Solutions to the Problems in Introduction to Commutative Algebra by M. F. Atiyah and I. G. MacDonald

J. David Taylor

October 20, 2018

Contents

1 Rings and Ideals 7 1.1...... 7 1.2...... 7 1.3...... 8 1.4...... 8 1.5...... 8 1.6...... 9 1.7...... 10 1.8...... 10 1.9...... 10 1.10...... 10 1.11...... 10 1.12...... 11 1.13...... 11 1.14...... 11 1.15...... 12 1.16...... 12 1.17...... 14 1.18...... 15 1.19...... 15 1.20...... 15 1.21...... 16 1.22...... 17 1.23...... 18 1.24...... 18

1 CONTENTS 2

1.25...... 19 1.26...... 20 1.27...... 20 1.28...... 20

2 Modules 22 2.1...... 22 2.2...... 22 2.3...... 23 2.4...... 23 2.5...... 24 2.6...... 24 2.7...... 25 2.8...... 25 2.9...... 25 2.10...... 26 2.11...... 26 2.12...... 27 2.13...... 27 2.14...... 27 2.15...... 27 2.16...... 28 2.17...... 29 2.18...... 29 2.19...... 29 2.20...... 30 2.21...... 30 2.22...... 31 2.23...... 31 2.24...... 31 2.25...... 31 2.26...... 32 2.27 Incomplete...... 32 2.28...... 33

3 Rings and Modules of Fractions 34 3.1 Current...... 34 3.2...... 34 CONTENTS 3

3.3...... 34 3.4...... 35 3.5...... 35 3.6...... 35 3.7...... 35 3.8...... 36 3.9...... 36 3.10...... 37 3.11...... 37 3.12...... 38 3.13...... 39 3.14...... 39 3.15...... 39 3.16...... 39 3.17...... 40 3.18...... 40 3.19...... 40 3.20...... 41 3.21...... 42 3.22...... 44 3.23...... 44 3.24...... 44 3.25...... 45 3.26...... 45

4 Primary Decomposition 46 4.1...... 46 4.2...... 46 4.3...... 46 4.4...... 46 4.5...... 46 4.6...... 46 4.7...... 47 4.8...... 47 4.9...... 47 4.10...... 48 4.11...... 48 CONTENTS 4

4.12...... 49 4.13...... 49 4.14...... 50 4.15...... 50 4.16...... 50 4.17...... 50 4.18...... 50 4.19...... 51 4.20...... 51 4.21...... 52 4.22...... 52 4.23...... 53

5 Integral Dependence and Valuations 54 5.1...... 54 5.2...... 54 5.3...... 54 5.4...... 54 5.5...... 55 5.6...... 55 5.7...... 55 5.8...... 55 5.9...... 56 5.10...... 56 5.11...... 56 5.12...... 56 5.13...... 57 5.14...... 57 5.15...... 57 5.16...... 57 5.17...... 58 5.18...... 58 5.19 Undone...... 58 5.20...... 58 5.21...... 58 5.22 Undone...... 59 5.23...... 59 CONTENTS 5

5.24...... 59 5.25...... 60 5.26...... 60 5.27...... 61 5.28...... 61 5.29...... 62 5.30...... 62 5.31...... 62 5.32 Undone...... 63 5.33 Undone...... 63 5.34 Undone...... 63 5.35 Undone...... 63

6 Chain Conditions 64 6.1...... 64 6.2...... 64 6.3...... 64 6.4...... 65 6.5...... 65 6.6...... 65 6.7...... 66 6.8...... 66 6.9 Undone...... 66 6.10 Undone...... 66 6.11 Undone...... 66 6.12 Undone...... 66

7 Noetherian Rings 67 7.1...... 67

8 Artin Rings 68

9 Discrete Valuation Rings and Dedekind Domains 69

10 Completions 70

11 Dimension Theory 71

12 Conventions 72 CONTENTS 6

13 Remarks 73 1 RINGS AND IDEALS 7

1 Rings and Ideals

1.1

Let A be a ring with nilpotent element x and unit element u. We will show that u + x is a unit. Take u = 1 to see that 1 + x is a unit. Constructive Proof: Let n > 0 be the least integer such that xn = 0 and define

n−1 X y := u−1 (−u−1x)i. i=0 Note that (−u−1x)n = (−u−1)nxn = 0. As a consequence, we see that

n−1 ! X y(u + x) = (−u−1x)i (1 + u−1x) = 1 − (−u−1x)n = 1. i=0 So u + x is a unit with inverse y. Non-Constructive Proof: By Proposition 1.8, the nilradical of A is the intersection of all prime ideals of A. Since every maximal ideal of A is prime, the nilradical of A is contained in the Jacobson radical. Therefore, x is contained in the Jacobson radical of A. Apply Proposition 1.9 with y = −u−1 to seee that, 1 + u−1x is a unit. Since the units of A form a group, u(1 + u−1x) = u + x is a unit as well.

1.2

n m Let f = a0 + a1x + ... + anx and g = b0 + b1x + ... + bmx be elements of A[x]. (i) Suppose that f is a unit with inverse g. Let p be a prime ideal of A. Since A/p is a domain and f · g = 1 in (A/p)[x], f must have degree 0 when reduced mod p. Thus, a1, . . . , an are contained in every prime ideal of A. On the other hand, Proposition 1.8 tells us that the nilradical of A is the intersection of all the prime ideals of A. So a1, . . . , an are nilpotent. As a consequence, a0 = f + (a0 − f) is the sum of a unit and a nilpotent element, hence a unit by Exercise 1.1.

If a0 is a unit and a1, . . . , an are nilpotent, then f − a0 is nilpotent and f = a0 + (f − a0) is the sum of a unit and a nilpotent element, hence a unit by Exercise 1.1. (ii) Let p be a prime ideal of A. If f k = 0, then f k = 0 in (A/p)[x]. Since A/p is a domain, f = 0 in (A/p)[x]. Thus, a0, . . . , an are contained in every prime ideal of A, and are nilpotent as a consequence of Proposition 1.8.

If a0, . . . , an are in the nilradical of A[x], then the A[x]-linear combination f is also in the nilradical because the nilradical is an ideal by Proposition 1.7. (iii) If f is a zero-divisor, let g be a non-zero annihilator of f in A[x] of least degree. If g has positive degree, then ang is an annihilator of f of degree less than g, hence ang = 0. If k is a positive integer such that ajg = 0 k n for j ≥ k, then (f − (akx + ... + anx ))g = 0 implies that ak−1g = 0. Therefore, the least such k is 0. That is, a0g = 0. Furthermore, we have shown that akbm = 0 for all k. Therefore, bmf = 0. (In fact, we have implicitly shown that g has degree 0) (iv) Let a, b, c be the ideals generated by the coefficients of f, g, and fg, respectively. The inclusion c ⊆ ab shows that f and g are primitive if fg is. On the other hand, if fg is not primitive, then c is contained in some maximal ideal, m, of A. Therefore fg = 0 in (A/m)[x]. Since A/m is a field, f = 0 or g = 0 in (A/m)[x]. This in turn implies that a ⊆ m or b ⊆ m. In either case, at least one of f or g is not primitive. 1 RINGS AND IDEALS 8

1.3

n Define Ar := A[x1, . . . , xr] for all positive integers r. Let f = a0 + ... + anxr be an element of Ar = Ar−1[xr] with a0, . . . , an elements of Ar−1.

Suppose that every unit of Ar−1, when viewed as a polynomial with coefficients in A, has unit constant term and all other coefficients nilpotent. Furthermore, suppose that every nilpotent element of Ar−1, when viewed as a polynomial with coefficients in A, has nilpotent coefficients.

If f is a unit, then by Exercise 1.2 (i), a0 is a unit and a1, . . . , an are nilpotent in Ar−1. By our hypotheses, the constant coefficient of f over A is a unit, and all other coefficients of f over A are nilpotent.

If f is nilpotent, then by Exercise 1.2 (ii), a0, . . . , an are nilpotent in Ar−1. By our hypothesis, all coefficients of f over A are nilpotent.

Pick a monomial order on the monomials in the indeterminates x1, . . . , xr. If f is a zero-divisor, let g ∈ Ar be a non-zero annihilator of f with minumum leading term. Let a ∈ A be the lead coefficient of f over A. Then ag is an annihilator of f of degree less than g, hence ag = 0. By successively subtracting off the leading terms of f, a similar induction on the chosen monomial order as in Exercise 1.2 (iii) shows that g annihilates every coefficient of f over A. Therefore, bf = 0, where b is the lead coefficient of g over A. (Remark1)

Let f, g ∈ Ar and let a, b, c be the ideals generated by the respective coefficients of f, g, and fg over A. The inclusion c ⊆ ab shows that f and g are primitive if fg is. On the other hand, if fg is not primitive, then c is contained in some maximal ideal, m, of A. Therefore fg = 0 in (A/m)[x1, . . . , xr]. Since A/m is a field, f = 0 or g = 0 in (A/m)[x1, . . . , xr]. This in turn implies that a ⊆ m or b ⊆ m. In either case, at least one of f or g is not primitive.

1.4

Since every maximal ideal is prime, the nilradical of any ring is contained in the Jacobson radical. If f is in the Jacobson radical of A[x], then 1 − fx is a unit by Proposition 1.9. By Exercise 1.2 (i), the coefficients of f are nilpotent. Therefore f is in the nilradical of A[x] by Exercise 1.2 (ii).

1.5

P n Let f = n≥0 anx be an element of A x . J K P n (i) If f is a unit with inverse g = n≥0 bnx , then fg = 1. On the other hand, fg has constant term a0b0. Therefore a0 is a unit. −1 −1 P n Suppose a0 is a unit. Define h := 1 − a0 f. Define g := a0 n≥0 h . Since h has positive order, each of the coefficients in the definition of g is a finite polynomial in the coefficients of h. Thus g is well defined. Furthermore,

−1 X n X n X n+1 0 X n X n fg = a0(1 − h)a0 h = h − h = h + h − h = 1. n≥0 n≥0 n≥0 n≥1 n≥1

Therefore, f is a unit. (ii) Let k be the order of f as a power series and suppose that f N = 0 for some positive integer N. The least N N k degree term of f has coefficient ak = 0. Thus ak is nilpotent. Since the nilradical is an ideal, f − akx is also nilpotent. By induction on k, we see that every coefficient of f is nilpotent. The converse is false: Define

p−1 p−2 p−3 p A := Fp[T,T ,T ,T ,...]/(T ).

Claim: For every n ≥ 0, the element T p−n has nilpotency order pn+1 in A. 1 RINGS AND IDEALS 9

n  −n p Proof: Clearly, T p = T . On the other hand, suppose

 −n a  −m  T p = T p · g T p for some integers a, m ≥ 0. Let N := max{m, n}. Take pN th powers to get

N−n N+1  N−m  T ap = T p · g T p in Fp[T ]. By comparing degrees we see that apN−n ≥ pN+1. In particular, a ≥ pn+1 as desired. P p−n n Claim: The element f := n≥0 T x of A x is not nilpotent. J K Proof: It is easy to see that  p p X p−n n X p1−n pn X p−n p(n+1) p p f(x) =  T x  = T x = T x = x f(x ). n≥0 n≥0 n≥0

By repeated application of this identity, we see that

n n n f(x)p = xp f xp
for all n ≥ 0. Since the right hand side has the same coefficients as the left (albeit with new indices), no power of f(x) is zero. (iii) By Proposition 1.9, f is in the Jacobson radical of A x if and only if 1 − fg is a unit of A x for all g J K J K in A x . By (i), this latter condition is equivalent to requiring that 1 − a0b is a unit in A for all b in A. Again, J K Proposition 1.9 shows that this is equivalent to letting a0 be an element of the Jacobson radical of A. (iv) Let m be a maximal ideal of A x . Since 0 is in the Jacobson radical of A, x is in the Jacobson radical of A x by (iii). Thus mc consists of the constantJ K terms of elements of m and m = mc + xA x . Furthermore, J K J K A x /m =∼ (A x /xA x )/(m/xA x ) =∼ A/mc J K J K J K J K shows that A/mc is a field. As a consequence, mc is maximal. (v) Let p be a prime ideal of A, and define

φ : A x → A → A/p J K by following the evaluation at 0 map with reduction modulo p. This map is obviously surjective. Furthermore, the kernel of φ consists of power series with constant term in p. That is, ker(φ) = p + xA x . By the First Isomorphism Theorem, A x /(p + xA x ) =∼ A/p. Since A/p is a domain, p + xA x is a prime ideal.J K On the other hand, p = (p + xA x )c showsJ K that p JisK the contraction of a prime ideal of A x .J K J K J K

1.6

Let A be a ring such that every ideal not contained in the nilradical contains a non-zero idempotent. Let e be an idempotent in the Jacobson radical. By Proposition 1.9, 1 − e is a unit. Since e is idempotent,

e(1 − e) = e − e2 = 0.

Thus e = 0. Since, the Jacobson radical contains no non-zero idempotents, it must be contained in the nilradical. 1 RINGS AND IDEALS 10

1.7

Let A be a ring such that every element is a root of a polynomial of the form xn − x with n a positive integer. Let p be a prime ideal and x an element of A not in p. For some positive n, xn − x = 0 is in p. Since p is a prime ideal not containing x, we have the equality xn−1 = 1 in A/p. We have now shown that the non-zero elements of A/p have multiplicative inverses. Therefore, A/p is a field and p is a maximal ideal.

1.8

Let Σ be the set of prime ideals of A. By Theorem 1.3, Σ is not empty. Order Σ by reverse inclusion and let ∞ T∞ {pn}n=1 be a chain. Define p := n=1 pn. If x, y are not elements of p, then there are i, j such that x is not in pi and y is not in pj. Let k be the greater of i, j. Then x and y are both not in pk. Since pk is a prime ideal, xy is ∞ not in pk. Thus xy is not in p. In particular, p is a prime ideal. Since p is an upper bound for the chain {pn}n=1, Zorn’s Lemma implies that Σ has maximal elements. Therefore, A has minimal prime ideals.

1.9 √ Let a be a proper ideal of A. If a = a, then a is the intersection of the prime ideals that contain it by Proposition 1.14. On the other hand, let a be the intersection of some collection of prime ideals Σ and let Σ0 be the set of all prime ideals containing a. Then the following inclusions \ \ √ a = ⊇ = a ⊇ a p∈Σ p∈Σ0 √ show that a = a.

1.10

Let A be a ring with nilradical R. (i)⇒(ii) Let p be the unique prime ideal of A. By Proposition 1.8, p = R. Furthermore, p is a maximal ideal by Corollary 1.4. If x is a non-unit, then it is contained in p and is nilpotent as a consequence. (i)⇒(iii) Let p be the unique prime ideal of A. By Proposition 1.8, p = R. Furthermore, p is a maximal ideal by Corollary 1.4. Thus A/R is a field. (ii)⇒(i) Since every element of A not in R is a unit, Proposition 1.6 (i) asserts that A is a local ring with unique maximal ideal R. On the other hand, Proposition 1.8 states that R is contained in every prime ideal of A. Therefore, R is the only prime ideal of A. (ii)⇒(iii) Since every element of A not in R is a unit, Proposition 1.6 (i) asserts that A is a local ring with unique maximal ideal R. In particular, A/R is a field. (iii)⇒(i) Note that every prime ideal of A contains R. By the Lattice Isomorphism Theorem, p is a prime ideal of A if and only if p/R is a prime ideal of A/R. Since A/R is a field, every prime ideal of A is equal to R. Thus A has exactly one prime ideal. (iii)⇒(ii) Let x be a non-nilpotent element of A. Let y be an element of A such that xy = 1 in A/R. In other words, xy − 1 is nilpotent. By Exercise 1.1, xy = 1 + (xy − 1) is a unit. So x is a unit with inverse y(xy)−1.

