to describe.V; V / purely in terms of A: ˘ V is the set of maximal ideals in A.

˘ For each f 2 A, let D.f / D fm j f … mg; the topology on V is that for which the O sets D.f / form a base. ˘ For f 2 Ah and m 2 D.h/, let f.m/ denote the image of f in Ah=mAh'k; in this V is the V / D Ah for all h 2 A. way Ah becomes identified with a k-algebra of functions D.h/! k, and unique sheaf of k-valued functions on V such that.D.h/; O O 3.16. When V is irreducible, all the rings attached to it can be identified with subrings of its function field k.V /. For example,.D.h/; V / D O.U hN n g h 2 k.V / ˇ 2 k.V / ˇ o ˇ g 2 kŒV ; N 2 N o ˇ h.P / ¤ 0 P P 2U O.D.hi /; V / if U D [ D.hi /: O Note that every element of k.V / defines a function on some dense open subset of V. Following tradition, we call the elements of k.V / rational functions on V.2 Examples 3.17. The ring of regular functions on A h.X1; : : : ; Xn/, the ring of regular functions on D.h/ is n is kŒX1; : : : ; Xn. For a nonzero polynomial n g hN 2 k.X1; : : : ; Xn/ ˇ ˇ g 2 kŒX1; : : : ; Xn; N 2 N ˇ o : For a point P D.a1; : : : ; an/, the local ring at P is O P D ˚ g 2 k.X1; : : : ; Xn/ j h.P / ¤ 0 D kŒX1; : : : ; Xn.X1a1;:::;Xnan/; h and its maximal ideal consists of those g= h with g.P / D 0: 2The terminology is similar to that of “meromorphic function”, which is also not a function on the whole space. 64 3. AFF

INE ALGEBRAIC VARIETIES 3.18. Let U D A 2, but it is not a basic open subset because its complement f.0; 0/g has dimension 0, and therefore can’t be of the form V..f // (see 2.64). Since U D D.X/ [ D.Y /, the ring of regular functions on U is 2 X f.0; 0/g. It is an open subset of A.U; OA2/ D kŒX; Y X \ kŒX; Y Y (intersection inside k.X; Y /). Thus, a regular function f on U can be expressed f D g.X; Y / X N D h.X; Y / Y M : We may assume that X − g and Y − h. On multiplying through by X N Y M, we find that g.X; Y /Y M D h.X; Y /X N : Because X doesn’t divide the left hand side, it can’t divide the right hand side either, and so N D 0. Similarly, M D 0, and so f 2 kŒX; Y. We have shown that every regular function on U extends uniquely to a regular function on A 2:.U; 2; OA2/ D kŒX; Y D.A OA2/: d. Morphisms of ringed spaces A morphism of k-ringed spaces.V; that O V /!.W; O W / is a continuous map 'W V! W such f 2 W.U / H) f ı'2 V.'1U / O O for all open subsets U of W. Then, for every pair of open subsets U W and U 0 V with '.U 0/ U, we get a homomorphism of k-algebras f 7! f ı 'W W.U 0/! O V.U /; O and these homomorphisms are compatible with restriction to smaller open subsets. Sometimes we write '.f / for f ı '. A morphism of ringed spaces is an isomorphism if it is bijective and its inverse is also a morphism of ringed spaces (in particular, it is a homeomorphism). If U is an open subset of V, then the inclusion U,! V is a morphism

of k-ringed spaces.U; V jU /!.V; V /. O A morphism of ringed spaces maps germs of functions to germs of functions. More O precisely, a morphism 'W.V; V /!.W; W / induces a k-algebra homomorphism O O OW;'.P /! V;P O for each P 2 V, which sends the germ represented by.U; f / to the germ represented by.'1.U /; f ı '/. In the interesting cases, V;P is a local ring with maximal ideal mP O consisting of the germs represented by pairs.U; f / with f.P / D 0. Therefore, the homoV;P defined by'maps m'.P / into mP : it is a local homomorphism morphism of local rings. OW ;'.P /! O e. Affine algebraic varieties 65 Examples 3.19. Let V and W be topological spaces endowed with their sheaves W of continuous real valued functions (3.2). Every continuous map 'W V! W is a morphism of ringed structures.V; V and V /!.W; O 3.20. Let V and W be open subsets of R function.a1; : : : ; an/ 7! ai W V! R. Recall from advanced calculus that a map n and R m respectively, and let xi be the coordinate W /. O O O 'W V! W R m defD xi ı 'W V! R is smooth. If is said to be smooth if each of its component functions 'i'is smooth, then f ı'is smooth for every smooth function f W W! R. Since a similar statement is true for functions f on open subsets of W, we see that a continuous map 'W V! W is smooth if and only if it is a morphism of the associated ringed spaces (3.3). 3.21. Same as 3.20, but replace R with C and “smooth” with “analytic”. e. Affine algebraic varieties We have just seen that every algebraic set V kn gives rise to a k-ringed space.V; V /. A k-ringed space isomorphic to one of this form is called an affine algebraic variety over k. We

usually denote an affine algebraic variety.V; V / by V. Let.V; V / and.W; W / be affine algebraic varieties. A map 'W V! W is regular (or a morphism of affine algebraic varieties) if it is a morphism of k-ringed spaces. With these definitions, the affine algebraic varieties become a category. We usually shorten “affine algebraic variety” to “affine variety”. O O O O In particular, the regular functions define the structure of an affine variety on every n as an affine variety. The affine varieties we have n. We now explain how to construct affine algebraic set. We usually regard A constructed so far have all been embedded in A varieties with no preferred embedding. An affine k-algebra is a reduced finitely generated k-algebra. For such an algebra A, there exist xi 2 A such that A D kŒx1; : : : ; xn, and the kernel of the homomorphism Xi 7! xi W kŒX1; : : : ; Xn! A is a radical ideal. Therefore 2.18 implies that the intersection of the maximal ideals in A is 0. Moreover, 2.12 implies that, for every maximal ideal m A, the map k! A! A=m is an isomorphism. Thus we can identify A=m with k. For f 2 A, we write f.m/ for the image of f in A=m D k, i.e., f.m/ D f (mod m/. This allows us to identify elements of A with functions spm.A/! k. We attach a ringed space.V; For f 2 A, let O V / to A by letting V be the set of maximal ideals in A. D.f / D fm j f.m/ ¤ 0g D fm j f … mg: Since D.fg/ D D.f / \ D.g/, there is a topology on V for which the D.f / form a base. A pair of elements g; h 2 A, h ¤ 0, defines a function For U an open subset of V, we define of this form in some neighbourhood of each point of U. O W D.h/! k: m 7! g.m/ h.