1.11

Let A be a Boolean ring. (i) Given x in A, 0 = (x + 1)2 − (x + 1) = x2 + 2x + 1 − x − 1 = 2x. 1 RINGS AND IDEALS 11

(ii) If x is not an element of a prime ideal p, then x(1 − x) = 0 implies that x = 1 in A/p. Thus A/p is the field with two elements and p is a maximal ideal. (iii) Let x, y be elements of an ideal a. Define z := x + y + xy and note that

xz = x(x + y + xy) = x2 + 2xy = x and yz = y(x + y + xy) = y2 + 2xy = y. Therefore, (x, y) = (z). If a is a finitely generated ideal with more than one generator, we may apply this process to reduce the number of generators by one. By repeated application, we can find a principal generator fora.

1.12

Let A, m be a local ring with an idempotent element x. If x is in m, then 1 − x is a unit because m is the Jacobson radical. Thus x(1 − x) = 0 implies that x = 0. If x is a unit, then x(1 − x) = 0 implies that x = 1. Therefore the only idempotents of A are 0 and 1.

1.13

Let K0 = K be a field. Given a non-negative integer n for which the field, Kn, is defined, let Σn be the set of monic irreducible elements of Kn[x] and let An be the polynomial ring over Kn generated by the set of indeterminants {xf |f ∈ Σn}. Define an as the ideal of An generated by the set {f(xf ) ∈ A|f ∈ Σn}. Since Kn is a field, An is a domain. Thus every element of an has positive degree and an does not contain 1. Let mn be a maximal ideal of An containing an and define Kn+1 = An/mn. The map

Kn → An → An/mn = Kn+1, given by the composition of the canonical inclusion and quotient maps, is a field homomorphism. Thus it is injective and we may identify Kn with a subfield of Kn+1. Let K = S K (really, K = lim K ). If x, y are in K, then they are contained in some subfields K ,K . n≥0 n −→ n n m Letting k be the greater of m and n, x and y are in Kk. Therefore the sum, difference, and product of x, y are in K. If y is non-zero, then the quotient x/y is in K as well. Since 0, 1 are in K, and any field arithmetic of K can be performed in a subfield, K is a field. Let f be an irreducible monic polynomial in K[x]. Since f has only finitely many coefficients, there is some n such that f is an irreducible monic polynomial in Kn[x]. By construction, f has a root in Kn+1, hence in K. By the Euclidean division, f must have degree 1. Therefore, K is algebraically closed.

By construction, the field extension Kn+1/Kn is algebraic for every n. Since the class algebraic extensions is a distinguished class of field extensions (Lang’s Algebra), Kn/K is an algebraic extension for every n ≥ 0. Therefore, every element of K is algebraic over K, and K is the algebraic closure of K up to isomorphism.

1.14

Let Σ be the set of ideals of the ring A which consist of solely zero-divisors. This set is non-empty because S it contains (0). Order Σ by inclusion and suppose that {an}n≥1 is a chain. Since a = n≥1 an is an ideal of A consisting solely of zero divisors, it is an upper bound for {an}n≥1 in Σ. By Zorn’s Lemma, Σ has a maximal element b. If x, y are not elements of b, then (x)+b, (y)+b each contain non-zero-divisors, because b is a maximal element of Σ. Since the product of two non-zero-divisors is a non-zero-divisor and ((x) + b)((y) + b) ⊆ (xy) + b, some element of (xy) + b is not a zero-divisor. Thus xy is not an element of b. Therefore, b is a prime ideal of A. 1 RINGS AND IDEALS 12

If d is a zero-divisor, then (d) is an element of Σ. Thus (d) is contained in a maximal element of Σ, which in turn is a prime ideal of A. Therefore, the union of the ideals that are maximal in Σ is the set of zero-divisors, and each of these is prime ideal.

1.15

Let X be the set of all prime ideals of the ring A. For any subset E of A, define V (E) as the set of prime ideals containing E. (i) If a is the ideal of A generated by E, then V (a) ⊆ V (E) by the definition of V . If p is a prime ideal containing E, then p contains a because a is the smallest ideal containing E. Thus V (E) ⊆ V (a) as well. √ √ a
a a Given that every ideal is contained√ in its radical, V ⊆ V ( ). Since is the intersection of all prime ideals containing a, V (a) ⊆ V a
. (ii) Since (0) is contained in every ideal of A, V (0) = X. On the other hand, no proper ideals of A contain 1, hence V (1) is empty. S (iii) Let {Ei}i∈I be a family of subsets of A and let p be a prime ideal. If p contains i∈I Ei, then p contains S
T each Ei. Thus V i∈I Ei ⊆ i∈I V (Ei). If p contains each Ei, then clearly p contains their union. Thus T S
i∈I V (Ei) ⊆ V i∈I Ei as well. (iv) If a, b are ideals of A, then √  √  V (ab) = V ab = V a ∩ b = V (a ∩ b) by Exercise 1.13 of the text. Given that ab is contained in a and b, V (a)∪V (b) ⊆ V (ab). If p is a prime ideal containing ab, then p contains a or b, by Proposition 1.11. Thus V (ab) ⊆ V (a) ∪ V (b).

1.16

• Because Z is a principal ideal domain, the prime ideals of Z are of the form (p), where p is a prime number, and (0). Therefore, Spec Z is a curve through all the principal ideals of Z of the form (p), with p a prime ideal number, and the point (0). For geometric reasons that will be clear later, draw (0)) off to the side of the curve. c Let U ⊆ Spec Z be an open set. By definition, U = V (a) for some ideal a of Z. Since Z is a principal ideal domain, a = (a) for some integer a. If a = 0, then V (a) = V (0) = Spec Z and U = ∅. If U is non-empty, Qk αi then a 6= 0. By the Fundamental Theorem of Arithmetic, a = ± i=1 pi for some prime numbers p1, . . . , pk √ Qk and positive integers α1, . . . , αk. Define a as i=1 pi. Since non-zero prime ideals of Z are maximal, we have k √ [ V (a) = V (a) = V ( a) = V (pi) = {(p1),..., (pk)}. i=1 c Therefore, U = {(p1),..., (pk)} . In summary, non-empty open subsets of Spec Z are cofinite sets containing (0).

• Since R is a field, Spec R is the one point space containing (0) under the trivial topology.

• As before, C[x] is a principal ideal domain. Thus its prime ideals are of the form (f), where f is a monic irreducible or f = 0. Since C is algebraically closed, the monic irreducible elements of C[x] are of the form x − a, where a is a complex number. Therefore, Spec C[x] can be drawn as the complex plane, where the point (x − a) is identified with the complex number a, with an additional point corresponding to the prime ideal (0). 1 RINGS AND IDEALS 13

c Let U ⊆ Spec C[x] be an open set. By definition, U = V (a) for some ideal a of C[x]. Since C[x] is a principal ideal domain, a = (f) for some monic polynomial f ∈ C[x]. If f = 0, then V (a) = V (0) = Spec C[x] and Qk αi U = ∅. If U is non-empty, then f 6= 0. By the Fundamental Theorem of Algebra, f = i=1(x − ai) √ Qk for some complex numbers a1, . . . , ak and positive integers α1, . . . , αk. Define f as i=1(x − ai). Since non-zero prime ideals of C[x] are maximal, we have

k p  [ V (a) = V (f) = V f = V (x − ai) = {(x − a1),..., (x − ak)}. i=1

c Therefore, U = {(x − a1),..., (x − ak)} . In summary, non-empty open subsets of Spec C[x] are cofinite sets containing (0).

• As before, R[x] is a principal ideal domain. Therefore, its prime ideals are of the form (f), where f is a monic irreducible polynomial or f = 0. Since C/R is a degree 2 extension and C is algebraically closed, the irreducible polynomials in R[x] are degree 1 or 2. Let (f) be a non-zero prime ideal of R[x] and let ι : R[x] → C[x] be the inclusion map. If deg(f) = 1, then ι(f) = x − r for some real number r, and ι∗((x − r)) = (f). If deg(f) = 2, then ι(f) = (x − a)(x − a) for some non-real complex number a, and ∗ ∗ ∗ ι ((x − a)) = ι ((x − a)) = (f). By Exercise 1.21 (i), ι : Spec C[x] → Spec R[x] is a continuous map. We have just shown that it is surjective and that Spec R[x] is a quotient of Spec C[x] by identifying conjugate points. So Spec R[x] can be drawn as a copy of the complex plane, folded over the real axis, with an additional point corresponding to the prime ideal (0). ∗−1 Let U ⊆ Spec R[x] be a non-empty open set. Then ι (U) is a non-empty open subset of Spec C[x]. By our ∗−1 c ∗ previous arguments, ι (U) = {(x − a1),..., (x − ak)} for some complex numbers a1, . . . , ak. Since ι is surjective, we know that

∗ ∗−1 ∗ ∗ c U = ι (ι (U)) = {ι ((x − a1)), . . . , ι ((x − ak))} .

In summary, non-empty open subsets of Spec R[x] are cofinite sets containing (0). ∗ • Let ι : Z → Z[x] be the inclusion map. By Exercise 1.21 (i), ι : Spec Z[x] → Spec Z is continuous. Claim: The map ι∗ is surjective. ∼ Proof: Let p be a prime number. The identification, Z[x]/pZ[x] = (Z/pZ)[x] = Fp[x] shows that pZ[x] is a ∗ ∗ ∗ prime ideal of Z[x]. Clearly, ι (pZ[x]) = (p). On the other hand, ι ((0)) = (0). So ι is surjective. ∗ Therefore, we can partition Spec Z[x] as a union of fibers of the map ι via   ∗−1 [ ∗−1 Spec Z[x] = ι ((0)) ∪  ι ((p)) . p prime number

−1 Let p be a non-zero prime ideal of Z[x]. Then the map Z/ι (p) → Z[x]/p obtained from ι : Z → Z[x] by ∗ passing to quotients is injective. In particular, if ι (p) = (c) with c ≥ 0, then c is the characteristic of Z[x]/p. Claim: The fiber ι∗−1((0)) consists of prime ideals of the form (f) where f is either an irreducible polynomial or 0. Proof: If ι∗(p) = (0), then every non-zero element of p has positive degree. Suppose that p 6= (0). Let f ∈ p be a irreducible polynomial with minimum degree. Let g be any element of p. Since Q[x] is a Euclidean Domain, there are q, r ∈ Q[x] such that g = qf + r and r has degree less than f. Since f, g are in p, r is in p as well. By our hypothesis on f, r is zero. By Gauss’ Lemma, q is in Z[x]. Therefore, p = (f). Claim: Let p be a prime number. The fiber ι∗−1((p)) consists of prime ideals of the form (p, f) where f is a polynomial that is irreducible mod p or f is 0. ∗ Proof: If ι (p) = (p), then the quotient map Z[x] → Z[x]/p factors through the quotient Z[x] → Fp[x]. So
Z[x]/p is a quotient of Fp[x] by some prime ideal q. Since Fp[x] is a Principal Ideal Domain, q = f for 1 RINGS AND IDEALS 14

some element f of Fp[x] which is irreducible or 0. By the Lattice Isomorphism Theorem, q lifts to the ideal (p, f) of Z[x], where f is any lift of f. In summary, the prime ideals of Z[x] break into four types: (f), where f ∈ Z[x] is an irreducible polynomial of positive degree, (p), where p ∈ Z is a prime number, (p, f), where p ∈ Z is a prime number and f ∈ Z[x] is irreducible modulo p, and (0). Therefore, Spec Z[x] can be drawn as a grid. The bottom of the grid consists of ideals (c) where c is either a prime number or 0. To maintain consistency with the description of Spec Z given above, draw (0) off to the side from the other ideals of this form. Above each ideal on the bottom row, place the ideals (c, f), where f is irreducible mod c. Each such f is irreducible, so you may as well put the ideal (p, f) in the same row as the ideal f. (Though f won’t necessarily be irreducible mod every prime. For instance, x2 + x + 1 is irreducible mod 2 but not mod 3.)

Claim: Every proper closed subset of Spec Z[x] is a finite union of sets of the form V (g1, . . . , gn), where g1, . . . , gn are irreducible elements of Z[x]. Proof: Since Z is a Euclidean Domain, it is Noetherian. By Theorem 7.5, Z[x] is Noetherian. By Exercises 6.7 and 6.8, V (a) is a finite union of irreducible closed subspaces of Spec Z[x]. If g1, g2 are elements of Z[x], then Exercise 1.15 (iv) implies V (g1g2) = V (g1) ∪ V (g2). Therefore, if V (g) is irreducible, then g is irreducible. By Exercise 1.18 (ii) and our taxonomy of the prime ideals in Z[x], the converse holds as well. By Exercise 1.17 (i), the sets Xg form a basis for the topology on X = Spec Z[x]. By Exercise 6.6, every open Qn Tn subset of Spec Z[x] is quasi-compact. If g has factorization into irreducibles i=1 gi, then Xg = i=1 Xgi . By De Morgan’s Laws and the preceding facts, every closed subset of Spec Z[x] is a finite union of sets of the form V (g1, . . . , gn), where g1, . . . , gn are irreducible elements of Z[x]. Claim: Every proper closed subset of Spec Z[x] is a finite union of sets V (p),V (f),V (p, f) for varying p, f, where p is a prime number and f is either irreducible (for (f)) or irreducible mod p (for (p, f)). Note that irreducible mod p implies irreducible.

Proof: Let g1, . . . , gn be distinct irreducible elements of Z[x]. Let p be a minimal prime ideal containing the ideal (g1, . . . , gn). If p = (f) for some irreducible f, then all the gi’s are multiples of f. If this case, g1 = f and n = 1. If n 6= 1, then (g1, . . . , gn) has only finitely many minimal primes, and each must be of the form (p, f) for p a prime number and f irreducible mod p. Note that V (p), for p a prime number, consists of (p) and all prime ideals (p, f) with f irreducible mod p. Furthermore, V (f), for f an irreducible polynomial of positive degree, consists of (f) and all prime ideals of the form (p, f) where p is a prime number such that f is irreducible mod p. Conclusion: Every non-empty open subset of Spec Z[x] can be formed by puncturing finitely many closed points and removing finitely many horizontal and vertical grid lines from the total space, and every non-empty open set contains (0).

1.17

c For each f in A, define Xf = V (f) in X = Spec A. Let U be an open set containing the point p of X. By definition of the Zariski topology, there is an ideal a such that U c = V (a). Let f be an element of a not contained c in p. Then U = V (a) ⊆ V (f) implies that Xf ⊆ U and p is in Xf . Thus the sets Xf form a basis for the Zariski topology. (i) By Exercise 1.15 (iv),

c c c c Xf ∩ Xg = V (f) ∩ V (g) = (V (f) ∪ V (g)) = V (fg) = Xfg.

(ii) An element of A is nilpotent if and only if it is contained in every prime ideal. Since Xf is the set of prime ideals not containing f, the equivalence follows. 1 RINGS AND IDEALS 15

(iii) By Corollary 1.5, every non-unit is contained in a maximal ideal. Conversely, no prime ideal contains a unit. Therefore, f is contained in no prime ideals precisely when f is a unit. (iv) Every prime ideal containing g contains f if and only if p(g) ⊇ p(f). The first condition is equivalent to Xf ⊆ Xg. Swapping the roles of f and g gives the desired equivalence.

(v) Let {Xf }f∈E be an open cover of X. Taking complements shows that V (E) is empty. Therefore, 1 is in the ideal generated by E. This in turn implies that there are f1, . . . , fn in E and a1, . . . , an in A such that Pn n 1 = i=1 aifi. Thus V (f1, . . . , fn) is empty. Taking complements again shows that {Xfi }i=1 is an open cover of X.

(vi) Let {Xg}g∈E be an open cover of Xf . Taking complements shows that V (f) ⊇ V (E). Therefore, f is in the radical ideal generated by E. This in turn implies that there are g1, . . . , gn in E, a1, . . . , an in A, and a m Pn positive integer m such that f = i=1 aigi. Thus V (f) ⊇ V (g1, . . . , gn). Taking complements again shows that n {Xgi }i=1 is an open cover of Xf . (vii) Let U be a quasi-compact open set of X. Since U is open, it is the union of a collection of basic open sets {Xf }f∈E. On the other hand, only finitely many such sets are needed to cover U because it is quasi-compact. Conversely, finite unions of quasi-compact sets are quasi-compact.

1.18

(i) By (ii), the closure of {x} is the set of prime ideals containing px. Thus {x} is closed if and only if px is not properly contained in any other prime ideals. In other words, {x} is closed if and only if px is maximal. (ii) The closure of {x} is the intersection of all closed sets containing x. Furthermore, x is in V (E) if and only if E is a subset of px. Thus every closed set containing x contains V (px). On the other hand, V (px) is a closed set containing x.

(iii) By (ii), y is in the closure of {x} if and only if py ⊇ px.

(iv) If x, y are distinct points in X such that y is in the closure of {x}, then V (py) ⊆ V (px) by (iii). Since x and y are distinct, x is not in V (py). Therefore, x is in the complement of the closure of {y}.

1.19

If Spec A is irreducible and fg is nilpotent, then Xf ∩ Xg = Xfg is empty by Exercise 1.17. Thus Xf or Xg is empty, and correspondingly f or g is nilpotent.

Suppose that the nilradical of A is prime ideal. Let Xf ,Xg be non-empty open sets. Then f, g are not nilpotent and neither is fg. Thus Xf ∩ Xg = Xfg is non-empty.

1.20

Let X be a topological space. (i) Let U, V be open subsets of X such that U ∩ Y and V ∩ Y are non-empty. By the definition of the closure, U ∩ Y and V ∩ Y are non-empty open sets of Y . Since Y is irreducible, (U ∩ Y ) ∩ (V ∩ Y ) is a non-empty subset of (U ∩ Y ) ∩ (V ∩ Y ). Thus Y is irreducible. (ii) Note that every point is irreducible. If Y is an irreducible subspace of X, then let Σ be the set of irreducible S subspaces of X containing Y , ordered by inclusion. Let {Zn}n≥1 be a chain in Σ and define Z = n≥1 Zn. Suppose that U, V are open sets intersecting Z. Then there are positive integers i, j such that U ∩ Zi and V ∩ Zj are non- empty. If k is the greater of i, j, then both U and V intersect Zk. Since Zk is irreducible, U ∩ V contains a point of Zk. Thus U ∩ V contains a point of Z. Therefore, Z is an upper bound for {Zn}n≥1 in Σ. By Zorn’s Lemma, there is a maximal irreducible subspace of X containing Y . 1 RINGS AND IDEALS 16

(iii) If Y is a maximal irreducible subspace of X, then it must equal its closure by (i). As in (ii), note that points are irreducible subspaces of X. By (ii), the maximal irreducible subspaces cover X. In a Hausdorff space, any subspace with more than one point has disjoint non-empty open sets, and is thus not irreducible. (iv) Let X = Spec A for some ring A. If Y is a closed irreducible subspace of X, then there is a radical ideal a such that Y = V (a). If f, g are not in a, then Xf ,Xg both meet Y , because a is its own radical. Since Y is irreducible, Xf ∩ Xg = Xfg intersects Y . Thus fg is also not in a, and we see that a is a prime ideal.

If p is a prime ideal, Y = V (p), and Xf ,Xg intersect Y , then f, g are not in p. Thus fg is not in p and Xf ∩ Xg = Xfg intersects Y . Therefore, Y is irreducible. For any prime ideals p, q, V (p) ⊆ V (q) if and only if p ⊇ q. Thus maximal irreducible components of X correspond to minimal prime ideals.

1.21

Let φ : A → B be a ring homomorphism, X = Spec A, and Y = Spec B. If q is a prime ideal of B, then A/φ−1(q) → B/q is injective by the First Isomorphism Theorem. Thus φ−1(q) is a prime ideal of A. Define φ∗ : Y → X by φ∗(q) = φ−1(q). −1 (i) If f is an element of A, then φ (q) is contained in Xf if and only if q is contained in Yφ(f). Therefore, ∗−1 φ (Xf ) = Yφ(f). (ii) If φ−1(q) ⊇ a, then q ⊇ qce ⊇ ae. Conversely, q ⊇ ae implies that φ−1(q) ⊇ aec ⊇ a. Therefore, φ∗−1 (V (a)) = V (ae). (iii) Let q be a prime ideal containing b and define p = φ−1(q). Then bc is contained in p. Thus φ∗ (V (b)) ⊆ V (bc). √ c Without√ loss of generality, b = b. Let p ∈ V (b ) and suppose f∈ / p. By the transitivity of containment, φ(f) ∈/ b = b. By Proposition 1.14, there exists q ∈ V (b) such that φ(f) ∈/ q. Therefore,

∗ ∗
∗ φ (q) ∈ φ Yφ(f) ∩ V (b) ⊆ Xf ∩ φ (V (b)).

Hence, V (bc) ⊆ φ∗(V (b)). (iv) If φ is surjective, then φ : A/ ker φ → B is an isomorphism by the First Isomorphism Theorem. By the Lattice Isomorphism Theorem, φ∗ : Y → V (ker φ) is bijective and continuous. Since φ is surjective, every basic ∗
∗ open set of Y can be put in the form Yφ(f) for some f in A. By (i), φ Yφ(f) = Xf ∩ V (ker φ). Therefore, φ is a homeomorphism of Y onto its image.

(v) Suppose that ker φ is the in the nilradical of A and let Xf be a non-empty open set of X. Since f is not in ∗ the nilradical of A, φ(f) is not nilpotent. Therefore, Yφ(f) contains some point q. By (i), φ (q) is in Xf . Therefore, φ∗(Y ) is dense in X. ∗ ∗−1 Suppose that φ (Y ) is dense in X and let f be an element of ker φ. Then Yφ(f) = φ (Xf ) is empty. Thus Xf is empty and f is nilpotent. (vi) If ψ : B → C is another ring homomorphism and r is a prime ideal of C, then

(ψ ◦ φ)∗(r) = (ψ ◦ φ)−1(r) = φ−1(ψ−1(r)) = (φ∗ ◦ ψ∗)(r).

(vii) Let A be a domain with exactly one non-zero prime ideal p, and let K be the field of fractions of A. Define B = (A/p) × K and φ : A → B by φ(x) = (x, x). The prime ideals of B are (A/p) × (0) and (0) × K. Furthermore, φ∗ is bijective due to φ∗((A/p) × (0)) = (0) and φ∗((0) × K) = p. On the other hand, Spec B has the discrete topology, but the closure of {(0)} in Spec A is the whole space. 1 RINGS AND IDEALS 17

1.22

Qn Let A := i=1 Ai be a direct product of the rings Ai, let πi : A → Ai be the canonical projections, let Qn ai := ker πi, and define Xi := V (ai). By the Chinese Remainder Theorem, ψ : A → i=1 A/ai is an isomorphism Qn that matches a prime ideal p to the product i=1(p + ai)/ai. Therefore,

n ∼ Y A/p = A/(p + ai) i=1 ∼ is a domain. Thus all but one factor ring must be zero. Calling the remainding index j, A/p = A/(p + aj) by the natural projection. So p must contain aj. We have now shown that Spec A is the disjoint union of the closed sets ∗ Xi. Since the complement of each Xi is a finite union of closed sets, each Xi is open. By Exercise 1.21 (iv), πi gives a homeomorphism from Spec Ai to Xi. (i)⇒(ii) If X = Spec A is disconnected, then there are proper non-zero ideals a, b such that X = V (a)∪V (b) = V (ab) and ∅ = V (a) ∩ V (b) = V (a + b). By Exercise 1.15, there are f, g in a, b and a positive integer n such that f + g = 1 and (fg)n = 0. Since (f, g) ⊆ p(f n, gn) and V (f, g) is empty, V (f n, gn) is also empty. Thus (f n) + (gn) = (1) and the Chinese Remainder Theorem implies that A → (A/(f n)) × (A/(gn)) is an isomorphism. Neither of f, g is a unit, because they are elements of the proper ideals a, b. (i)⇒(iii) Suppose that A is reduced. If X = Spec A is disconnected, then there are proper non-zero ideals a, b such that X = V (a) ∪ V (b) = V (ab) and ∅ = V (a) ∩ V (b) = V (a + b). By Exercise 1.15, there are f, g in a, b such that f + g = 1 and fg = 0. Therefore,

f 2 = f(1 − g) = f − fg = f.

Furthermore, f 6= 0, 1 because a, b are both proper ideals. (Remark2) (ii)⇒(i) This was shown above. −1 (ii)⇒(iii) Let φ : A → A1 ×A2 be an isomorphism, where neither of A1,A2 are the zero ring. Then φ ((1, 0)) is a non-trivial idempotent of A. (iii)⇒(i) If e is a non-trivial idempotent of A, then

V (e) ∩ V (1 − e) = V (1) = ∅ and V (e) ∪ V (1 − e) = V (e − e2) = V (0) = Spec A is a separation of Spec A. (iii)⇒(ii) Let e be a non-trivial idempotent of A. Then (e) + (1 − e) = (1) and (e)(1 − e) = (0). By the Chinese Remainder Theorem, A =∼ A/(e) × A/(1 − e). (iii)⇒(ii) (Alternative) Let e be a non-trivial idempotent of A. The ideal Ae is a ring with identity e. Since

(1 − e)2 = 1 − 2e + e2 = 1 − e,

1 − e is also a non-trivial idempotent. Let φ : A → Ae × A(1 − e) be the ring homomorphism given by

φ(x) = (xe, x(1 − e)).

If x ∈ ker φ, then x = xe + x(1 − e) = 0. So φ is injective. On the other hand, let y, z be any elements of A. Then φ(ye + z(1 − e)) = (ye, z(1 − e)). So φ is surjective and an isomorphism. Let A be a local ring. By Exercise 1.12, A contains no non-trivial idempotent. Therefore, Spec A is connected by (iii). 1 RINGS AND IDEALS 18

1.23

Let A be a Boolean ring and X = Spec A.

(i) If f is an element of A, then Xf ∩ X(1−f) = X0 is empty and

c c c Xf ∪ X(1−f) = (V (f) ∩ V (1 − f)) = V ((f) + (1 − f)) = V (1) = X.

c Therefore, Xf = X(1−f) is closed.

(ii) If f1, . . . , fn are elements of A, then by (i)

n n n ! [ [ Y Xfi = V (1 − fi) = V (1 − fi) = Xf , i=1 i=1 i=1 Qn where f = 1 − i=1(1 − fi). (iii) Let Y be a closed and open subset of X. Because Y is open, it is a union of basic open sets. Since Y is closed and X is quasi-compact, Y is also quasi-compact. Thus Y is a finite union of basic open sets. By (ii), Y is a basic open set.

(iv) If x, y are distinct points in X, then, without loss of generality, there is some f in A such that Xf contains x but not y, by Exercise 1.18 (iv). By (i), y is in X(1−f) and the two sets are disjoint. X is quasi-compact by Exercise 1.17 (v).

1.24

Let L be a Boolean algebra. Then L has a greatest element 1 and a least element 0, each of ∨, ∧ distribute over each other, and each element a has a unique complement a0. Let A(L) be the set of L under the operations

a + b = (a ∧ b0) ∨ (a0 ∧ b), a · b = a ∧ b.

Let a, b, c be elements of A(L). First, +, · are commutative because ∨, ∧ are. The additive identity is 0 by

a + 0 = (a ∧ 1) ∨ (a0 ∧ 0) = a ∨ 0 = a.

The multiplicative identity is given by 1. The additive inverse of a is a by

a + a = (a ∧ a0) ∨ (a0 ∧ a) = 0 ∨ 0 = 0.

Addition is associative by

(a + b) + c = ((a ∧ b0) ∨ (a0 ∧ b)) ∧ c0 ∨ ((a0 ∨ b) ∧ (a ∨ b0)) ∧ c = (a ∧ b0 ∧ c0) ∨ (a0 ∧ b ∧ c0) ∨ (a0 ∧ b0 ∧ c) ∨ (a ∧ b ∧ c) = a + (b + c), where the second line follows from the first by distributing c, c0 and further shows that the expression is symmetric in a, b, and c. Multiplication is associative because ∧ is. Finally, the distributive law holds:

a(b + c) = a ∧ (b ∧ c0) ∨ (b0 ∧ c)
= [(a ∧ b) ∧ c0] ∨ [b0 ∧ (a ∧ c)] = [(a ∧ b) ∧ (a0 ∨ c0)] ∨ [(a0 ∨ b0) ∧ (a ∧ c)] = ab + ac.

Therefore, A(L) is a Boolean ring. 1 RINGS AND IDEALS 19

Let A be a Boolean ring with elements a, b and define a ∨ b = a + b + ab, a ∧ b = ab, and a ≤ b if and only if ab = a. First, we show that ≤ is a partial order. If a ≤ b and b ≤ a, then a = ab = ba = b. So ≤ is anti-symmetric. If a ≤ b ≤ c, then ac = abc = ab = a, and a ≤ c. So ≤ is transitive. Second, we show that A is a lattice under this ordering. a(a ∨ b) = a(a + b + ab) = a and a(a ∧ b) = aab = a ∧ b shows that a ∧ b ≤ a ≤ a ∨ b. By symmetry, a ∧ b ≤ b ≤ a ∨ b as well. If c ≤ a, b, then c(a ∧ b) = cab = c shows that c ≤ a ∧ b. Thus a ∧ b is the infimum of a and b. If a, b ≤ c, then c(a ∨ b) = c(a + b + ab) = a ∨ b shows that a ∨ b ≤ c. Thus a ∨ b is the supremum of a and b. Third, we show that this is a Boolean lattice. In general, a ∨ 1 = a + 1 + a · 1 = 1 and a ∧ 0 = a · 0 = 0. For distributivity, (a ∧ b) ∨ c = ab + c + abc = (a + c + ac)(b + c + bc) = (a ∧ c) ∨ (b ∧ c) and (a ∨ b) ∧ c = ac + bc + abc = ac + bc + acbc = (a ∧ c) ∨ (b ∧ c). Finally, if we define a0 = 1 + a, then a ∧ a0 = a(1 + a) = a + a = 0 and a ∨ a0 = a + (1 + a) + a(1 + a) = 1. Furthermore, if x is a complement of a, then 0 = a ∧ x = ax and 1 = a ∨ x = a + x + ax. These equations imply that x = 1 + a = a0. So complements are unique. Thus A, ≤ is a Boolean lattice.