m/ V.U / to be the set of functions f W U! k that are 66 3. AFFINE ALGEBRAIC VARIETIES PROPOSITION 3.22. The pair.V; Ah for each h 2 A X f0g. O V / is an affine algebraic variety with.D.h/; V /'O PROOF. Represent A as a quotient kŒX1; : : : ; Xn=a D kŒx1; : : : ; xn. Then.V; morphic to the k-ringed space attached to the algebraic set V.a/ (see 3.15). V / is iso- O We write spm.A/ for the topological space V, and Spm.A/ for the k-ringed space.V; V /. O ASIDE 3.23. We have attached to an affine k-algebra A an affine variety whose underlying topological space is the set of maximal ideals in A. It may seem strange to be describing a topological space in terms of maximal ideals in a ring, but the analysts have been doing this for more than 70 years. Gel’fand and Kolmogorov in 19393 proved that if S and T are compact topological spaces, and the rings of real-valued continuous functions on S and T are isomorphic (just as rings), then S and T are homeomorphic. The proof begins by showing that, for such a space S, the map P 7! mP defD ff W S! R j f.P / D 0g is one-to-one correspondence between the points in the space and maximal ideals in the ring. f. The category of affine algebraic varieties For each affine k-algebra A, we have an affine variety Spm.A/, and conversely, for each V /. We now make this affine variety.V; correspondence into an equivalence of categories. V /, we have an affine k-algebra kŒV D.V; O O Let ˛W A! B be a homomorphism of affine k-algebras. For every h 2 A, ˛.h/ is invertible in B˛.h/, and so the homomorphism A! B! B˛.h/ extends to a hom

omorphism g hm 7! ˛.g/ ˛.h/m W Ah! B˛.h/: For every maximal ideal n of B, m D ˛1.n/ is maximal in A because A=m! B=n D k is an injective map of k-algebras which implies that A=m D k. Thus ˛ defines a map 'W spm B! spm A; '.n/ D ˛1.n/ D m: For m D ˛1.n/ D '.n/, we have a commutative diagram: A A=m ˛'B B=n: Recall that the image of an element f of A in A=m'k is denoted f.m/. Therefore, the commutativity of the diagram means that, for f 2 A, f.'.n// D ˛.f /.n/, i.e., f ı'D ˛ ı f: (*) Since '1D.f / D D.f ı '/ (obviously), it follows from (*) that '1.D.f // D D.˛.f //; 3On rings of continuous functions on topological spaces, Doklady 22, 11-15. See also Allen Shields, Banach Algebras, 1939–1989, Math. Intelligencer, Vol 11, no. 3, p15. g. Explicit description of morphisms of affine varieties 67 and so'is continuous. Let f be a regular function on D.h/, and write f D g= hm, g 2 A. Then, from (*) we see that f ı'is the function on D.˛.h// defined by ˛.g/=˛.h/m. In particular, it is regular, and so f 7! f ı'maps regular functions on D.h/ to regular functions on D.˛.h//. It follows that f 7! f ı'sends regular functions on any open subset of spm.A/ to regular functions on the inverse image of the open subset. Thus ˛ defines a morphism of ringed spaces Spm.B/! Spm.A/. Conversely, by definition, a morphism of 'W.V; W / of affine algebraic varieties

defines a homomorphism of the associated affine k-algebras kŒW! kŒV. Since these maps are inverse, we have shown: V /!.W; O O PROPOSITION 3.24. For all affine algebras A and B, Homk-alg.A; B/ '! Mor.Spm.B/; Spm.A//I for all affine varieties V and W, Mor.V; W / '! Homk-alg.kŒW ; kŒV /: In terms of categories, Proposition 3.24 can be restated as: PROPOSITION 3.25. The functor A Spm A is a (contravariant) equivalence from the category of affine k-algebras to the category of affine algebraic varieties over k, with quasi-inverse.V; V /.V; V /. O O g. Explicit description of morphisms of affine varieties PROPOSITION 3.26. Let V km and W kn be algebraic subsets. The following conditions on a continuous map 'W V! W are equivalent: (a)'is a morphism of ringed spaces.V; V /!.W; W /; O O (b) the components '1; : : : ;'m of'are regular functions on V ; (c) f 2 kŒW H) f ı'2 kŒV. PROOF. (a) H) (b). By definition 'i D yi ı ', where yi is the coordinate function.b1; : : : ; bn/ 7! bi W W! k: Hence this implication follows directly from the definition of a regular map. (b) H) (c). The map f 7! f ı'is a k-algebra homomorphism from the ring of all functions W! k to the ring of all functions V! k, and (b) says that the map sends the coordinate functions yi on W into kŒV. Since the yi generate kŒW as a k-algebra, this implies that it sends kŒW into kŒV. (c) H) (a). The map f 7! f ı'is a homomorphism ˛W kŒW! k�

�V. It therefore defines a map spm.kŒV /! spm.kŒW /, and it remains to show that this coincides with'when we identify spm.kŒV / with V and spm.kŒW / with W. Let P 2 V, let Q D '.P /, and let mP and mQ be the ideals of elements of kŒV and kŒW that are zero at P and Q respectively. Then, for f 2 kŒW, ˛.f / 2 mP ” f.'.P // D 0 ” f.Q/ D 0 ” f 2 mQ: Therefore ˛1.mP / D mQ, which is what we needed to show. 68 3. AFFINE ALGEBRAIC VARIETIES The equivalence of (a) and (b) means that 'W V! W is a regular map of algebraic sets (in the sense of Chapter 2) if and only if it is a regular map of the associated affine algebraic varieties. Now consider equations Y1 D f1.X1; : : : ; Xm/ : : : Yn D fn.X1; : : : ; Xm/: On the one hand, they define a regular map 'W A m! A n, namely,.a1; : : : ; am/ 7!.f1.a1; : : : ; am/; : : : ; fn.a1; : : : ; am//: On the other hand, they define a homomorphism ˛W kŒY1; : : : ; Yn! kŒX1; : : : ; Xm of kalgebras, namely, that sending Yi to fi.X1; : : : ; Xm/. This map coincides with g 7! g ı ', because ˛.g/.P / D g.: : : ; fi.P /; : : :/ D g.'.P //: Now consider closed subsets V.a/ A m and V.b/ A n with a and b radical ideals. I claim that'maps V.a/ into V.b/ if and only if ˛.b/ a. Indeed, suppose '.V.a// V.b/, and let g 2 b;

for Q 2 V.a/, ˛.g/.Q/ D g.'.Q// D 0; and so ˛.g/ 2 I V.a/ D a. Conversely, suppose ˛.b/ a, and let P 2 V.a/; for f 2 b, f.'.P // D ˛.f /.P / D 0; and so '.P / 2 V.b/. When these conditions hold,'is the morphism of affine varieties V.a/! V.b/ corresponding to the homomorphism kŒY1; : : : ; Yn=b! kŒX1; : : : ; Xm=a defined by ˛. Thus, we see that the regular maps are all of the form V.a/! V.b/ P 7!.f1.P /; : : : ; fn.P //; fi 2 kŒX1; : : : ; Xm: In particular, they all extend to regular maps A m! A n. Examples of regular maps 3.27. Let R be a k-algebra. To give a k-algebra homomorphism kŒX! R is the same as giving an element (the image of X under the homomorphism): Homk-alg.kŒX; R/'R: Therefore Mor.V; A 1/ 3.24' Homk-alg.kŒX; kŒV /'kŒV : In other words, the regular maps V! A would hope). 1 are simply the regular functions on V (as we g. Explicit description of morphisms of affine varieties 69 0 D Spm k. Then A 0 consists of a single point and.A 0; 0! V from A 0 to an affine variety, sends A 0 to a point of V, and so Mor.A OA0/ D k. Every 0; V /'V. 3.28. Let A map A Alternatively, Mor.A 0; V /'Homk-alg.kŒV ; k/'V; where the last map sends ˛W kŒV! k to the point corresponding to the maximal ideal Ker.˛/. 3.29. Consider the regular map t 7!.t 2; t 3/W A bijective onto its image