1.25

Given a Boolean lattice L, define

φ : L → P(Spec A(L)) : f 7→ Xf .

By Exercise 1.24, f ≤ g in L if and only if fg = f. This in turn implies that Xf ∩ Xg = Xfg = Xf , which yields Xf ⊆ Xg.

If Xf = Xg, then c Xf = Xg = X(1+g) by the solution to Exercise 1.23 (i). So Xf ∩ X(1+g) = Xf(1+g) is empty. Therefore, f(1 + g) is nilpotent. It is easy to see that every Boolean ring is reduced. So f(1 + g) = 0. In particular, f = fg. So f ≤ g. Swap the roles of f and g to see that g ≤ f. Since the ordering on L is reflexive, f = g and φ is injective. On the other hand, the image of φ is precisely the class of open-and-closed subspaces of the compact Hausdorff space Spec A(L) by Exercise 1.23. 1 RINGS AND IDEALS 20

1.26

We will use the notation of the problem. (i) Shown in the text. (ii) Shown in the text. (iii) Since µ : X → X˜ is bijective,

µ(Uf ) = µ({x ∈ X : f(x) 6= 0}) = {µ(x): x ∈ X and f(x) 6= 0}

= {mx : x ∈ X and f∈ / mx} = {m ∈ Xe : f∈ / m}

= Uef .

Let U be an open subset of X. If x ∈ U, then {x} and U c are disjoint closed sets. By Urysohn’s Lemma there is a continuous function f : X → R such that f(x) = 1 and f(y) = 0 for all y∈ / U. Therefore, x ∈ Uf ⊆ U. So the sets Uf form a basis for the topology on X. Let U˜ be an open subset of X˜. Since X˜ ⊆ Spec C(X) has the subspace topology, there is an open set W˜ of Spec C(X) such that U˜ = X˜ ∩ W˜ . Let m ∈ U˜ ⊆ W˜ . By Exercise 1.17, there is f ∈ C(X) such that m ∈ (Spec C(X))f ⊆ W˜ . Note that X˜ ∩ (Spec C(X))f = U˜f . Therefore,

m ∈ U˜f = X˜ ∩ (Spec C(X))f ⊆ X˜ ∩ W˜ = U.˜

So the sets U˜f form a basis for the topology on X˜.

1.27

We will use the notation of the problem. We will prove that the map µ : X → Xe is surjective. If m is a maximal ideal of P (X), then P (X)/m is a finitely generated k-algebra that is also a field. By Corollary 5.24, P (X)/m is a finite field extension of k. Since k is algebraically closed, k → P (X)/m is an isomorphism of k-algebras. For each i, let xi be the image of ξi in k. Then the elements ξ1 − x1, . . . , ξn − xn each map to 0 in k. Therefore, m = (ξ1 − x1, . . . , ξn − xn) = mx, and µ is surjective.

1.28

Let X,Y be affine algebraic varieties in kn, km respectively, let φ : X → Y be a regular map, let P (X) = k[t1, . . . , tn]/I(X) and P (Y ) = k[s1, . . . , sm]/I(Y ). Furthermore, let ξ1, . . . , ξn and η1, . . . , ηm be the respective images of t1, . . . , tn and s1, . . . , sm in P (X) and P (Y ). ∗ ∗ ∗ (Inverse Functor) Given a k-algebra homomorphism φ : P (Y ) → P (X), the elements φ η1, . . . , φ ηm lift m to polynomials g1, . . . , gm in k[t1, . . . , tn]. Let ψ : X → k be the regular morphism with coordinate functions ∗ g1, . . . , gm. Since all lifts of g1, . . . , gm are congruent mod I(X), the regular morphism ψ only depends on φ and not on the choice of g1, . . . , gm. Suppose that h ∈ I(Y ). Then

∗ 0 = φ (h(η1, . . . , ηm)) = h(g1, . . . , gm) mod I(X) = (h ◦ ψ)(ξ1, . . . , ξn). Therefore, im ψ ⊆ Y . By abuse of notation, we call the corestricted morphism ψ : X → Y . (φ 7→ φ∗ is injective) If φ∗ : P (Y ) → P (X) is induced by some regular morphism φ : X → Y , then

∗ fj mod I(X) = φ ηj = gj mod I(X) 1 RINGS AND IDEALS 21 for all j. So ψ = φ as functions from X to Y . ∗ ∗ ∗ ∗ (φ 7→ φ is surjective) Furthermore, ψ ηj = gj mod I(X). So ψ = φ as k-algebra homomorphisms. 2 MODULES 22

2 Modules

2.1

Direct Proof: If m, n are coprime, then there are integers x, y such that mx + ny = 1. For any simple tensor, f ⊗ g, we get f ⊗ g = (mx + ny)(f ⊗ g) = x(mf ⊗ g) + y(f ⊗ ng) = 0.

Thus Z/mZ ⊗Z Z/nZ = 0. Proof via Localization: (Remark3) If p is a prime not dividing m, then the Euclidean algorithm provides integers q, r such that p = qm + r. Since any common factor of m and r divides p, r is coprime to m. Therefore, p is a unit in Z/mZ. In particular, (Z/mZ)(p) = 0. Let X = Spec Z. If m, n are coprime, then (m) + (n) = (1). Therefore,

c c Xm ∪ Xn = V (m, n) = V (1) = X.

Let p be a prime. If (p) ∈ Xm, then

(Z/mZ ⊗ Z/nZ)(p) = (Z/mZ)(p) ⊗ (Z/nZ)(p) = 0 ⊗ (Z/nZ)(p) = 0.

Swap the roles of m and n to see that Z/mZ ⊗ Z/nZ localizes to 0 at every closed point of X. By Proposition 3.8, Z/mZ ⊗ Z/nZ = 0. Proof via Deferment: Set A = Z, a = (m), and M = Z/nZ. Apply Exercise 2.2 and note that aM = M in this case.

2.2

If a is an ideal of A and M is an A-module, then

0 → a → A → A/a → 0 is a short exact sequence of A-modules. Tensoring with M yields

a ⊗A M → A ⊗A M → (A/a) ⊗A M → 0, another exact sequence of A-modules, since the tensor product is right exact. If we define f : a × M → aM by f(a, m) = am, then f is obviously A-bilinear and surjective. By the universal property of the tensor product, ⊗ ⊗ there is a unique A-linear map f : a ⊗A M → aM that satisfies f (a ⊗ m) = am. Furthermore, define g :(A/a) × M → M/aM by g(a, m) = am. This map is obviously A-bilinear and surjective as well. By the ⊗ universal property of the tensor product, there is a unique A-linear map g :(A/a) ⊗A M → M/aM such that g⊗(a ⊗ m) = am. On the other hand, 0 → aM → M → M/aM → 0 is a short exact sequence of A-modules. Putting this together gives the following commutative diagram with exact rows, where central vertical arrow is the canonical isomorphism.

a ⊗A M A ⊗A M (A/a) ⊗A M 0

f ⊗ g⊗ 0 aM M M/aM 0 2 MODULES 23

Applying the Snake Lemma to this diagram yields another exact sequence,

0 → ker g⊗ → coker f ⊗ = 0.

Therefore, g⊗ is injective. Given that g⊗ is surjective as well,

⊗ g :(A/a) ⊗A M → M/aM is an isomorphism of A-modules.

2.3

Let A be a local ring with maximal ideal m and finitely generated A-modules M,N such that M ⊗A N = 0. Additionally, let κ = A/m be the residue field of A. By Proposition 2.14 and Exercise 2.15,

Mκ ⊗κ Nκ = (κ ⊗A M) ⊗κ (κ ⊗A N)

= [(κ ⊗A M) ⊗κ κ] ⊗A N

= (κ ⊗A M) ⊗A N

= κ ⊗A (M ⊗A N)

= (M ⊗A N)κ = 0, where the equalities are the canonical identifications. Since M,N are finitely generated A-modules, Mκ,Nκ are finitely generated κ-vector spaces by Proposition 2.17. If d, e are the respective dimensions of Mκ,Nκ, then d e de Mκ ⊗κ Nκ is isomorphic to κ ⊗κ κ , which can further be identified with κ by Proposition 2.14. This shows that d or e must by zero, because Mκ ⊗κ Nκ = 0. Without loss of generality, d = 0 and Mκ = 0. On the other hand, Mκ = M/mM by Exercise 2.2. Finally, Nakayama’s Lemma implies that M = 0.

2.4

Let {Mi}i∈I be a family of A-modules with direct sum M. L Claim: The module M ⊗A N is canonically isomorphic to i∈I (Mi ⊗A N) for regardless of the A-module N.

Proof 1: (Construct Maps) By the universal property of the tensor product, the inclusions ιi : Mi → M induce L maps Mi ⊗A N → M ⊗A N. These in turn induce a unique A-linear map f : i∈I (Mi ⊗A N) → M ⊗A N such that L P P f(mj ⊗ n) = mj ⊗ n. On the other hand, the map M × N → i∈I (Mi ⊗A N) given by ( i mi, n) 7→ i mi ⊗ n is A-bilinear. By the universal property of the tensor product, it induces a unique A-linear map g : M ⊗A N → L P P L i∈I (Mi ⊗A N) such that g( i mi ⊗n) = i mi ⊗n. By construction, f ◦g = idM⊗AN and g ◦f = id (Mi⊗AN) L i∈I and M ⊗A N is canonically isomorphic to i∈I (Mi ⊗A N). Proof 2: (Yoneda) We will make use of the adjunction preceding Proposition 2.18 and the following charac- terization of the direct sum:

“There are morphisms ιi : Mi → M such that for any A-module N and any family of A-linear maps {fi : Mi → N}i∈I , there is a unique A-linear map f : M → N such that f ◦ ιi = fi for all i ∈ I.”

In particular, the map Y homA(M,N) → homA(Mi,N): f 7→ (f ◦ ιi)i∈I i∈I is a bijection, natural in N. 2 MODULES 24

Let N,P be any A-modules. We have the following natural isomorphisms,

homA(M ⊗A N,P ) = homA(M, homA(N,P )) Y = homA(Mi, homA(N,P )) i∈I Y = homA(Mi ⊗A N,P ) i∈I ! M = homA (Mi ⊗A N) ,P i∈I

L By Yoneda’s Lemma, M ⊗A N is naturally isomorphic to i∈I (Mi ⊗A N). (Remark4) Solution of Problem 1: (Element Chase) Let φ : N 0 → N be injective, let M be a flat A-module, and let η be an element of ker (idMi ⊗ φ). Then

0 0 = (ιi ⊗ idN )(idMi ⊗ φ)(η) = (idM ⊗ φ)(ιi ⊗ idN )(η).

Since M is flat, idM ⊗ φ is injective. Furthermore, for each i ∈ I, the map ιi ⊗ idN 0 is injective because it is 0 inclusion into the ith coordinate of the direct sum M ⊗A N . Therefore, η = 0 and idMi ⊗ φ is injective. By Proposition 2.19, the A-modules Mi are all flat. P Now suppose that every Mi is a flat A-module and let i θi be an element of ker (idM ⊗ φ). Then

0 (ιi ⊗ idN )(idMi ⊗ φ)(θi) = (idM ⊗ φ)(ιi ⊗ idN )(θi) = 0 P for every i. Since idMi ⊗ φ and ιi ⊗ idN are injective, θi = 0. Thus i θi = 0, idM ⊗ φ is injective, and M is a flat A-module. Solution of Problem 2: (Exactness) Let f : N 0 → N be injective. The map

0 idM ⊗f : M ⊗A N → M ⊗A N has the property that for each i ∈ I, this map takes the ith coordinate into the ith coordinate. In particular, it is injective if and only if it is injective in each coordinate. By Proposition 2.19, the equivalence follows.

2.5

L P P n Define f : n≥0 A → A[x] by f( n≥0 anen) = n≥0 anx . The map f is obviously an A-linear isomorphism. L By Exercise 2.4, n≥0 A is a flat A-module due to each A being a flat A-module. Therefore, A[x] is a flat A-algebra because it is A-linearly isomorphic to a flat A-module.

2.6

Let M be an A module and define M[x] as the space of polynomials with coefficients in M. Let E be the endomorphism ring of M and let E0 be the endomorphism ring of M[x] (as abelian groups). Letting E act on M[x] diagonally gives an injective ring homomorphism E → E0. Since M is an A-module, we already have a ring homomorphism A → E. Therefore, M[x] is an A-module via the composition A → E → E0. Extend A → E0 to A[x] → E0 be letting x · mxn := mxn+1 and extending by linearity. By the Universal Property of Polynomial Rings, this map is uniquely determined – and it makes M[x] into an A[x]-module. We have the following identifications   M n M n M n A[x] ⊗A M =  Ax  ⊗A M = (Ax ⊗A M) = Mx = M[x]. n≥0 n≥0 n≥0 2 MODULES 25

2.7

If p is a prime ideal of A, then (A/p)[x] is a domain. By Exercise 2.6, this is isomorphic to A[x] ⊗A (A/p) as an A[x]-module. Furthermore, Exercise 2.2 shows that this is isomorphic to A[x]/p[x] as an A-module. The P n P n composite map f :(A/p)[x] → A[x]/p[x] is an A-linear map such that f( n≥0 anx ) = n≥0 anx . Thus f is an A[x]-algebra isomorphism, A[x]/p[x] is a domain, and p[x] is a prime ideal of A[x].

On the other hand, (2) is a maximal ideal of Z, but

Z[x]/(2)[x] = (Z/2Z)[x] = F2[x] is not a field (x is not a unit).

2.8

(i) Let 0 → P 0 → P be exact, and let M,N be flat A-modules. Tensoring with M shows that

0 0 → P ⊗A M → P ⊗A M is exact. Now tensor with N to show that

0 0 → (P ⊗A M) ⊗A N → (P ⊗A M) ⊗A N is exact. By the Proposition 2.14 (ii),

0 0 → P ⊗A (M ⊗A N) → P ⊗A (M ⊗A N) is exact. Thus M ⊗A N is a flat A-module. (ii) Let 0 → P 0 → P be an exact sequence of A-modules, let B be a flat A-algebra, and let N be a flat B-module. Extend scalars by B to get the exact sequence of B-modules

0 0 → B ⊗A P → B ⊗A P. Tensoring with N yields another exact sequence of B-modules,

0 0 → N ⊗B (B ⊗A P ) → N ⊗B (B ⊗A P ). By Exercise 2.15, 0 0 → (N ⊗B B) ⊗A P → (N ⊗B B) ⊗A P is an exact sequence of A-modules. Thus N = N ⊗B B is a flat A-module.

2.9

Let f g 0 → M 0 −→ M −→ M 00 → 0 be an exact sequence of A-modules with M 0,M 00 finitely generated. 0 00 Constructive Proof: Let x1, . . . , xn be generators for M and g(y1), . . . , g(ym) be generators for M . Given Pm Pm z in M, there are b1, . . . , bm in A such that g(z) = i=1 big(yi). Thus z − i=1 biyi is in the kernel of g. Pn Pm By exactness of the sequence, there are a1, . . . , an in A such that z = j=1 ajf(xj) + i=1 biyi. Therefore, f(x1), . . . , f(xn), y1, . . . , ym is a finite generating set for M. Non-Constructive Proof: Let α : A⊕n → M 0 and β : A⊕m → M 00 be surjective A-linear maps. Since A⊕m is a free A-module, it is projective. So there is an A-linear map γ : A⊕m → M such that g ◦ γ = β. Define

δ = f ◦ α ⊕ γ : A⊕(n+m) = A⊕n ⊕ A⊕m → M via the universal property of the direct limit. Then we have the following commutative diagram with exact rows: 2 MODULES 26

0 A⊕n A⊕(n+m) A⊕m 0

α δ β 0 M 0 M M 00 0

Apply the Snake Lemma to get the exact sequence

0 = coker α → coker δ → coker β = 0.

Therefore, δ is surjective.

2.10

Let A be a ring, let a be an ideal in its Jacobson radical, and let u : M → N be a homomorphism of A-modules such that N is finitely generated. Let v, w be the maps induced from u by restriction and passing to the quotient. Then the diagram

0 aM M M/aM 0

v u w 0 aN N N/aN 0 is commutative and has exact rows. Applying the Snake Lemma yields an exact sequence

f coker v −→ coker u → coker w.

Note that im f = a coker u. Suppose that w is surjective. Then f is surjective,

coker u = im f = a coker u, and coker u is finitely generated because it is a quotient of N. By Nakayama’s Lemma, coker u = 0. Therefore, u is surjective.

2.11

Let A be a non-zero ring and let φ : A⊕m → A⊕n be a homomorphism of free A-modules. If φ is surjective, then the sequence of A-modules,

φ A⊕m −→ A⊕n → 0 is exact. Tensor this sequence with κ = A/m for some maximal ideal m of A to get an exact sequence

κ⊕m → κ⊕n → 0 of κ-vector spaces. Since m, n are the respective dimensions of κ⊕m and κ⊕n, m ≥ n. On the other hand, suppose that m > n and embed A⊕n as the submodule of the first n coordinates of A⊕m. The containment, ⊕m ⊕n ⊕m φ(A ) ⊆ A ( A , yields the polynomial equation, d d−1 φ + a1φ + ... + ad−1φ + ad = 0, 2 MODULES 27 by the Cayley-Hamilton Theorem. Let d be the minimal degree for such an equation and let π : A⊕m → A⊕(m−n) be the projection onto the last m − n coordinates. Then

d d−1 0 = π(φ + a1φ + ... + ad−1φ + ad) = adπ

⊕m implies that ad = 0. Since d is minimal, there is an x in A such that

d−1 d−2 y = (φ + a1φ + ... + ad−1)(x) is non-zero, but φ(y) = 0. Therefore, if φ is injective, then m ≤ n. Putting the above results together shows that if φ is an isomorphism, then m = n.

2.12

n Let M be a finitely generated A-module, φ : M → A a surjective A-module homomorphism, and e1, . . . , en n be the standard basis of A . Let f1, . . . , fn be lifts of e1, . . . , en to M via φ, and define the A-linear map n ψ : A → M : ei 7→ fi. Since φ ◦ ψ = idAn , M is isomorphic to im(ψ ◦ φ) ⊕ ker(φ) by the map

m 7→ ((ψ ◦ φ)(m), m − (ψ ◦ φ)(m)).

Therefore, ker φ is a quotient of the finitely generate A-module M, and is itself finitely generated as a consequence. (Remark5)

2.13

Let f : A → B be a ring homomorphism and let N be a B-module. Give N an A-module structure by restriction of scalars, define g : N → NB : n 7→ 1 ⊗ n, and define p : NB → N : b ⊗ y 7→ by.

Since p ◦ g = idN , g is injective and NB is isomorphic to im(g) ⊕ ker(p) by the map

b ⊗ y 7→ ((g ◦ p)(b ⊗ y), b ⊗ y − (g ◦ p)(b ⊗ y)).

(Remark5)

2.14

We need only show that µi = µj ◦ µij. If xi ∈ Mi, then

(µj ◦ µij)(xi) = µij(xi) + D = µij(xi) + (xi − µij(xi)) + D = xi + D = µi(xi).

(Remark6)

2.15

P Let x ∈ M. By the construction of M, x = i∈I xi + D for some xi ∈ Mi, and I is finite. Since I is finite, its elements have a common upper bound, k. Then ! X X X X X x = xi + D = xi − (xi − µik(xi)) + D = µik(xi) + D = µk µik(xi) . i∈I i∈I i∈I i∈I i∈I 2 MODULES 28

Pn If µi(xi) = 0, then xi ∈ D. Let xi = l=1(xjl − µjlkl (xjl )) be a representation of xi as a sum of elements of D with jl < kl for each l. Without loss of generality, j1 is minimal. If j1 = j2, then

(xj1 − µj1k1 (xj1 )) + (xj2 − µj2k2 (xj2 )) = ((xj1 + xj2 ) − µj1k1 (xj1 + xj2 )) + µj1k1 (xj2 ) − µj2k2 (xj2 ).

That is, µj1k1 (xj2 ) − µj2k2 (xj2 ) ∈ D and is 0 in coordinate j1. So we may assume that j1 < jl for all l > 1. If j1 6= i, then xj1 = 0 by the independence of coordinates in the direct sum.

Without loss generality, j2 is minimal in I \{i}. If j2 6= k1, then xj2 = 0 by the independence of coordinates.

Therefore, xj2 = µik1 (xi), and n X xi = (xjl − µjlkl (xjl )) l=1 n X = xi − µik1 (xi) + xj2 − µj2k2 (xj2 ) + (xjl − µjlkl (xjl )) l=3 n X = xi − µk1k2 (µik1 (xi)) + (xjl − µjlkl (xjl )) l=3 n X = xi − µik2 (xi) + (xjl − µjlkl (xjl )). l=3 By repeated application of the above argument we arrive at

xi = xi − µikn (xi).

Therefore, µikn (xi) = 0 in Mkn . (Remark7) Alternative Proof: We will use the construction of the direct limit in Remark6. The first statement follows ` from the construction of M as a quotient of the disjoint union i∈I Mi. The second statement follows directly from the definition of ∼.

2.16

Let αi : Mi → N be a family of A-module homomorphisms such that

αj ◦ µij = αi for i ≤ j, and let M = lim M . If i, j ≤ k such that µ (x ) = µ (x ), then −→i∈I i i i j j

0 = µi(xi) − µj(xj) = µk(µik(xi)) − µk(µjk(xj)) = µk(µik(xi) − µjk(xj)).

By Exercise 2.15 there is l ≥ k such that µil(xi) = µjl(xj). Thus

αi(xi) = αl(µil(xi)) = αl(µjl(xj)) = αj(xj).

Therefore, we may define α : M → N by α(µi(xi)) = αi(xi). The preceding calculations have show that this is well defined and satisfies the requirement that α ◦ µi = αi. Furthermore, this requirement forces the map to be unique. (Remark8) Alternative Proof: We will use the construction of the direct limit in Remark6. Define

α : M → N : [(x, i)] 7→ αi(x).

Suppose that µik(x) = µjk(y). Then

αi(x) = αk(µik(x)) = αk(µjk(y)) = αj(y).

So α is well-defined. It is easy to verify that α is A-linear. Additionally, α(µi(x)) = αi(x) by construction. Therefore, this condition uniquely determines α. 2 MODULES 29

2.17

Let {Mi}i∈I be a family of submodules of an A-module, such that for each i, j there exists k such that Mi + Mj ⊆ Mk. Define i ≤ j to mean Mi ⊆ Mj and define µij : Mi → Mj as the inclusion map. S P S S P Since Mj ⊆ i∈I Mi for every j, Mi ⊆ Mi. On the other hand, Mi is a union of subsets of Mi. Thus S P Mi ⊆ Mi. S S For each j, let αj : Mj → Mi be the inclusion map, and let α : lim Mi → Mi be the unique induced S −→ A-module homomorphism. If x ∈ Mi, then there is j such that x ∈ Mj. Thus, α(µj(x)) = αj(x) = x, and α is surjective. On the other hand, α(µj(xj)) = αj(xj) = xj shows that α is injective. Therefore, α is an isomorphism.

Take {Mi}i∈I to be the family of finitely generated submodules of M. Order I by i ≤ j, if Mi ⊆ Mj. Then [ M = M = lim M . i −→ i i∈I

2.18

Let M = (Mi, µij) and N = (Ni, νij) be direct systems of A modules over the same directed set, I. Let M,N be the respective direct limits and µi : Mi → M, νi : Ni → N be associated homomorphisms. A homomorphism Φ: M → N of direct systems is defined to be a collection of A-module homomorphisms φi : Mi → Ni such that

φj ◦ µij = νij ◦ φi for all i ≤ j.

Let Φ : M → N be a homomorphism of direct systems. For each i, define the map αi = νi ◦ φi : Mi → N. We check that the collection of maps αi are compatible by this derivation:

αj ◦ µij = νj ◦ φj ◦ µij = νj ◦ νij ◦ φi = νi ◦ φi = αi.

By the universal property of the direct limit there is a unique map φ : M → N such that φ ◦ µi = αi = νi ◦ φi for every i.

2.19

g Let M −→f N −→ P be an exact sequence of direct systems over a common index set I. Let M,N,P be their respective direct limits and let f : M → N, g : N → P be the unique maps obtained from f, g as in Exercise 2.18. Then we have g ◦ f ◦ µi = g ◦ νi ◦ fi = ρi ◦ gi ◦ fi = 0 for every i. By Exercise 2.15, the images of the maps µi cover M. Therefore f ◦ g = 0.

Let n ∈ ker(g). By Exercise 2.15 there is ni ∈ Ni such that νi(ni) = n. Thus, by Exercise 2.18,

0 = g(n) = (g ◦ νi)(ni) = (ρi ◦ gi)(ni).

We again cite Exercise 2.15 to obtain j ≥ i such that

0 = (ρij ◦ gi)(ni) = (gj ◦ νij)(ni).

f g Since νij(ni) ∈ ker(gj) and M −→ N −→ P is exact, there is mj ∈ Mj such that fj(mj) = νij(ni). Let m = µj(mj). We have obtained the following:

f(m) = (f ◦ µj)(mj) = (νj ◦ fj)(mj) = (νj ◦ νij)(ni) = νi(ni) = n.

Therefore, ker(g) ≤ im(f). 2 MODULES 30

2.20

0 0 Let M,D be as in the statement of Exercise 2.14, and let M ,D be their analogues for the system (Mi ⊗A N, µij ⊗ idN ). So we have two short exact sequences of A-modules

0 → D → M → lim M → 0 (1) −→ i and 0 → D0 → M 0 → P → 0. (2)

Note that ! M M 0 M ⊗A N = Mi ⊗A N = (Mi ⊗A N) = M , i∈I i∈I and 0 D ⊗A N → D :(xi − µij(xi)) ⊗ n 7→ xi ⊗ n − (µij ⊗ idN )(xi ⊗ n) is clearly surjective. Tensor (1) with N to obtain the following commutative diagram with exact rows (The right vertical arrow is induced from the middle vertical arrow by passing to the quotient):

D ⊗ N M ⊗ N lim M
⊗ N 0 A A −→ i A

0 D0 M 0 P 0

Apply the Snake Lemma and recall that the middle arrow is an isomorphism and the left arrow is surjective to see that the rightmost vertical arrow has trivial kernel and cokernel.

2.21

Let i ∈ I. For each j ≥ i, ring homomorphism αij : Ai → Aj makes Aj into an Ai-algebra. Therefore (Remark 9), lim A = lim A = A −→ j −→ j j∈Ji j∈I has a canonical Ai-module structure, and this is true for every i ∈ I. Let αi(x), αj(y) ∈ A. Define

αi(x) · αj(y) := x · αj(y), where the operation on the right is scalar multiplication of A as an Ai-module. The associative and distributive properties are inherited from the Ai-module axioms. Let us check that this is well-defined and commutative. Let k ≥ i, j. Then αi(x) = αk(αik(x)) and αj(y) = αk(αjk(y)). Furthermore, the definition of scalar multiplication yields x · αj(y) = x · αk(αjk(y)) = αk(αik(x)αjk(y)). The expression on the far right is obviously commutative and independent of the choice of x, y as representatives. By construction, the maps Ai → A are ring homomorphisms.

Suppose that A = 0. Let i ∈ I. Then αi(1) = 0. By Exercise 2.15, there is j ≥ i such that 1 = µij(1) = 0. Therefore, Aj = 0. 2 MODULES 31

2.22

Let “ρ” be the symbol for the structure maps on the system of Ri’s. Let i, j ∈ I with i ≤ j. Since Rj is an Aj-module and Aj is an Ai-algebra, we can view Rj as an Ai-module via restriction of scalars. Therefore, lim R = lim R =: R −→ j −→ i j∈Ji i∈I is an Ai-module. Furthermore, ιj : Rj → Aj is Ai-linear for all j ≥ i. By Exercise 2.18, there is a unique Ai-linear map ι : R → A such that ι ◦ ρi = ιi for all i ∈ I. These Ai-linear maps are compatible for varying i ∈ I. Therefore, R → A is A-linear. By Exercise 2.19, ι is injective. So R may be canonically identified with an ideal of A. By Exercise 2.15, every element of R is nilpotent, so R is contained in the nilradical of A. Suppose that x ∈ A n n n and x = 0 for some n ≥ 0. Let x = αi(ξ) for some i ∈ I. Let j ≥ i such that 0 = αij(ξ ) = αij(ξ) . Then αij(ξ) = ιj(y) for some y ∈ Rj. Then we have

x = αj(αij(ξ)) = αj(ιj(y)) = ι(ρj(y)). So every element of the nilradical of A is in the image of ι.

Suppose that Ai is an integral domain for each i ∈ I. Let x, y ∈ A such that xy = 0. By Exercise 2.16, x, y lift to a common index k such that xy = 0 in Ak. If x 6= 0 in A, then x 6= 0 in Ak. Since Ak is an integral domain, y = 0 in Ak; and y = 0 in A as a consequence. Therefore, A is an integral domain.

2.23

This Exercise asks no questions. I’m suppose the reader is expected to verify the assertions. These follow from Exercises 2.14, 2.15, 2.16, 2.18, 2.19, and 2.21.

2.24

(Remark 10)

(i)⇒(ii) Let N be an A-module. Let F∗ be a free resolution of N. Since M is flat, the functor M⊗A is exact. A So the complex M ⊗A F∗ is exact. Therefore, Torn (M,N) = Hn(M ⊗A F∗) = 0. (ii)⇒(iii) This is immediate. (iii)⇒(i) Let 0 → N 0 → N → N 00 → 0 A be a short exact sequence of A-modules. Apply the functor Tor∗ (M, ·) to get the exact sequence A 00 0 00 0 = Tor1 (M,N ) → M ⊗A N → M ⊗A N → M ⊗A N → 0. By Proposition 2.19, M is flat.

2.25

Let 0 → N 0 → N → N 00 → 0 be a short exact sequence of A-modules, let N 00 be flat, and let f : M 0 → M be an injective A-linear map. The Tor long exact sequence yields the following commutative diagram with exact rows:

A 0 00 0 0 0 0 00 0 = Tor1 (M ,N ) M ⊗A N M ⊗A N M ⊗A N 0

f⊗idN0 f⊗idN f⊗idN00 A 00 0 00 0 = Tor1 (M,N ) M ⊗A N M ⊗A N M ⊗A N 0 2 MODULES 32

By the Snake Lemma, the sequence

0 → ker(f ⊗ idN 0 ) → ker(f ⊗ idN ) → ker(f ⊗ idN 00 ) = 0

0 is exact. Therefore, the map ker(f ⊗ idN 0 ) → ker(f ⊗ idN ) is an isomorphism and N is flat if and only if N is flat.