, 1! A 2. This is V W Y 2 D X 3; but it is not an isomorphism onto its image because the inverse map is not regular. In view of 3.25, to prove this it suffices to show that t 7!.t 2; t 3/ does not induce an isomorphism on the 1 D kŒT and kŒV D rings of regular functions. We have kŒA kŒX; Y =.Y 2 X 3/ D kŒx; y. The map on rings is x 7! T 2; y 7! T 3; kŒx; y! kŒT ; which is injective, but its image is kŒT 2; T 3 ¤ kŒT. In fact, kŒx; y is not integrally closed:.y=x/2 x D 0, and so.y=x/ is integral over kŒx; y, but y=x … kŒx; y (it maps to T under the inclusion k.x; y/,! k.T //. 3.30. Let k have characteristic p ¤ 0, and consider the regular map.a1; : : : ; an/ 7!.ap 1 ; : : : ; ap n /W A n! A n: This is a bijection, but it is not an isomorphism because the corresponding map on rings, Xi 7! X p i W kŒX1; : : : ; Xn! kŒX1; : : : ; Xn; is not surjective. This is the famous Frobenius map. Take k to be the algebraic closure of Fp, and write F for the map. Recall that for each m 1 there is a unique subfield Fpm of k of degree m over Fp, and that its elements are the solutions of X pm D X (FT 4.23). The fixed points of F m n with coordinates in Fpm. Let f.X1; : : : ; Xn/ be a polynomial are precisely the points of A with coefficients in Fpm, say, f D X ci1inX i1 1 X in n ; ci1in 2 Fpm: If f.a1; : : : ; an/ D 0, then X 0 D c˛ai1

Spm.Ah/I PROOF. The map A! Ah defines a morphism spm.Ah/! spm.A/. The image is D.h/, and it is routine (using (1.13)) to verify the rest of the statement. If V D V.a/ A n, then.a1; : : : ; an/ 7!.a1; : : : ; an; h.a1; : : : ; an/1/W D.h/! A nC1; defines an isomorphism of D.h/ onto V.a; 1 hXnC1/. For example, there is an isomorphism of affine varieties a 7!.a; 1=a/W A 1 X f0g! V A 2; i. Properties of the regular map Spm.˛/ 71 with V equal to the subvariety XY D 1 of A 2, By an open affine (subset) U of an affine algebraic variety V, we mean an open subset V jU / is an affine algebraic variety. Thus, the proposition says that, for U such that.U; all nonzero h 2.V; V /, the open subset of V, where h is nonzero is an open affine. An open affine subset of an irreducible affine algebraic variety V is irreducible with the same dimension as V (2.52). O O REMARK 3.33. We have seen that all closed subsets and all basic open subsets of an affine variety V are again affine varieties with their natural ringed structure, but this is not true for all open subsets of V. For an open affine subset U, the natural map U! spm.U; V / is a bijection. However, for O U defD A 2 X f.0; 0/g D D.X/ [ D.Y / A 2; OA2/ D kŒX; Y (see 3.18), but U! spm kŒX; Y is not a bijection, we know that.U; OA2jU / is a union of affine because the ideal.X; Y / is not in the image. Clearly.U; algebraic varieties, and in Chapter 5 we shall recognize it as a (nonaffine) algebraic variety.

i. Properties of the regular map Spm.˛/ PROPOSITION 3.34. Let ˛W A! B be a homomorphism of affine k-algebras, and let 'W Spm.B/! Spm.A/ be the corresponding morphism of affine varieties. (a) The image of'is dense for the Zariski topology if and only if ˛ is injective. (b) The morphism'is an isomorphism from Spm.B/ onto a closed subvariety of Spm.A/ if and only if ˛ is surjective. PROOF. (a) Let f 2 A. If the image of'is dense, then f ı'D 0 H) f D 0: XY=1 72 3. AFFINE ALGEBRAIC VARIETIES On the other hand, if the image of'is not dense, then the closure of its image is a proper closed subset of Spm.A/, and so there is a nonzero function f 2 A that is zero on it. Then f ı'D 0. (See 2.40.) (b) If ˛ is surjective, then it defines an isomorphism A=a! B, where a is the kernel of ˛. This induces an isomorphism of Spm.B/ with its image in Spm.A/. The converse follows from the description of the closed subvarieties of Spm.A/ in the last section. A regular map 'W V! W of affine algebraic varieties is said to be a dominant if its image is dense in W. The proposition then says that:'is dominant ” f 7! f ı 'W.W; W /!.V; O V / is injective. O A regular map 'W V! W of affine algebraic varieties is said to be a closed immersion if it is an isomorphism of V onto a closed subvariety of W. The proposition then says that'is a closed immersion ” f 7! f ı 'W.W; W /!.V; O O V / is surjective. j. Affine space without coordinates Let E be a vector space over k of dimension n. The set A.E/ of points of E has a natural structure of an algebraic variety:

the choice of a basis for E defines a bijection A.E/! A n, and the inherited structure of an affine algebraic variety on A.E/ is independent of the choice of the basis (because the bijections defined by two different bases differ by an automorphism of A n). We now give an intrinsic definition of the affine variety A.E/. Let V be a finite- dimensional vector space over a field k. The tensor algebra of V is T V D M i 0 V ˝i D k ˚ V ˚.V ˝ V / ˚.V ˝ V ˝ V / ˚ with multiplication defined by.v1 ˝ ˝ vi /.v0 1 ˝ ˝ v0 j / D v1 ˝ ˝ vi ˝ v0 1 ˝ ˝ v0 j : It is a noncommutative k-algebra, and the choice of a basis e1; : : : ; en for V defines an isomorphism e1 ei 7! e1 ˝ ˝ ei W kfe1; : : : ; eng! T.V / to T V from the k-algebra of noncommuting polynomials in the symbols e1; : : : ; en. The symmetric algebra S.V / of V is defined to be the quotient of T V by the two- sided ideal generated by the elements v ˝ w w ˝ v; v; w 2 V: This algebra is generated as a k-algebra by commuting elements (namely, the elements of V D V ˝1), and so is commutative. The choice of a basis e1; : : : ; en for V defines an isomorphism e1 ei 7! e1 ˝ ˝ ei W kŒe1; : : : ; en! S.V / to S.V / from the commutative polynomial ring in the symbols e1; : : : ; en. This shows that S.V / is an affine k-algebra. The pair.S.V /; i/ consisting of S.V / and the natural k. Birational equivalence 73 k-linear map i W V! S.V / has the following universal

property: every k-linear map V! A from V into a k-algebra A extends uniquely to a k-algebra homomorphism S.V /! A: V i S.V / k-linear 9Š k-algebra A: (17) As usual, this universal property determines the pair.S.V /; i/ uniquely up to a unique isomorphism. We now define A.E/ to be Spm.S.E_//, where E_ is the dual vector space. For an affine k-algebra A, Mor.Spm.A/; A.E//'Homk-algebra.S.E_/; A/'Homk-linear.E_; A/'E ˝k A.3.24/.17/.linear algebra/: In particular, A.E/.k/'E: Moreover, the choice of a basis e1; : : : ; en for E determines a (dual) basis f1; : : : ; fn of E_, and hence an isomorphism of k-algebras kŒf1; : : : ; fn! S.E_/. The map of algebraic varieties defined by this homomorphism is the isomorphism n A.E/! A whose map on the underlying sets is the isomorphism E! kn defined by the basis of E. k. Birational equivalence Recall that if V is irreducible, then kŒV is an integral domain, and we write k.V / for its field of fractions. If U is an open affine subvariety of V, then kŒV kŒU k.V /, and so k.V / is also the field of fractions of kŒU. DEFINITION 3.35. Two irreducible affine algebraic varieties over k are birationally equivalent if their function fields are isomorphic over k. PROPOSITION 3.36. Irreducible affine varieties V and W are birationally equivalent if and only if there exist open affine subvarieties UV and UW of V and W respectively such that UV UW. PROOF. Let V and W be birationally equivalent irreducible affine varieties, and let A D kŒV and B D kŒW. We use the isomorphism to identify k