2.26

(⇒) This follows from Exercise 2.24.

(⇐) Let (Mi)i∈I be a directed system of A-modules with direct limit M and let N be an A-modules. Let F∗ → N be a flat resolution of N. By Exercises 2.19 and 2.20,

TorA(M,N) = H (M ⊗ F ) = H lim (M ⊗ F )
= lim H (M ⊗ F ) = lim TorA(M ,N). (3) n n A ∗ n −→ i A ∗ −→ n i A ∗ −→ n i

A Suppose that N is an A-module such that Tor1 (A/a,N) = 0 for all finitely generated ideals a of A. Let b be an ideal of A and let Σ be the directed family of all finitely generated subideals of b (as in Exercise 2.17). By the exactness of direct limits and (3), ! TorA (A/b,N) = TorA lim A/a,N = lim TorA(A/a,N) = 0. 1 1 −→ −→ 1 a∈Σ a∈Σ

A So Tor1 (·,N) vanishes on cyclic A-modules. Pn Pk Let M = i=1 Ami be a finitely generated A-module, and define Mk := i=1 Ami for 1 ≤ k ≤ n. Note that A → Mk/Mk−1 : 1 7→ mk is surjective, so Mk/Mk−1 is cyclic. The Tor functor yields exact sequences

A A A Tor1 (Mk−1,N) → Tor1 (Mk,N) → Tor1 (Mk/Mk−1,N) = 0, where the equality is due to Mk/Mk−1 being a cyclic A-module. By induction on k, there is a surjective map from A A A A Tor1 (M1,N) onto Tor1 (Mn,N) = Tor1 (M,N). Since M1 is a cyclic A-module, Tor1 (M,N) = 0.

Let M be any A-module and let (Mi)i∈I be the directed family of its finitely generated submodules. By (3), A Tor1 (M,N) = 0. Since M was arbitrary, Exercise 2.24 implies that N is flat. (Remark 2.26)

2.27 Incomplete

(i)⇒(ii) Let A be absolutely flat with element a. Since the inclusion (a) → A is injective and A/(a) is flat, the tensored map, 2 (a)/(a) = (a) ⊗A A/(a) → A ⊗A A/(a) = A/(a) is also injective. However, xa 7→ 0 for all x ∈ A. Therefore, (a)/(a)2 = 0. (ii)⇒(iii) If a is an element of A, then (a) = (a)2 implies that there is an element b such that a = a2b. Furthermore, (ab)2 = a2b · b = ab shows that ab is idempotent, and a · ab = a2b = a shows that (a) = (ab). Thus every principal ideal is generated by an idempotent element of A. If e, f are idempotent elements of A, then

e(e + f − ef) = e + ef − ef = e and (e + f − ef)f = ef + f − ef = f show that (e, f) = (e + f − ef). Additionally,

(e + f − ef)2 = e(e + f − ef) + f(e + f − ef) − ef(e + f − ef) = e + f − ef. 2 MODULES 33

By induction, every finitely generated ideal is principal, generated by an idempotent element. Thus, if a is a finitely generated ideal, there is an idempotent g such that a = (g). Then A = (g) ⊕ (1 − g) shows that a is a direct summand of A. (iii)⇒(i) Let N be an A-module, a a finitely generated ideal, and A = a ⊕ b. Additionally, let ι : a → A be the natural inclusion and let π : A → a be the natural projection. Then π ⊗ idN is a left inverse for ι ⊗ idN , and A ι ⊗ idN is injective. Thus Tor1 (A/a,N) = 0. By Exercise 2.26, N is flat and A is absolutely flat. (Remark 12)

2.28

If a is an element in a boolean ring A, then a2 = a implies that (a) = (a)2. Thus A is absolutely flat by Exercise 2.27. n(x) n(x)−2 2 Let A be a ring with a map n : A → Z such that n(x) ≥ 2 and x = x for every x. Then x · x = x implies that (x) ⊆ (x)2 ⊆ (x). Thus A is absolutely flat. Let f : A → B be a ring homomorphism with A absolutely flat. If y is in the image of f, then there are a, x in A such that x = ax2 and y = f(x) = f(ax2) = f(a)y2. Thus (y) = (y)2 and the image of f is absolutely flat. Let A be an absolutely flat local ring with maximal ideal m. Let x be an element of m and a an element of A such that x = ax2. Then ax is a non-unit idempotent in a local ring. By Exercise 1.12, ax is zero. Therefore, x = ax · x = 0. As a consequence, m is the zero ideal and A is a field. Let A be absolutely flat with x = ax2. Then x(1 − ax) = 0. If x is not a zero-divisor, then 1 = ax and x is a unit. 3 RINGS AND MODULES OF FRACTIONS 34

3 Rings and Modules of Fractions

(Remark 13)

3.1 Current

Pk Let S be a multiplicatively closed subset of the ring A and let M = i=1 Axi be an A-module. By Corollary −1 −1 3.4, S M = 0 if and only if S Axj = 0 for every j. This is equivalent to the requirement that S ∩ AnnA(Axj) is non-empty for each j. On the other hand, the existence of s in S such that sM = 0 is equivalent to the condition that S ∩ AnnA(M) is non-empty.

The following inclusions show that S ∩ AnnA(M) is non-empty precisely when every S ∩ AnnA(Axj) is non- empty. k k Y \ (S ∩ AnnA(Axi)) ⊆ (S ∩ AnnA(Axj)) i=1 j=1  k  \ = S ∩  AnnA(Axj) j=1

= S ∩ AnnA(M)

⊆ S ∩ AnnA(Axj), where the step from the second to third line uses Exercise 2.2 (i).

3.2

a −1 Let a be an ideal of A, and define S = 1 + a. If 1+b is an element of S a, then a 1 + b − a 1 − = 1 + b 1 + b is a unit in S−1A. Thus S−1a is in the Jacobson radical of S−1A. Let M be a finitely generated A-module such that aM = M. Then (S−1a)(S−1M) = S−1M and S−1a is in the Jacobson radical of S−1A. By Nakayama’s Lemma, S−1M = 0. By Exercise 3.1 there is some s in S = 1 + a such that sM = 0.

3.3

Let S, T be multiplicatively closed subsets of A, let f : A → S−1A be the natural map, and let U = f(T ). Note −1 s that ST is a multiplicatively closed subset of A and let g : A → (ST ) A be the natural map. Since g(s) = 1 is a unit in (ST )−1A for each s in S, g has a unique lift h : S−1A → (ST )−1A such that h ◦ f = g by Proposition 3.1. If u = f(t) is in U = f(T ), then t h(u) = h(f(t)) = g(t) = 1 −1 −1 −1 is a unit in (ST ) A. If 0 = h(a/s) = g(a)g(s) = a/s in (ST ) A, then there are elements s0, t0 in S, T such that s0t0a = 0 in A. Thus 1 a 0 = · f(s0t0a) = f(t0) s0 1 −1 a −1 in S A. Finally, for every element st of (ST ) A, a = g(a)g(st)−1 = (g(a)g(s)−1)g(t)−1 = h(a/s)h(f(t))−1. st By Corollary 3.2, h lifts to an isomorphism from U −1(S−1A) to (ST )−1A. 3 RINGS AND MODULES OF FRACTIONS 35

3.4

Let f : A → B be a ring homomorphism, S a multiplicatively closed subset of A, and T = f(S). Consider B −1 −1 b b0 −1 as an A-module by restriction of scalars, and form the S A-module S B. Let s = s0 in S B. Then f(s00)(f(s0)b − f(s)b0) = s00 · (s0 · b − s · b0) = 0

−1 −1 b b in B. Thus, the map g : S B → T B : s 7→ f(s) is well defined and bijective. We show that it is additive and S−1A-homogeneous of degree 1. The calculation,

b b0  s0 · b + s · b0  s0 · b + s · b0 f(s0)b + f(s)b0 g + = g = = s s0 ss0 f(ss0) f(s)f(s0) b b0 b b0  = + = g + g , f(s) f(s0) s s0 shows that g is additive. On the other hand,

a b  a · b a · b f(a) b a  b  a  b  g · = g = = = (S−1f) g = · g s s0 ss0 f(ss0) f(s) f(s0) s s0 s s0 shows that g is S−1A-homogeneous of degree 1. Therefore, g is an isomorphism of S−1A-modules.

3.5

By Corollary 3.12, localization commutes with taking nilradicals. If A is a ring such that (R(A))p = R(Ap) = 0 for every prime ideal p, then R(A) = 0 by Proposition 3.8.

Let A = Z/6Z. Then (2), (3) are maximal ideals, and 2 · 3 = 0 shows that every prime ideal contains 2 or 3. Thus Spec A = {(2), (3)}. Furthermore, 2 · 3 = 0 implies that A(2) = F2 and A(3) = F3. Yet, A is not a domain.

3.6

Let A be a non-zero ring and let Σ be the set of multiplicatively closed subsets of A that do not contain 0. S Since A is not the zero ring, {1} is an element of Σ. Let {Si}i∈I be a chain in Σ and define S = i∈I Si. Then 1 ∈ S and if x, y ∈ S, then there is some j ∈ I such that x, y ∈ Sj. Thus xy ∈ Sj ⊆ S. Furthermore, if 0 were in S, then it would have to be in some Si. Since S is a multiplicatively closed subset of A not containing 0, it is an element of Σ. By Zorn’s Lemma, Σ has maximal elements. Let S be a maximal element of Σ and let x be an element of A. Since S is maximal in Σ, x is not in S if and only if the multiplicative sub-monoid of A generated by x and S contains 0. This in turn is equivalent to the condition that x is a nilpotent element of S−1A. Therefore, A \ S is the contraction of the nilradical of S−1A, hence an ideal. Since S is multiplicatively closed, p = A \ S is a prime ideal not meeting S. By Proposition 3.11 −1 (iv), S p = R(Ap) is a minimal prime ideal, and thus p is a minimal prime ideal. −1 Let p be a minimal prime ideal of A, x an element, and define S = A \ p. Then S p = R(Ap). Thus x is not in S if and only if x is nilpotent in Ap. This in turn is equivalent to the condition that the multiplicative sub-monoid of A generated by x and S contains 0. Thus S is a maximal element of Σ.

3.7

Let S be a multiplicatively closed subset of the ring A, 3 RINGS AND MODULES OF FRACTIONS 36

(i) If S is saturated, then A× ⊆ S because 1 is in S. If 0 is in S then S = A, and A \ S is the empty union of −1 prime ideals. Suppose that 0 is not in S, and let {mi}i∈I be the set of maximal ideals of S A. If a/s is a unit in S−1A, then there is some s0 in S such that s0(a − s) = 0 in A. In other words, s0a = s0s is in S. Thus, a is in S, S c S c because S is saturated. Therefore, A \ S ⊆ i∈I mi . Conversely, i∈I mi ⊆ A \ S by Proposition 3.11 (iv). S Now, suppose that A \ S = i∈I pi where the pi are prime ideals. If xy is in S, then xy is in the complement of each prime ideal pi. By primality, x, y are each in the complement of each prime ideal pi. Thus x, y are both in S, and S is saturated. (ii) Let Σ be the set of X such that S ⊆ X ⊆ A and X is saturated. Since A is in Σ, it is non-empty. Order Σ T by reverse inclusion, let {Xi}i∈I be a chain, and let Xb = i∈I Xi. Then Xb is obviously a multiplicatively closed S set containing S. Furthermore, A \ Xb = i∈I A \ Xi is a union of prime ideals by (i). Thus Xb is saturated and an element of Σ. By Zorn’s Lemma, Σ has minimal elements. If S1,S2 are elements of Σ, then S1 ∩ S2 is in Σ. Thus there is a unique minimal element, S in Σ. −1 S c
Let {mi}i∈I be the set of maximal ideals in S A. Then A \ i∈I mi is a saturated subset of A, by (i), that S c
S c
S contains S. Thus A \ i∈I mi is in Σ and S ⊆ A \ i∈I mi . Since S is saturated, A \ S = j∈J qj where the S S c qj are prime ideals of A not meeting S. Therefore, j∈J qj ⊇ i∈I mi . On the other hand, if p is a prime ideal of A not meeting S, then p does not meet S, and S−1p is a prime ideal −1 S S c of S A. Thus j∈J qj ⊆ i∈I mi holds as well. Let a be an ideal of A and let S = 1 + a. Note that every factor of an element of S is in S, and that the collection of these factors is a multiplicatively closed subset of A. Thus S is the set of factors of elements of S. We make this more concrete as follows: If x is in S, then there is y in S such that xy ≡ 1 mod a. Conversely, if xy ≡ 1 mod a, then there is a in a such that xy = 1 + a, and x, y are in S. Thus

S = π−1 (A/a)×
, where π : A → A/a is the natural projection.

3.8

Let S ⊂ T be multiplicatively closed subsets of the ring A, and let

φ : S−1A → T −1A be the lift of f : A → T −1A as given by Proposition 3.1. −1 1
t −1 1 t
−1 (i)⇒(ii) If t be an element of T , then φ t · 1 = φ t · 1 = 1 in S A. x t −1 xt 1 −1 0 (ii)⇒(iii) Let s be the inverse of 1 in S A. The equation s = 1 in S A implies that there is s in S such that s0(xt − s) = 0 in A. Hence, (s0x)t = ss0 is in S. (iii)⇒(iv) For each t in T there is x in A such that xt is in S. Thus t is in S. (iv)⇒(v) Since T ⊆ S, A \ T contains every prime ideal not meeting S, by Exercise 3.7. Thus every prime ideal meeting T also meets S. (v)⇒(i) Note that φ : S−1A → T −1A is a homomorphism of S−1A-modules. If p is a prime ideal of S−1A, c −1 −1 then S ⊆ T ⊆ A \ p . By Exercise 3.3,(S A)p = Apc = (T A)p, and φp is the identity on Ap, hence bijective. By Proposition 3.9, φ is bijective.

3.9

Let S0 be the set of all non-zero-divisors in the non-zero ring A. Then S0 is obviously a saturated multiplica- tively closed subset and its complement D is a union of prime ideals by Exercise 3.7. If p is a minimal prime ideal, 3 RINGS AND MODULES OF FRACTIONS 37 then S = A \ p is a saturated, maximal multiplicatively closed set not containing 0, by Problems 3.6 and 3.7. If x is not in S, then 0 is in the multiplicative monoid generated by S and x. Therefore x is a zero divisor, and p = A \ S ⊆ D. (i) Let S be a multiplicatively closed subset of A containing a non-zero, zero-divisor, d, annihiliated by c. Then c cd 0 = = = 0 1 d d −1 −1 shows that A → S A is not injective. On the other hand, if a/1 = 0 in S0 A, then there is s in S0 such that −1 sa = 0 in A. Since s is not a zero-divisor, a = 0. Thus, the canonical map A → S0 A is injective. d c cd (ii) If d is a zero-divisor of A, annihilated by c, then s · 1 = s = 0 for every s in S0. Therefore, if a/s is not a zero-divisor, then a is in S0 = A \ D, and a/s is a unit with inverse s/a. × × −1 (iii) If A = A ∪ D, then S0 = A . Let f : A → S0 A be the canonical map. The f is surjective, since a as−1 as−1 = = = f(as−1), s ss−1 1 a −1 and f is injective, since 0 = f(a) = 1 implies that there is s in S such that as = 0 in A. Thus 0 = ass = a. Therefore, f is an isomorphism.