.V / and k.W /. This allows us to suppose that A and B have a common field of fractions K. Let x1; : : : ; xn generate B as k-algebra. As K is the field of fractions of A, there exists a d 2 A such that dxi 2 A for all i ; then B Ad. The same argument shows that there exists an e 2 B such that Ad Be. Now B Ad Be H) Be Ade.Be/e D Be; and so Ade D Be. This shows that the open subvarieties D.de/ V and D.e/ W are isomorphic. We have proved the “only if” part, and the “if” part is obvious. 74 3. AFFINE ALGEBRAIC VARIETIES THEOREM 3.37. Every irreducible affine algebraic variety of dimension d is birationally equivalent to a hypersurface in A d C1. PROOF. Let V be an irreducible variety of dimension d. According to (3.38) below, there exist rational functions x1; : : : ; xd C1 on V such that k.V / D k.x1; : : : ; xd ; xd C1/. Let f 2 kŒX1; : : : ; Xd C1 be an irreducible polynomial satisfied by the xi, and let H be the hypersurface f D 0. Then k.V / k.H / and so V and H are birationally equivalent. We review some definitions from FT, Chapter 2. Let F be a field. A polynomial f 2 F ŒX is separable if it has no multiple roots. Equivalent condition: gcd.f; df dX / D 1. When f is irreducible, this just says that df dX < deg f. An element dX of an algebraic extension E of F is separable over F if its minimal polynomial over F is separable, and E is separable over F if all its elements are separable over F. ¤ 0 because deg df PROPOSITION 3.38. Let ˝ be a finitely generated field extension of k of transcendence degree d. If k is perfect, then there exist x1; : : : ; xd C1 2 �

� such that ˝ D k.x1; : : : ; xd C1/. After renumbering, fx1; : : : ; xd g will be a transcendence basis for ˝ over k and xd C1 will be separable over k.x1; : : : ; xd /. PROOF. Let ˝ D k.x1; : : : ; xn/. After renumbering, we may suppose that x1; : : : ; xd are algebraically independent, hence a transcendence basis (1.63). If F has characteristic zero, then xd C1; : : : ; xn are separable over k.x1; : : : xd /, and so the primitive element theorem (FT 5.1) shows that there exists a y 2 ˝ for which ˝ D k.x1; : : : ; xd ; y/. Thus, we may suppose that k has characteristic p ¤ 0. Because k is perfect, every polynomial in with coefficients in k is a pth power in kŒX1; : : : ; Xn: X ai1inX i1p 1 : : : X inp n D X 1 p i1in a X i1 1 : : : X in n p : (18) Let.x1; : : : ; xn/ be a generating set for ˝ over k with the fewest elements. We shall assume that n > d C 1 and obtain a contradiction. As before, we may suppose that x1; : : : ; xd are algebraically independent. Then f.x1; : : : ; xd C1/ D 0 for some nonzero irreducible polynomial f.X1; : : : ; Xd C1/ with coefficients in k. Not all polynomials @f =@Xi are zero, for otherwise f would be a polynomial in X p d C1, and hence a pth power. After renumbering, we may suppose that @f =@Xd C1 ¤ 0. Now xd C1 is separably algebraic over k.x1; : : : ; xd / and xd C2 is algebraic over k.x1; : : : ; xd C1/ (hence over k.x1

; : : : ; xd /). According to the primitive element theorem (FT 5.1), there exists a y 2 ˝ such that k.x1; : : : ; xd C2/ D k.x1; : : : ; xd ; y/. Now ˝ D k.x1; : : : ; xd ; y; xd C3; : : : ; xn/, contradicting the minimality of n. 1 ; : : : ; X p We have shown that ˝ D k.z1; : : : ; zd C1/ for some zi 2 ˝. The argument in the last paragraph shows that, after renumbering, zd C1 will be separably algebraic over k.z1; : : : ; zd /, and this implies that fz1; : : : ; zd g is a transcendence basis for ˝ over k (1.63). l. Noether Normalization Theorem DEFINITION 3.39. The dimension of an affine algebraic variety is the dimension of the underlying topological space (2.48). DEFINITION 3.40. A regular map 'W W! V of affine algebraic varieties is finite if the map 'W kŒV! kŒW makes kŒW a finite kŒV -algebra. m. Dimension 75 THEOREM 3.41. Let V be an affine algebraic variety of dimension n. Then there exists a finite map V! A n. PROOF. Immediate consequence of (2.45). m. Dimension By definition, the dimension d of an affine variety V is the maximum length of a chain V0 V1 of distinct closed irreducible affine subvarieties. In this section, we prove that it is the length of every maximal chain of such subvarieties. THEOREM 3.42. Let V be an irreducible affine variety, and let f be a nonzero regular function on V. If f has a zero in V, then its zero set is of pure codimension 1. The Noether normalization theorem allows us to deduce this from the special case V D A proved in 2.64. n, PROOF. 4Let Z1; : : : ; Zn be the irreducible components of V.f /.

We have to show that dim Zi D dim V 1 for each i. There exists a point P 2 Zi not contained in any other Zj. Because the Zj are closed, there exists an open affine neighbourhood U of P in V not meeting any Zj with j ¤ i. Now V.f jU / D Zi \ U, which is irreducible. Therefore, on replacing V with U, we may assume that V.f / is irreducible. As V.f / is irreducible, the radical of.f / is a prime ideal p in kŒV. According to the d,! kŒV realizing d and f divides d. We claim that, in Noether normalization theorem (2.45), there exists an inclusion kŒA kŒV as a finite kŒA f0 in kŒV (see 1.45). Hence f0 2.f / p, and so rad.f0/ p \ kŒA fact, d -algebra. Let f0 D Nmk.V /=k.Ad / f. Then f0 2 kŒA d. d. Then g 2 p defD rad.f /, and so gm D f h for some h 2 kŒV, m 2 N. rad.f0/ D p \ kŒA Let g 2 p \ kŒA Taking norms, we find that gme D Nm.f h/ D f0 Nm.h/ 2.f0/; where e D Œk.V / W k.A The inclusion kŒA n/, and so g 2 rad.f0/, as claimed. d,! kŒV therefore induces an inclusion kŒA d = rad.f0/,! kŒV =p: This makes kŒV =p into a finite algebra over kŒA of these two k-algebras have the same transcendence degree: d = rad.f0/, and so the fields of fractions dim V.p/ D dim V.f0/: Clearly f ¤ 0 ) f0 ¤ 0, and f0 2 p ) f0 is nonconstant. Therefore dim V.f0/ D d 1 by (2.64). 4This proof was found by John Tate. 76 3. AFFINE AL

GEBRAIC VARIETIES We can restate Theorem 3.42 as follows: let V be a closed irreducible subvariety of A and let F 2 kŒX1; : : : ; Xn; then n V \ V.F / D 8 < : V ; pure codimension 1 otherwise. if F is identically zero on V if F has no zeros on V COROLLARY 3.43. Let V be an irreducible affine variety, and let Z be a maximal proper irreducible closed subset of V. Then dim.Z/ D dim.V / 1. PROOF. Because Z is a proper closed subset of V, there exists a nonzero regular function f on V vanishing on Z. Let V.f / be the zero set of f in V. Then Z V.f / V, and Z must be an irreducible component of V.f / for otherwise it wouldn’t be maximal in V. Thus Theorem 3.42 shows that dim Z D dim V 1. COROLLARY 3.44. Let V be an irreducible affine variety. Every maximal (i.e., nonrefinable) chain V D V0 V1 Vd (19) of distinct irreducible closed subsets of V has length d D dim.V /. PROOF. The last set Vd must be a point and each Vi must be maximal in Vi 1, and so, from 3.43, we find that dim V0 D dim V1 C 1 D dim V2 C 2 D D dim Vd C d D d: COROLLARY 3.45. Let V be an irreducible affine variety, and let f1; : : : ; fr be regular functions on V. Every irreducible component Z of V.f1; : : : fr / has codimension at most r: codim.Z/ r: For example, if the fi have no common zero on V, so that V.f1; : : : ; fr / is empty, then there are no irreducible components, and the statement is vacuously true. PROOF. We use induction on r. Because Z is an irreducible closed subset of V.f1; : : : ; fr1/, it is contained in some irreducible component Z

0 of V.f1; : : : fr1/. By induction, codim.Z0/ r 1. Also Z is an irreducible component of Z0 \ V.fr / because Z Z0 \ V.fr / V.f1; : : : ; fr / and Z is a maximal irreducible closed subset of V.f1; : : : ; fr /. If fr vanishes identically on Z0, then Z D Z0 and codim.Z/ D codim.Z0/ r 1; otherwise, the theorem shows that Z has codimension 1 in Z0, and codim.Z/ D codim.Z0/ C 1 r. EXAMPLE 3.46. In the setting of 3.45, the components of V.f1; : : : ; fr / need not all have the same dimension, and it is possible for all of them to have codimension < r without any of the fi being redundant. For example, let V be the cone in A 4. Then V.X1/ \ V is the union of two planes: X1X4 X2X3 D 0 V.X1/ \ V D f.0; 0; ; /g [ f.0; ; 0; /g: m. Dimension 77 Both of these have codimension 1 in V (as required by 3.42). Similarly, V.X2/ \ V is the union of two planes, V.X2/ \ V D f.0; 0; ; /g [ f.; 0; ; 0/g: However V.X1; X2/ \ V consists of a single plane f.0; 0; ; /g: it still has codimension 1 in V, but it requires both X1 and X2 to define it. PROPOSITION 3.47. Let Z be an irreducible closed subvariety of codimension r in an affine variety V. Then there exist regular functions f1; : : : ; fr on V such that Z is an irreducible component of V.f1; : : : ; fr / and all irreducible components of V.f1; : : : ; fr / have codimension r. PROOF. We know that there exists a chain of irreducible closed subsets V Z1 Zr D Z with cod

im Zi D i. We shall show that there exist f1; : : : ; fr 2 kŒV such that, for all s r, Zs is an irreducible component of V.f1; : : : ; fs/ and all irreducible components of V.f1; : : : ; fs/ have codimension s. We prove this by induction on s. For s D 1, take any f1 2 I.Z1/, f1 ¤ 0, and apply Theorem 3.42. Suppose f1; : : : ; fs1 have been chosen, and let Y1; Y2; : : : ; Ym, be the irreducible components of V.f1; : : : ; fs1/, numbered so that Zs1 D Y1. We seek an element fs that is identically zero on Zs but is not identically zero on any Yi — for such an fs, all irreducible components of Yi \ V.fs/ will have codimension s, and Zs will be an irreducible component of Y1 \ V.fs/. But no Yi is contained in Zs because Zs has smaller dimension than Yi, and so I.Zs/ is not contained in any of the ideals I.Yi /. Now the prime i I.Yi /, and this is avoidance lemma (see below) tells us that there exist an fs 2 I.Zs/ X S the function we want. LEMMA 3.48 (PRIME AVOIDANCE LEMMA). If an ideal a of a ring A is not contained in any of the prime ideals p1; : : : ; pr, then it is not contained in their union. PROOF. We may assume that none of the prime ideals pi is contained in a second, because then we could omit it. For a fixed i, choose an fi 2 a X pi and, for each j ¤ i, choose an j D1 fj lies in each pj with j ¤ i and a, but not in pi (here we fj 2 pj X pi. Then hi use that pi is prime). The element Pr i D1 hi is therefore in a but not in any pi. defD Qr EXAMPLE 3.49. When V is an affine variety whose coordinate ring is a unique factorization domain, every closed subset Z

of codimension 1 is of the form V.f / for some f 2 kŒV (see 2.66). The condition that kŒV be a unique factorization domain is definitely needed. Again consider the cone, in A 4 and let Z and Z0 be the planes V W X1X4 X2X3 D 0 Z D f.; 0; ; 0/g Z0 D f.0; ; 0; /g: Then Z \ Z0 D f.0; 0; 0; 0/g, which has codimension 2 in Z0. If Z D V.f / for some regular function f on V, then V.f jZ0/ D f.0; : : : ; 0/g, which has codimension 2, in violation of 3.42. Thus Z is not of the form V.f /, and so kŒX1; X2; X3; X4=.X1X4 X2X3/ is not a unique factorization domain. 78 3. AFFINE ALGEBRAIC VARIETIES Restatement in terms of affine algebras We restate some of these results in terms of affine algebras. 3.50. Theorem 3.42 says the following: let A be an affine k-algebra; if A is an integral domain and f 2 A is neither zero nor a unit, then every prime ideal p minimal among those containing.f / has height 1 (principal ideal theorem). 3.51. Corollary 3.44 says the following: let A be an affine k-algebra; if A is integral domain, then every maximal chain pd pd 1 p0 of distinct prime ideals has length equal to the Krull dimension of A. In particular, every maximal ideal in A has height dim.A/. 3.52. Let A be an affine k-algebra; if A is an integral domain and every prime ideal of height 1 in A is principal, then A is a unique factorization domain. In order to prove this, it suffices to show that every irreducible element f of A is prime (1.26). Let p be minimal among the prime ideals containing.f /. According to 3.50, p has height 1, and so it is principal, say p D.g/. As.f /.g

/, f D gq for some q 2 A. Because f is irreducible, q is a unit, and so.f / D.g/ D p — the element f is prime. 3.53. Proposition 3.47 says the following: let A be an affine k-algebra, and let p be a prime ideal in A. If p has height r, then there exist elements f1; : : : ; fr 2 A such that p is minimal among the prime ideals containing.f1; : : : ; fr /. ASIDE 3.54. Statements 3.50 and 3.53 are true for all noetherian rings (CA 21.3, 21.8). However, 3.51 may fail. For example, as we noted on p. 16 a noetherian ring may have infinite Krull dimension. Moreover, a noetherian ring may have finite Krull dimension d without all of its maximal ideals having height d. For example, let A D RŒX, where R D kŒt.t/ is a discrete valuation ring with maximal ideal.t/. The Krull dimension of A is 2, and.t; X/.t/.0/ is a maximal chain of prime ideals, but the ideal.tX 1/ is maximal (because A=.tX 1/'Rt, see 1.13) and of height 1 (because it is in kŒt; X and A is obtained from kŒt; X by inverting the elements of kŒt X.t/). 3 is an irreducible component of V.f1; f2/ ASIDE 3.55. Proposition 3.47 shows that a curve C in A for some f1, f2 2 kŒX; Y; Z. In fact C D V.f1; f2; f3/ for suitable polynomials f1; f2, and f3 — this is an exercise in Shafarevich 1994 (I 6, Exercise 8; see also Hartshorne 1977, I, Exercise 2.17). 3 — see Apparently, it is not known whether two polynomials always suffice to define a curve in A 3 can’t be defined by two polynomials (ibid. Kunz 1985, p136.5 The union of two skew lines in P 3 can be defined by two polynomials