3.10

Let A be a ring. (i) Suppose A is absolutely flat, S is a multiplicatively closed subset, and N is an S−1A-module. Then N −1 is a flat A-module by restriction of scalars. Now extend scalars to get NS−1A = S A ⊗A N, which is a flat −1 −1 −1 S A-module by Exercise 2.20. Furthermore, NS−1A = S N by Proposition 3.5. Finally, S N = N because N is an S−1A-module. Thus N is a flat S−1A-module, and S−1A is absolutely flat.

(ii) If A is absolutely flat, then Am is an absolutely flat local ring for each maximal ideal m, hence a field. On the other hand, let A be a ring that localizes to a field at each maximal ideal, m, and let N be an A- module. Then Nm is an Am-vector space, hence a free Am-module. Therefore, Nm is a flat Am-module. Since m was arbitrary, N is a flat A-module by Proposition 3.10.

3.11

Let A be a ring.

(iv)⇒(iii) Every Hausdorff space is T1. (iii)⇒(ii) Closed points of Spec(A) correspond to maximal ideals by Exercise 1.18 (i). (ii)⇒(i) By the Lattice Isomorphism Theorem, we may identify the maximal ideals of A and A/R(A). Let m be such an ideal. Then

(A/R(A))m = Am/R(Am) = Am/m is a field, and A/R(A) is absolutely flat. (i)⇒(iv) By Exercise 1.21 (iv), Spec(A) and Spec(A/R(A)) are naturally homeomorphic. Without loss of generality, R(A) = 0 and A is absolutely flat. Let p be a prime ideal of A contained in a maximal ideal m. Since Am is a field, it has only one prime ideal, the zero ideal. Thus p = m by Proposition 3.11 (iv), and every prime ideal of A is maximal. Suppose that m, n are distinct maximal ideals. Then there is f ∈ m and g ∈ n such that f + g = 1. On the other hand, the elements of m vanish in Am. So there is s not in m such that sf = 0. Together, we have s = sf + sg = sg is an element of n and f = 1 − g is not. Therefore, m ⊆ Xs, n ⊆ Xf and Xs ∩ Xf = Xsf = ∅. So Spec(A) is Hausdorff. 3 RINGS AND MODULES OF FRACTIONS 38

In general, Spec(A) is quasi-compact by Exercise 1.17 (v). If it is Hausdorff, then it is compact. Let X be a subset of Spec(A) containing distinct points x, y, and let Xf contain x but not y. Since Xf is a quasi-compact subset of a Hausdorff space, it is closed. Thus Xf ∩ X,X \ Xf is a disjoint cover of X by non-empty, relatively open sets, and X is disconnected. Therefore, the only connected subsets of Spec(A) are singletons.

3.12

Let A be a domain, M an A-module, and T (M) the set of torsion elements of M. If x, y ∈ T (M), a ∈ A, and b, c are respectively non-zero elements of AnnA(x), AnnA(y), then

bc(x + ay) = c(bx) + ab(cy) = 0, and bc 6= 0 because A is a domain. Thus T (M) is a submodule of M.

(i) Let x ∈ M, 0 6= a ∈ A such that ax ∈ T (M). Then there is some non-zero b ∈ AnnA(ax), and (ba)x = b(ax) = 0. Since A is a domain, ba is a non-zero element of AnnA(x), and x ∈ T (M). Thus T (M/T (M)) = 0, and M/T (M) is torsion-free.

(ii) Let f : M → N be an A-module homomorphism, x ∈ T (M), and a a non-zero element of AnnA(x). Then af(x) = f(ax) = f(0) = 0. Thus f(T (M)) ⊆ T (N). (iii) Suppose f g 0 → M 0 −→ M −→ M 00 is an exact sequence of A-modules, and consider the induced sequence

T (f) T (g) 0 → T (M 0) −−−→ T (M) −−−→ T (M 00).

Since it is the restriction of an injective function, T (f) is injective. Similarly, T (g) ◦ T (f) = 0, because g ◦ f = 0. If x ∈ T (M) such that T (g)(x) = 0, then g(x) = 0. Since the first sequence is exact, and T (M) is a submodule 0 of M, there is y ∈ M such that f(y) = x. If a is a non-zero element of AnnA(x), then f(ay) = af(y) = ax = 0. Since f is injective, ay = 0. Thus y ∈ T (M 0) and T (f)(y) = x. Therefore the induced sequence on torsion is exact.

(iv) Let K be the field of fractions of A and define φ : M → K ⊗A M by f(x) = 1 ⊗ x. If x is in T (M), then it is annihilated by some a in A. Thus φ(x) = 1 ⊗ x = a−1 ⊗ ax = 0, and x is in the kernel of φ. The following demonstrates the converse.

The family {Aξ}ξ∈K is a family of A-submodules of K. Given ξ, η ∈ K, there are non-zero a, b in A such that 1 aξ, bη are in A. Thus Aξ + Aη ⊆ A ab . By Exercise 2.17, [ K = Aξ = lim Aξ. −→ ξ∈K ξ∈K

By Exercise 2.20, we may identify K ⊗ M with lim (Aξ ⊗ M). A −→ξ∈K A −1 If φ(x) = 0, then there is some ξ in K such that 1 ⊗ x = 0 in Aξ ⊗A M, by Exercise 2.15. Since 1 is in Aξ, ξ is in A. Furthermore, A → Aξ : a 7→ aξ is an isomorphism of A-modules because K is a field. Therefore,

ψ : Aξ ⊗A M → A ⊗A M → M : aξ ⊗ x 7→ ax is an A-module isomorphism. Finally, we look at the composition

0 = ψ(0) = (ψ ◦ φ)(x) = ψ(1 ⊗ x) = ξ−1x.

Thus x is in T (M). 3 RINGS AND MODULES OF FRACTIONS 39

3.13

Let S be a multiplicatively closed subset of the domain A. For any x in M and s in S,

Ann(x/s) = Ann(S−1Ax) = S−1 Ann(Ax) = S−1 Ann(x)

x −1 −1 by Proposition 3.14. Thus s is in T (S M) if and only if it is in S T (M).

In particular, T (Mp) = T (M)p for all prime ideals p of A. By Proposition 3.8, (i), (ii), and (iii) are equivalent.

3.14

Let a be an ideal of A and let M be an A-module such that Mm = 0 for all maximal ideals m containing a. Since localization is exact, M/aM is an A/a-module such that (M/aM)m = Mm/(aM)m = 0 for all maximal ideals m of A/aA. By Proposition 3.8, M/aM = 0. Thus M = aM.

3.15

n Let F = A for some ring A. Let x1, . . . , xn be a set of generators for F , let e1, . . . , en be the canonical basis, and let φ : F → F be the map given by φ(ei) = xi. The map φ is surjective because x1, . . . , xn generate F . The following demonstrates that φ is injective. Without loss of generality, we may assume A is a local ring by Proposition 3.9. Let m be the maximal ideal of A, let k be its residue field, and let N be the kernel of φ. Then the sequence

φ 0 → N → F −→ F → 0 is exact. Since F is a flat A-module,

φb 0 → k ⊗A N → k ⊗A F −→ k ⊗A F → 0

n is exact. Furthermore, k ⊗A F is naturally isomorphic to k . Thus φb is a surjective k-linear map of n-dimensional vector spaces, and bijective as a consequence. By Exercise 2.12, N is a finitely generated A-module. Since k ⊗A N = 0, Exercise 2.3 implies that N = 0.

3.16

Let f : A → B be a flat A-algebra. (i)⇒(ii) Let p be a point in Spec(A). By Proposition 3.16 there is q in Spec(B) such that f ∗(q) = qc = p. (ii)⇒(iii) Let m be a maximal ideal of A and let q be a prime ideal of B such that f ∗(q) = m. Then me = qce ⊆ q 6= (1). (iii)⇒(iv) Let x be a non-zero element of M and let a = Ann(x). Since x is non-zero, a is a proper ideal of A. By hypothesis, ae is then a proper ideal of B. By Exercise 2.2 and the First Isomorphism Theorem, we have the following isomorphisms of B-modules

e ∼ ∼ B/a = B ⊗A (A/a) = B ⊗A Ax.

Since B is flat, the map B ⊗A Ax → B ⊗A M induced by the inclusion Ax → M is injective. Thus MB is a non-zero B-module. 0 (iv)⇒(v) Let M be the kernel of the natural map M → B ⊗A M so that we have a short exact sequence of A-modules 0 0 → M → M → MB → 0. 3 RINGS AND MODULES OF FRACTIONS 40

Since B is flat, we may tensor with B ⊗A − to get a short exact sequence of B-modules

0 0 → MB → MB → (MB)B → 0.

0 0 By Exercise 2.13, the map on the right is injective. Therefore MB = 0. By hypothesis, this implies that M = 0. Thus M → B ⊗A M is injective. (v)⇒(i) By hypothesis, the natural map

∼ e A/a → (A/a)B = B/a is injective. Thus a ⊆ aec ⊆ a.

3.17

f g Let A −→ B −→ C be a pair of ring homomorphisms such that g ◦ f is flat and g is faithfully flat, and let h : M → N be an injective homomorphism of A-modules. Let K be the kernel of the extension hB : MB → NB. Now tensor with C to get the commutative diagram

0 K ⊗B C (M ⊗A B) ⊗B C (N ⊗A B) ⊗B C

0 M ⊗A C N ⊗A C, where the right and center vertical arrows are the canonical isomorphisms. The bottom row is exact because g ◦ f is flat, and the top row is exact because g is flat. Thus K ⊗B C = 0. Since g is faithfully flat, K = 0 by Exercise 3.16 (iv). Therefore, f : A → B is a flat A-algebra.

3.18

Let f : A → B be a flat homomorphism of rings, let q be a prime ideal of B, and let p = qc. By Proposition e ce 3.10, Bp is flat over Ap and Bq is flat over Bp. By Exercise 2.8 (ii), Bq is flat over Ap. Since p = q ⊆ q, fb : Ap → Bq satisfies the conditions of Exercise 3.16 (iii). Therefore, Bq is a faithfully flat Ap-algebra, and ∗ f : Spec(Bq) → Spec(Ap) is surjective.

3.19

Let M be an A-module and define the support of M by

Supp(M) = {p ∈ Spec(A)|Mp 6= 0}.

−1 −1 We will use the result that S A = 0 if and only if 0 ∈ S. In particular, S (A/a) = 0 if and only if S ∩ a 6= ∅.

(i) My Proposition 3.8, M = 0 if and only if Mp = 0 for all prime ideals p. This is equivalent to Supp(M) = ∅.

(ii) A given prime ideal p contains a if and only if (A \ p) ∩ a = ∅. Thus p is in V (a) if and only if (A/a)p 6= ∅. (iii) Let 0 → M 0 → M → M 00 → 0 be an exact sequence of A-modules. By Proposition 3.3,

0 00 0 → Mp → Mp → Mp → 0

0 00 is an exact sequence of Ap-modules for every prime ideal p in A. Thus Mp = 0 if and only if Mp = Mp = 0. 3 RINGS AND MODULES OF FRACTIONS 41

(iv) By (ii), Supp(Mi) ⊆ Supp(M) for each i. Conversely, suppose that x/s is a non-zero element of Mp. Then

 k  k X X x/s ∈  Mij  = (Mij )p j=1 j=1 p S for some indices i1, . . . , ik. Therefore, (Mi)p 6= 0 for some i, and Supp(M) ⊆ i∈I Supp(Mi) as a consequence. (v) A prime ideal p contains Ann(M) if and only if (A \ p) ∩ Ann(M) = ∅. This is equivalent to the condition that (A/ Ann(M))p = Ap/ Ann(M)p is non-zero. Since Mp = 0 if and only if Ap = Ann(Mp), we are done.

(vi) By Proposition 3.7, (M ⊗A N)p = Mp ⊗Ap Np as Ap-modules. If Mp = 0 or Np = 0, then Mp ⊗ap Np = 0. Thus Supp(M ⊗A N) ⊆ Supp(M) ∩ Supp(N).

By Corollary 3.4, Mp,Np are finitely generated Ap-modules. Since Ap is a local ring, if Mp ⊗Ap Np = 0, then Mp = 0 or Np = 0 by Exercise 2.3. Therefore,

Supp(M ⊗A N) ⊇ Supp(M) ∩ Supp(N).

(vii) By Exercise 2.2, M/a = (A/a) ⊗A M. Thus, parts (ii), (v), and (vi) imply that

Supp(M/aM) = Supp((A/a) ⊗A M) = Supp(A/a) ∩ Supp(M) = V (a) ∩ V (Ann(M)) = V (a + Ann(M)).

By Proposition 3.3 and Exercise 2.2,

(M/aM)p = Mp/apMp = (Ap/ap) ⊗Ap Mp = (A/a)p ⊗Ap Mp.

By Corollary 3.4, Mp is a finitely generated Ap-module. Therefore, (M/aM)p = 0 if and only if (A/a)p = 0 or Mp = 0. (viii) Let q be a point in Spec(B) and let p = f ∗(q). By Proposition 3.5,

(B ⊗A M)q = Bq ⊗B (B ⊗A M)

= Bq ⊗A M

= (Bq ⊗Ap Ap) ⊗A M

= Bq ⊗Ap Mp.

∗−1 Therefore, Supp(B ⊗A M) ⊆ f (Supp(M)).

Let κp = Ap/pAp and κq = Bq/qBq be the respective residue fields of p and q, and assume that 0 = (B ⊗A M)q. Tensor with κq over Bq to get

0 = κq ⊗Bq (B ⊗A M)q = κq ⊗Bq Bq ⊗Ap Mp = κq ⊗Ap Mp.

Since the tensor product is right exact, κq ⊗Ap Mp surjects onto κq ⊗Ap (Mp/pMp). Therefore, the κp-vector space

κq ⊗Ap (Mp/pMp) is zero. Since κq is a field, Mp/pMp = 0. Furthermore, Mp is a finitely generated Ap-module ∗−1 and p is the Jacobson radical of Ap. By Nakayama’s Lemma, Mp = 0. Thus Supp(B ⊗A M) ⊇ f (Supp(M)).

3.20

Let f : A → B be a ring homomorphism. 3 RINGS AND MODULES OF FRACTIONS 42

(i) If f ∗ : Spec(B) → Spec(A) is surjective, then every prime ideal of A is a contracted ideal. Conversely, let p = bc for some ideal b in B then pec = bcec = bc = p. By Proposition 3.16, p is the contraction of a prime ideal of B, and f ∗ is surjective. e e (ii) Suppose that every prime ideal of B is an extended ideal, and let q1 = a1, q2 = a2 be prime ideals of B. If ∗ ∗ f (q1) = f (q2), then e ece ∗ e ∗ e ece e q1 = a1 = a1 = f (q1) = f (q2) = a2 = a2 = q2, and f ∗ is injective. For a counterexample to the converse, let

2 f : A = C → C[x]/(x ) = B be the natural inclusion map. Since x2 = 0 in B, every prime ideal of B contains x. On the other hand, (x) is a maximal ideal, so B has exactly one prime ideal. Similarly, A has only the prime ideal (0) because it is a field. Thus f ∗ : Spec(B) → Spec(A) is the injective map between one point sets. Finally,

ce e (x) = (0A) = (0B) shows that (x) cannot be an extended ideal.