. p. 140), but it is unknown whether all connected curves in P Macaulay (the man, not the program) showed that for every r 1, there is a curve C in A 3 such that I.C / requires at least r generators (see the same exercise in Hartshorne for a curve whose ideal can’t be generated by 2 elements).6 5Kunz, Ernst Introduction to commutative algebra and algebraic geometry. Birkh¨auser Boston, Inc., Boston, MA 6In 1882 Kronecker proved that every algebraic subset in P n can be cut out by n C 1 polynomial equations. 3 which he claimed was not In 1891 Vahlen asserted that the result was best possible by exhibiting a curve in P the zero locus of 3 equations. It was only 50 years later, in 1941, that Perron gave 3 equations defining Vahlen’s curve, thus refuting Vahlen’s claim which had been accepted for half a century. Finally, in 1973 Eisenbud and n (mo35468 Evans proved that n equations always suffice to describe (set-theoretically) an algebraic subset of P Georges Elencwajg). m. Dimension 79 In general, a closed variety V of codimension r in A n/ is said to be a set-theoretic complete intersection if there exist r polynomials fi 2 kŒX1; : : : ; Xn (resp. homogeneous polynomials fi 2 kŒX0; : : : ; Xn/ such that n (resp. P V D V.f1; : : : ; fr /: Such a variety is said to be an ideal-theoretic complete intersection if the fi can be chosen so that I.V / D.f1; : : : ; fr /. Chapter V of Kunz’s book is concerned with the question of when a variety is a complete intersection. Obviously there are many ideal-theoretic complete intersections, but most of the varieties one happens to be interested in turn out not to be. For example, no abelian variety of dimension > 1 is an ideal-theoretic complete intersection (being an ideal-theoretic complete intersection imposes constraints on the cohomology of the variety, which are not fulfilled in the case of abelian varieties). Let P be a point on an irreducible

variety V A n. Then 3.47 shows that there is a neighbourhood n and functions f1; : : : ; fr on U such that U \ V D V.f1; : : : ; fr / (zero set in U /. Thus U of P in A U \ V is a set-theoretic complete intersection in U. One says that V is a local complete intersection n such that the ideal I.V \ U / can be at P 2 V if there is an open affine neighbourhood U of P in A generated by r regular functions on U. Note that ideal-theoretic complete intersection ) local complete intersection at all p: It is not difficult to show that a variety is a local complete intersection at every nonsingular point (cf. 4.36). Exercises 3-1. Show that a map between affine varieties can be continuous for the Zariski topology without being regular. 3-2. Let V D Spm.A/, and let Z D Spm.A=a/ Spm.A/. Show that a function f on an open subset U of Z is regular if and only if, for each P 2 U, there exists a regular function f 0 on an open neighbourhood U 0 of P in V such that f and f 0 agree on U 0 \ U. 3-3. Find the image of the regular map.x; y/ 7!.x; xy/W A 2! A 2 and verify that it is neither open nor closed. 3-4. Show that the circle X 2 C Y 2 D 1 is isomorphic (as an affine variety) to the hyperbola XY D 1, but that neither is isomorphic to A 1. (Assume char.k/ ¤ 2:/ 3-5. Let C be the curve Y 2 D X 2 C X 3, and let'be the regular map t 7!.t 2 1; t.t 2 1//W A 1! C: Is'an isomorphism? CHAPTER 4 Local Study Geometry is the art of drawing correct conclusions from incorrect figures. (La g´eom´etrie est l’art de raisonner juste sur des figures fausses.) Descartes In this chapter, we examine the structure of an affine algebraic variety near a point. We begin with the case of a plane curve, since the ideas in the general case are the same but

the proofs are more complicated. a. Tangent spaces to plane curves Consider the curve V in the plane defined by a nonconstant polynomial F.X; Y /, V W F.X; Y / D 0: We assume that F.X; Y / has no multiple factors, so that.F.X; Y // is a radical ideal and I.V / D.F.X; Y //. We can factor F into a product of irreducible polynomials, F.X; Y / D Q Fi.X; Y /, and then V D S V.Fi / expresses V as a union of its irreducible components (see 2.29). Each component V.Fi / has dimension 1 (by 2.64) and so V has pure dimension 1. If F.X; Y / itself is irreducible, then kŒV D kŒX; Y =.F.X; Y // D kŒx; y is an integral domain. Moreover, if F ¤ X c, then x is transcendental over k and y is algebraic over k.x/, and so x is a transcendence basis for k.V / over k. Similarly, if F ¤ Y c, then y is a transcendence basis for k.V / over k. Let.a; b/ be a point on V. If we were doing calculus, we would say that the tangent space at P D.a; b/ is defined by the equation.a; b/.X a/ C @F @Y @F @X.a; b/.Y b/ D 0: (20) This is the equation of a line unless both @F the equation of a plane. @X.a; b/ and @F @Y.a; b/ are zero, in which case it is We are not doing calculus, but we can define @ X aij X i Y j D X iaij X i 1Y j ; @ @X 81 @ @Y @X and @ X @Y by aij X i Y j D X jaij X i Y j 1, 82 4. LOCAL STUDY and make the same definition. DEFINITION 4.1. The tangent space TP V to V at P D.a; b/ is the algebraic subset defined by equation (20). If @F @X.a;

b/ and @F @Y.a; b/ are not both zero, then TP.V / is a line through.a; b/, and we say that P is a nonsingular or smooth point of V. Otherwise, TP.V / has dimension 2, and we say that P is singular or multiple. The curve V is said to be nonsingular or smooth if all its points are nonsingular. Examples For each of the following examples, the reader is invited to sketch the curve. Assume that char.k/ ¤ 2; 3. 4.2. X m C Y m D 1. The tangent space at.a; b/ is given by the equation mam1.X a/ C mbm1.Y b/ D 0: All points on the curve are nonsingular unless the characteristic of k divides m, in which case X m C Y m 1 has multiple factors, X m C Y m 1 D X m0p C Y m0p 1 D.X m0 C Y m0 1/p: 4.3. Y 2 D X 3 (sketched in 4.12 below). The tangent space at.a; b/ is given by the equation 3a2.X a/ C 2b.Y b/ D 0: The only singular point is.0; 0/. 4.4. Y 2 D X 2.X C 1/ (sketched in 4.10 below). Here again only.0; 0/ is singular. 4.5. Y 2 D X 3 C aX C b. In 2.2 we sketched two nonsingular examples of such curves, and in 4.10 and 4.11 we sketch two singular examples. The singular points of the curve are the common zeros of the polynomials Y 2 X 3 aX b; 2Y; 3X 2 C a, which consist of the points.c; 0/ with c a common zero of X 3 C aX C b; 3X 2 C a. As 3X 2 C a is the derivative of X 3 C aX C b, we see that V is singular if and only if X 3 C aX C b has a multiple root. 4.6. V D V.F G/ where F G has no multiple factors (so F and G are coprime). Then V D V.F / [ V.G/, and a

point.a; b/ is singular if and only if it is ˘ a singular point of V.F /, ˘ a singular point of V.G/, or ˘ a point of V.F / \ V.G/. This follows immediately from the product rule: @.F G/ @X D F @G @X C @F @X G; @.F G/ @Y D F @G @Y C @F @Y G: b. Tangent cones to plane curves 83 The singular locus PROPOSITION 4.7. The nonsingular points of a plane curve form a dense open subset of the curve. PROOF. Let V D V.F /, where F is a nonconstant polynomial in kŒX; Y without multiple factors. It suffices to show that the nonsingular points form a dense open subset of each irreducible component of V, and so we may assume that V (hence F ) is irreducible. It suffices to show that the set of singular points is a proper closed subset. Since it is the set of common zeros of the polynomials F ; @F @X ; @F @Y ; it is obviously closed. It will be proper unless @F=@X and @F=@Y are both identically zero on V, and hence both multiples of F, but, as they have lower degree than F, this is impossible unless they are both zero. Clearly @F=@X D 0 if and only if F is a polynomial in Y (k of characteristic zero) or is a polynomial in X p and Y (k of characteristic p/. A similar remark applies to @F=@Y. Thus if @F=@X and @F=@Y are both zero, then F is constant (characteristic zero) or a polynomial in X p, Y p, and hence a pth power (characteristic p, see (18)). These are contrary to our assumptions. Thus the singular points form a proper closed subset, called the singular locus. ASIDE 4.8. In common usage, “singular” means uncommon or extraordinary as in “he spoke with singular shrewdness”. Thus the proposition says that singular points (mathematical sense) are singular (usual sense). b. Tangent cones to plane curves A polyn

omial F.X; Y / can be written (uniquely) as a finite sum F D F0 C F1 C F2 C C Fm C (21) with each Fm a homogeneous polynomial of degree m. The term F1 will be denoted F` and called the linear form of F, and the first nonzero term on the right of (21) (the homogeneous summand of F of least degree) will be denoted F and called the leading form of F. If P D.0; 0/ is on the curve V defined by F, then F0 D 0 and (21) becomes F D aX C bY C higher degree terms, and the equation of the tangent space is aX C bY D 0: DEFINITION 4.9. Let F.X; Y / be a polynomial without square factors, and let V be the curve defined by F. If.0; 0/ 2 V, then the geometric tangent cone to V at.0; 0/ is the zero set of F. The tangent cone is the pair.V.F/; kŒX; Y =F/. To obtain the tangent cone at any other point, translate to the origin, and then translate back. Note that the geometric tangent cone at a point on a curve always has dimension 1. While the tangent space tells you whether a point is nonsingular or not, the tangent cone also gives you information on the nature of a singularity. 84 4. LOCAL STUDY In general we can factor F as F.X; Y / D cX r0 Y.Y ai X/ri : i Then deg F D P ri is called the multiplicity of the singularity, multP.V /. A multiple point is ordinary if its tangents are nonmultiple, i.e., ri D 1 all i. An ordinary double point is called a node. There are many names for special types of singularities — see any book, especially an old book, on algebraic curves. Examples The following examples are adapted from Walker, Robert J., Algebraic Curves. Princeton Mathematical Series, vol. 13. Princeton University Press, Princeton, N. J., 1950 (reprinted by Dover 1962). 4.10. F.X. The tangent cone at.0; 0/ is defined by Y 2 X 2. It is the pair

of lines Y D ˙X, and the singularity is a node. 4.11. F.X. The origin is an isolated point of the real locus. It is again a node, but the tangent cone is defined by Y 2 C X 2, which is the pair of lines Y D ˙iX. In this case, the real locus of the tangent cone is just the point (0,0). 4.12. F.X; Y / D X 3 Y 2. Here the origin is a cusp. The tangent cone is defined by Y 2, which is the X-axis (doubled). 4.13. F.X; Y / D 2X 4 3X 2Y C Y 2 2Y 3 C Y 4. The origin is again a double point, but this time it is a tacnode. The tangent cone is again defined by Y 2. 4.14. F.X; Y / D X 4 C X 2Y 2 2X 2Y XY 2 Y 2. The origin is again a double point, but this time it is a ramphoid cusp. The tangent cone is again defined by Y 2. c. The local ring at a point on a curve 85 4.15. F.X; Y / D.X 2 C Y 2/2 C 3X 2Y Y 3. The origin is an ordinary triple point. The tangent cone is defined by 3X 2Y Y 3, which is the triple of lines Y D 0, Y D ˙ p 3X. 4.16. F.X; Y / D.X 2 C Y 2/3 4X 2Y 2. The origin has multiplicity 4. The tangent cone is defined by 4X 2Y 2, which is the union of the X and Y axes, each doubled. 4.17. F.X; Y / D X 6 X 2Y 3 Y 5. The tangent cone is defined by X 2Y 3 C Y 5, which consists of a triple line Y 3 D 0 and a pair of lines Y D ˙iX. ASIDE 4.18. Note that the real locus of the algebraic curve in 4.17 is smooth even though the curve itself is singular. Another example of such a curve is Y 3 C 2X 2Y X 4 D 0. This is singular at.0; 0/, but its real

locus is the image of R under the analytic map t 7!.t 3 C 2t; t.t 3 C 2//, which is injective, 2 with closed image. See Milnor, J., Singular proper, and immersive, and hence an embedding into R points of complex hypersurfaces. PUP, 1968, or mo98366 (Elencwajg). c. The local ring at a point on a curve PROPOSITION 4.19. Let P be a point on a plane curve V, and let m be the corresponding maximal ideal in kŒV. If P is nonsingular, then dimk.m=m2/ D 1, and otherwise dimk.m=m2/ D 2. PROOF. Assume first that P D.0; 0/. Then m D.x; y/ in kŒV D kŒX; Y =.F.X; Y // D kŒx; y. Note that m2 D.x2; xy; y2/, and m=m2 D.X; Y /=.m2 C F.X; Y // D.X; Y /=.X 2; XY; Y 2; F.X; Y //: In this quotient, every element is represented by a linear polynomial cx C dy, and the only relation is F`.x; y/ D 0. Clearly dimk.m=m2/ D 1 if F` ¤ 0, and dimk.m=m2/ D 2 otherwise. Since F` D 0 is the equation of the tangent space, this proves the proposition in this case. The same argument works for an arbitrary point.a; b/ except that one uses the variables X 0 D X a and Y 0 D Y b; in essence, one translates the point to the origin. We explain what the condition dimk.m=m2/ D 1 means for the local ring P D kŒV m. Let n be the maximal ideal mkŒV m of this local ring. The map m! n induces an isomorphism m=m2! n=n2 (see 1.15), and so we have O P nonsingular ” dimk m=m2 D 1 ” dimk n=n2 D 1: Nakayama’s lemma (1.3) shows that

the last condition is equivalent to n being a principal ideal. As P has Krull dimension one (2.64), for its maximal ideal to be principal means that it is a regular local ring of dimension 1 (see 1.6). Thus, for a point P on a curve, O P nonsingular ” P regular. O 86 4. LOCAL STUDY PROPOSITION 4.20. Every regular local ring of dimension one is a principal ideal domain. PROOF. Let A be such a ring, and let m D./ be its maximal ideal. According to the Krull intersection theorem (1.8), T r0 mr D.0/. Let a be a proper nonzero ideal in A. As a is finitely generated, there exists an r 2 N such that a mr but a 6 mrC1. Therefore, there exists an a D c r 2 a such that a … mrC1. The second condition implies that c … m, and so it is a unit. Therefore. r / D.a/ a. r /, and so a D. r / D mr. We have shown that all ideals in A are principal. By assumption, there exists a prime ideal p properly contained in m. Then A=p is an integral domain. As … p, it is not nilpotent in A=p, and hence not nilpotent in A. Let a and b be nonzero elements of A. There exist r; s 2 N such that a 2 mr X mrC1 and b 2 ms X msC1. Then a D u r and b D v s with u and v units, and ab D uv rCs ¤ 0. Hence A is an integral domain. It follows from the elementary theory of principal ideal domains that the following conditions on a principal ideal domain A are equivalent: (a) A has exactly one nonzero prime ideal; (b) A has exactly one prime element up to associates; (c) A is local and is not a field. A ring satisfying these conditions is called a discrete valuation ring. THEOREM 4.21. A point P on a plane algebraic curve is nonsingular if and only if regular, in which case it is a discrete valuation ring. P is O PROOF. The statement summarizes the above discussion. d. Tangent spaces to algebraic subsets of Am Before defining tangent spaces at points of an algebraic

subset of A terminology from linear algebra (which should be familiar from advanced calculus). m we review some LINEAR ALGEBRA For a vector space km, let Xi be the i th coordinate function a 7! ai. Thus X1; : : : ; Xm is the dual basis to the standard basis for km. A linear form P ai Xi can be regarded as an element of the dual vector space.km/_ D Hom.km; k/. Let A D.aij / be an n m matrix. It defines a linear map ˛W km! kn, by 0 B @ a1 ::: am 1 C A 7! A 1 C A D 0 B @ a1 ::: am 0 Pm B @ j D1 a1j aj ::: j D1 anj aj Pm 1 C A : Write X1; : : : ; Xm for the coordinate functions on km and Y1; : : : ; Yn for the coordinate functions on kn. Then Yi ı ˛ D m X j D1 aij Xj : This says that the i th coordinate of ˛.a/ is m X j D1 aij.Xj a/ D m X j D1 aij aj : d. Tangent spaces to algebraic subsets of A m 87 TANGENT SPACES DEFINITION 4.22. Let V km be an algebraic subset of km, and let a D I.V /. The tangent space Ta.V / to V at a point a D.a1; : : : ; am/ of V is the subspace of the vector space with origin a cut out by the linear equations m X i D1 @F @Xi ˇ ˇ ˇ ˇa.Xi ai / D 0; F 2 a. (22) In other words, Ta.A the subspace of Ta.A m/ defined by the equations (22). m/ is the vector space of dimension m with origin a, and Ta.V / is Write.dXi /a for.Xi ai /; then the.dXi /a form a basis for the dual vector space m/_. As in m/_ to Ta.A m/ — in fact, they are the coordinate functions on Ta.A Ta.A advanced calculus, we define the differential of a polynomial F 2 kŒX1

; : : : ; Xm at a by the equation:.dF /a D.dXi /a: m X i D1 @F @Xi ˇ ˇ ˇ ˇa It is again a linear form on Ta.A m/ defined by the equations: Ta.A m/. In terms of differentials, Ta.V / is the subspace of.dF /a D 0; F 2 a: (23) I claim that, in (22) and (23), it suffices to take the F to lie in a generating subset for a. The product rule for differentiation shows that if G D P j Hj Fj, then.dG/a D X Hj.a/.dFj /a C Fj.a/.dHj /a: j If F1; : : : ; Fr generate a and a 2 V.a/, so that Fj.a/ D 0 for all j, then this equation becomes.dG/a D X Hj.a/.dFj /a: j Thus.dF1/a; : : : ;.dFr /a generate the k-vector space f.dF /a j F 2 ag. DEFINITION 4.23. A point a on an algebraic set V is nonsingular (or smooth) if it lies on a single irreducible component W of V and the dimension of the tangent space at a is equal to the dimension of W ; otherwise it is singular (or multiple). Thus, a point a on an irreducible algebraic set V is nonsingular if and only if dim Ta.V / D dim V. As in the case of plane curves, a point on V is nonsingular if and only if it lies on a single irreducible component of V, and is nonsingular on it. Let a D.F1; : : : ; Fr /, and let J D Jac.F1; : : : ; Fr / D D @Fi @Xj F1 @X1 ::: @Fr @X1 ; : : : ; 1 C C A : @F1 @Xm ::: @Fr @Xm Then the equations defining Ta.V / as a subspace of Ta.A linear algebra shows that m/ have matrix J.a/. Therefore, dimk Ta.V / D

m rank J.a/; 88 4. LOCAL STUDY and so a is nonsingular if and only if the rank of Jac.F1; : : : ; Fr /.a/ is equal to m dim.V /. For example, if V is a hypersurface, say I.V / D.F.X1; : : : ; Xm//, then Jac.F /.a/ D @F @X1.a/; : : : ;.a/ ; @F @Xm and a is nonsingular if and only if not all of the partial derivatives @F @Xi We can regard J as a matrix of regular functions on V. For each r, vanish at a. fa 2 V j rank J.a/ rg is closed in V, because it is the set where certain determinants vanish. Therefore, there is an open subset U of V on which rank J.a/ attains its maximum value, and the rank jumps on closed subsets. Later (4.37) we shall show that the maximum value of rank J.a/ is m dim V, and so the nonsingular points of V form a nonempty open subset of V. e. The differential of a regular map Consider a regular map 'W A m! A n; a 7!.P1.a1; : : : ; am/; : : : ; Pn.a1; : : : ; am//: We think of'as being given by the equations Yi D Pi.X1; : : : ; Xm/; i D 1; : : : ; n: It corresponds to the map of rings 'W kŒY1; : : : ; Yn! kŒX1; : : : ; Xm sending Yi to Pi.X1; : : : ; Xm/, i D 1; : : : ; n. Let a 2 A n/ to be the map such that m, and let b D '.a/. Define.d'/aW Ta.A.d Yi /b ı.d'/a D X @Pi @Xj m/! Tb.A ˇ ˇ ˇ ˇa.dXj /a; i.e., relative to the standard bases,.d'/a is the map with matrix Jac.P1; : : : ; Pn/.a/

D 0 B B @ @P1 @X1 @Pn @X1.a/; :::.a/; : : : ;.a/ @P1 @Xm ::: : : : ; @Pn @Xm.a/ 1 C C A : For example, suppose a D.0; : : : ; 0/ and b D.0; : : : ; 0/, so that Ta.A kn, and m/ D km and Tb.A n/ D Pi D cij Xj C.higher terms), i D 1; : : : ; n: m X j D1 Then Yi ı.d'/a D P i.e., it is simply t 7!.cij /t. j cij Xj, and the map on tangent spaces is given by the matrix.cij /, Let F 2 kŒX1; : : : ; Xm. We can regard F as a regular map A m! A 1, whose differential will be a linear map m/! Tb.A.dF /aW Ta.A 1/ with k, we obtain an identification of the differential of F (F b D F.a/: When we identify Tb.A regarded as a regular map) with the differential of F (F regarded as a regular function). 1/; f. Tangent spaces to affine algebraic varieties 89 LEMMA 4.24. Let 'W A km into W D V.b/ kn, then.d'/a maps Ta.V / into Tb.W /, b D '.a/. n be as at the start of this subsection. If'maps V D V.a/ m! A PROOF. We are given that and have to prove that f 2 b ) f ı'2 a; f 2 b ).df /b ı.d'/a is zero on Ta.V /: The chain rule holds in our situation: @f @Xi D n X j D1 @f @Yj @Yj @Xi ; Yj D Pj.X1; : : : ; Xm/; f D f.Y1; : : : ; Yn/: If'is the map given by the equations Yj D Pj.X1; : : : ; Xm/; j D 1; : : : ; n; then the chain rule

implies d.f ı '/a D.df /b ı.d'/a; b D '.a/: Let t 2 Ta.V /; then.df /b ı.d'/a.t/ D d.f ı '/a.t/; which is zero if f 2 b because then f ı'2 a. Thus.d'/a.t/ 2 Tb.W /. We therefore get a map.d'/aW Ta.V /! Tb.W /. The usual rules from advanced calculus hold. For example,.d /b ı.d'/a D d. ı '/a; b D '.a/: f. Tangent spaces to affine algebraic varieties The definition (4.22) of the tangent space at a point on an algebraic set uses the embedding n. In this section, we give an intrinsic definition of the tangent of the algebraic set into A space at a point of an affine algebraic variety that makes clear that it depends only on the local ring at the point. Dual numbers For an affine algebraic variety V and a k-algebra R (not necessarily an affine k-algebra), we define V.R/ to be Homk-alg.kŒV ; R/. For example, if V A n and a D I.V /, then V.R/ D f.a1; : : : ; an/ 2 Rn j f.a1; : : : ; an/ D 0 for all f 2 ag: A homomorphism R! S of k-algebras defines a map V.R/! V.S/ of sets. The ring of dual numbers is kŒ" defD kŒX=.X 2/, where " D X C.X 2/. Thus kŒ" D k ˚ k" as a k-vector space, and.a C b"/.a0 C b0"/ D aa0 C.ab0 C a0b/"; a; b; a0; b0 2 k: Note that there is a k-algebra homomorphism " 7! 0W kŒ"! k. 90 4. LOCAL STUDY DEFINITION 4.25. Let P be a point on an affine algebraic variety V over k. The tangent space to V