3.21

(i) Let S be a multiplicatively closed subset of the ring A, and let φ : A → S−1A be the canonical homomor- phism. By Proposition 3.11 (iv), φ∗ : Spec(S−1A) → Spec(A) = X is bijective onto its image, S−1X. Furthermore, −1 −1 T S X consists of prime ideals of A not meeting S. Thus S X = s∈S Xs. ∗ −1 Now we will show that φ is an open map. Let Yf/g be an open set in Spec(S A). Exercise 1.21 (i) allows us to calculate that

c c ∗−1 Yf/g = V (f/g) = V (fs) = Yfs = Yφ(fs) = φ (Xfs) ∗−1 ∗−1 ∗−1 = φ (Xf ∩ Xs) = φ (Xf ) ∩ φ (Xs), because gs is a unit in S−1A, for every s in S. Thus

\ ∗−1 ∗−1 Yf/g = (φ (Xf ) ∩ φ (Xs)) s∈S ! ∗−1 ∗−1 \ = φ (Xf ) ∩ φ Xs s∈S ∗−1 ∗−1 −1 = φ (Xf ) ∩ φ (S X).

∗ −1 −1 ∗ −1 ∗ ∗ Since φ : Spec(S A) → S X is bijective, φ (Yf/g) = Xf ∩ S X, and φ is an open map. Therefore, φ is a homeomorphism onto its image. −1 T If S is the monoid generated by f, then S X = n≥0 Xf n = Xf . (ii) Let f : A → B be a ring homomorphism, X = Spec(A),Y = Spec(B), and f ∗ : Y → X the associated mapping. The diagram, 3 RINGS AND MODULES OF FRACTIONS 43

f A B

S−1f S−1A S−1B, obviously commutes. By Proposition 3.11 (iv), we may identify the prime ideals of A and B not meeting S with their extensions to S−1A and S−1B, and (S−1f)−1(S−1q) = S−1p if and only if f −1(q) = p, by the commutativity of the diagram. Thus the restriction of f ∗ to S−1Y has image in S−1X, and can be identified with the map (S−1f)∗ : Spec(S−1B) → Spec(S−1A).

If f ∗(q) = p, then S−1(A/p) → S−1(B/q) is injective, because localization is exact. Thus, p must meet S, if q does. Consequently, the only points of Y mapping into S−1X by f ∗ are those of S−1Y . In other words, S−1Y = f ∗−1(S−1X). (iii) Let a be an ideal of A with extension b to B, and let f : A/a → B/b be the map obtained from f : A → B by passing to the quotient. If q is a prime ideal of B containing b, then qc ⊇ bc = aec ⊇ a. Thus f ∗ restricted to V (b) has image in V (a). By Exercise 1.21 (iv), the natural quotient maps A → A/a,B → B/b induce homeomorphisms Spec(A/a) → V (a), Spec(B/b) → V (b). We must show that the diagram,

∗ f Spec(B/b) Spec(A/a)

f ∗ V (b) V (a), commutes. But the diagram,

f A B

f A/a B/b, already commutes. Pulling back a prime ideal from the lower right corner to the upper left along the two routes shows that the previous diagram of spectra commutes. (iv) Let p be a prime ideal of A, let S = A \ p, let f : A → B be a ring homomorphism, and X = Spec(A),Y = Spec(B). Since V (p) is the set of prime ideals containing p and S−1X is the set of prime ideals contained in p, {p} = V (p) ∩ S−1X. Therefore, f ∗−1(p) = f ∗−1(V (p) ∩ S−1X) = V (pB) ∩ S−1Y,

−1 by Exercise 1.21 (ii), and (iii) above. By (iii), we may identify S Y with Spec(Bp). Under this identification, −1 V (pB) ∩ S Y = V (pBp). The latter space is canonically homeomorphic to Spec(Bp/pBp) by Exercise 1.21 (iv). On the other hand,

Bp/pBp = Ap/pAp ⊗Ap Bp

= κp ⊗Ap Ap ⊗A B

= κp ⊗A B. ∗−1 Therefore, f (p) = Spec(κp ⊗A B). 3 RINGS AND MODULES OF FRACTIONS 44

3.22

Let p be a prime ideal of the ring A. Every open neighborhood U of p contains a basic open set Xs with s in S = A \ p. Thus, −1 \ \ Spec(Ap) = S X = Xs = U, s∈S p∈U by Exercise 3.21 (i).

3.23

Let X = Spec(A) for some ring A and let U = Xf for some f in A. p p (i) Suppose that U = Xg. By Exercise 1.17 (iv), (f) = (g). Thus there are positive integers m, n and m n elements a, b of A such that f = ag and g = bf. Let φ : A → Af be the canonical map. We will use Corollary 3.2 to get our desired canonical isomorphism. First,

f mn = (ag)n = anbf

m shows that a, b are units in Af , and moreover ag = f shows that g is a unit in Af . Second, φ(x) = 0 implies that f kx = 0 for some k ≥ 0. Multiply through by bk to get 0 = bkf kx = gnkx. Third, x = φ(x)φ(f k)−1 = φ(x)φ(bk)φ(bkf k)−1 = φ(xbk)φ(gnk)−1. f k

Therefore, there is a unique isomorphism Ag → Af lifting the canonical homomorphism A → Af . Flipping the roles of f and g yields the inverse map. Thus A(U) is a well defined ring. 0 p (ii) Now let Xg = U ⊆ Xf = U. By Exercise 1.17 (iv), g ∈ (f). Therefore, there is an integer n ≥ 0 and a in A such that gn = af. As a consequence, f is a unit in A(U 0). By Proposition 3.1, there is a unique ring homomorphism ρ : A(U) → A(U 0) lifting the canonical map A → A(U 0). (iii) If U = U 0, then the uniqueness of ρ implies that it is the identity map. (iv) The three arrows are all unique maps lifting the identity by Proposition 3.1. The uniqueness forces the composition of the lower to maps to equal the top map.

(v) Let x = p be a point of X. If x ∈ U and U = Xf , then f ∈ A \ p. Thus there is a unique map µU : A(U) → Ap lifting the canonical homomorphism A → Ap. Furthermore, µU = µU 0 ◦ ρUU 0 by uniqueness. By Exercise 2.16 there is a unique map ρ : lim A(U) → A lifting the maps µ . −→x∈U p U x x Given f ∈ Ap. We have that f is in the image of µU for U = Xf . Thus ρ is surjective.

If ρ(a) = 0, then there is f ∈ A \ p such that fa = 0 in A. Let U = Xf , and note that this implies that a = 0 in A(U), and thus in lim A(U) as well. Therefore, ρ is injective. −→x∈U

3.24

Let {Ui}i∈I be a covering of X by basic open sets and let si ∈ A(Ui) for each i such that si = sj in A(Ui ∩ Uj) xi for each i, j. For each i, let U = X and s = m . Since s = s in A(U ∩ U ), there are integers n ≥ 0 i fi i f i i j i j ij m i nij j mi such that (fifj) (xifj − xjfi ) = 0 in A for all i, j. On the other hand, there are finitely many f’s such that (f1, . . . , fk) = (1) because X is quasi-compact. Raising the ideal (f1, . . . , fk) to a sufficiently large power yields Pk mi maxi6=j (nij ) Pk i=1 aifi = 1 for some a1, . . . , ak in A such that ai ∈ (fi) . Define s = i=1 aixi. Then, in A(Uj), we 3 RINGS AND MODULES OF FRACTIONS 45 have

k X s = aixi i=1 mj X = ajsjfj + aixi i6=j  

X mi X = sj 1 − aifi  + aixi i6=j i6=j

X mi = sj + ai(xi − sjfi ) i6=j

1 X mj mi = sj + ai(xif − xjf ) f mf j i j i6=j

= sj, for j = 1, . . . , k.

Now, suppose that i ∈ I is arbitrary. For each j = 1, . . . , k, s = sj = si in A(Ui ∩ Uj). Therefore, there is mi n n ≥ 0 such that (fi s − xi)(fifj) = 0 in A for each j. Taking powers of (f1, . . . , fk) as before, we may obtain an Pk n A-linear combination j=1 bjfj = 1. Summing the equations

mi n bj(fi s − xi)(fifj) = 0

mi n in A yields (fi s − xi)(fi) = 0. Thus s = si in A(Ui). 0 Now, suppose that s ∈ A also has image si in A(Ui) for every i ∈ I. Given the f1, . . . , fk as before, we get 0 n the systems of equations (s − s )fj = 0 for j = 1, . . . , k. Raising (f1, . . . , fk) to a sufficiently high power and then summing our equations yields s = s0.

3.25

Let f : A → B, g : A → C, h : A → B ⊗A C be ring homomorphisms such that h(x) = f(x) ⊗ g(x), and let X,Y,Z,T be the respective prime ideal spectra of A, B, C, B ⊗A C. We will make multiple uses of Exercise 3.21 (iv). For any prime ideal p of A, p ∈ h∗(T ) if and only if

∗−1 h (p) = Spec(κp ⊗ (B ⊗A C)) = Spec((κp ⊗A B) ⊗κp (κp ⊗A C)) is non-empty. This is equivalent to the condition that the κp-vector space (κp ⊗A B) ⊗κp (κp ⊗A C) is non-zero. In turn, this is equivalent to the condition that both κp ⊗A B and κp ⊗A C are non-zero κp-vector spaces. We ∗−1 ∗−1 may flip this again, to the condition that f (p) = Spec(κp ⊗A B) and g (p) = Spec(κp ⊗A C) are non-empty. Finally, this is equivalent to p ∈ f ∗(Y ) ∩ g∗(Z).

3.26

Let B be the direct limit of the direct system of rings Bα, gαβ, let fα : A → Bα be A-algebras such that ∗−1 gαβ ◦fα = fβ for all α ≤ β, and let f : A → B be the induced map. By Exercise 3.21 (iv), f (p) = Spec(κp ⊗A B) is empty if and only if κp ⊗A B = 0. On the other hand,

κ ⊗ B = lim(κ ⊗ B ) p A −→ p A α α by Exercise 2.20. By Exercise 2.15, a direct limit is 0 if and only if every element is eventually zero. In the case ∗−1 of a ring, this is equivalent to 1 = 0 for some α. Now, κp ⊗A Bα = 0 if and only if fα (p) = Spec(κp ⊗A Bα) is ∗ ∗ empty. In summary, p is in the image of f if and only if it is in the image of fα for every α. 4 PRIMARY DECOMPOSITION 46

4 Primary Decomposition

4.1

Suppose a is a decomposable ideal of A. The irreducible components of Spec(A/a) are the sets V (p), where p is a minimal prime ideal over a, by Exercise 1.20 (iv). By Proposition 4.6, these are the isolated prime ideals of a. Furthermore, Theorem 4.5 tells us that there are only finitely many such prime ideals.

4.2

√ Tn Suppose a is a decomposable ideal of A such that a = a. Let i=1 qi be an irredundant primary decompo- √ Tn √ Tn sition of a with associated prime ideals p1,..., pn. Then a = a = i=1 qi = i=1 pi yields another primary decomposition of a. Since the initial decomposition was assumed to be irredundant, this decomposition is as well. Therefore a has no embedded prime ideals.

4.3

Suppose A is absolutely flat and q is a p-primary ideal. By Exercise 2.28, A/q is absolutely flat, and non-units are zero-divisors. Since q is primary, non-units are nilpotent. Therefore, A/q is an absolutely flat local ring. By Exercise 2.28, A/q is a field, and q is a maximal ideal as a consequence.

4.4

Define m = (2, t), q = (4, t) as ideals of [t]. Since [t]/m =∼ , m is a maximal ideal. Note that 22, t ∈ q. √ Z Z F2 Thus m ⊆ q, and we have equality because m is maximal. By Proposition 4.2, q is m-primary. On the other hand, t ∈ q \ m2. So mk ⊆ m2 ⊂ q ⊂ m for all integers k ≥ 2. Therefore, q is not a power of m.

4.5

Define p1 = (x, y), p2 = (x, z), m = (x, y, z) in the polynomial ring K[x, y, z], where K is a field and x, y, z are ∼ ∼ ∼ indeterminates. Since K[x, y, z]/p1 = K[z], K[x, y, z]/p2 = K[y], and K[x, y, z]/m = K, we know that p1, p2 are prime ideal and m is maximal. 2 Define a = p1p2 = (x , xy, xz, yz). Clearly p1 and p2 are respectively p1-primary and p2-primary. By Proposi- 2 2 2 2 2 tion 4.2, m is m-primary. Since p1 ∩ p2 = (x, yz) and m = (x , xy, xz, yz, y , z ), we know that

2 p1 ∩ p2 ∩ m = a

2 is a primary decomposition. Furthermore, the primary components p1, p2, m have distinct associated prime ideals, and 2 2 2 2 2 x ∈ p1 ∩ p2 \ m , y ∈ p1 ∩ m \ p2, z ∈ p2 ∩ m \ p1 shows that the decomposition is irredundant. Let p be a prime ideal containing a, then x ∈ p because x2 ∈ a ⊆ p. Similarly, yz ∈ p implies that y ∈ p or z ∈ p. Thus p1, p2 are the isolated prime ideals of a and m is an embedded prime ideal.

4.6

Tn Let X be a compact Hausdorff space where (0) is a decomposable ideal of C(X), and let (0) = i=1 qi be an irredundant primary decomposition of (0). 4 PRIMARY DECOMPOSITION 47

Suppose a is an ideal contained in distinct maximal ideals mx, my. Since X is Hausdorff, x and y have disjoint open neighborhoods U, V . By Urysohn’s Lemma, there are f, g ∈ C(X) such that f(x) = g(y) = 1, f|X\U = 0, and g|X\V = 0. Thus, fg = 0 ∈ a, but f∈ / my and g∈ / mx. Therefore, each primary ideal of C(X) is contained in a unique maximal ideal.

For each i, let mi be the maximal ideal over qi. By Exercise 1.26, X is homeomorphic to n ! [ Max(C(X)) = Max(C(X)) ∩ V (0) = Max(C(X)) ∩ V (qi) = {m1,..., mn}. i=1 Therefore, X must be finite.

4.7

Let A[x] be the ring of polynomials in one indeterminate over the ring A, and let a[x] be the ideal of polynomials with coefficients in the ideal a of A. (i) Clearly, ae ⊆ a[x]. On the other hand, axn ⊆ ae for every non-negative integer n. Thus, ae = a[x]. (ii) Lift the natural map A → A/p to obtain a map A[x] → (A/p)[x] with kernel p[x]. Therefore, A[x]/p[x] =∼ (A/p)[x], and p[x] is a prime ideal of A[x] whenever p is a prime ideal of A. (iii) Let q be a p-primary ideal of A. As in (ii), (A/q)[x] =∼ A[x]/q[x]. Let f ∈ (A/q)[x] be a zero-divisor. By Exercise 1.2, there is b ∈ A \ q such that bf ∈ q. Therefore, the coefficients of f are zero-divisors in A/q. Since q is primary, the coefficients of f are nilpotent in A/q. Again we cite Exercise 1.2 to see that f is nilpotent. Thus q[x] is a primary ideal of A[x]. Furthermore, f k ∈ q[x] for some positive integer k if and only if the coefficients of √ f are in q = p. So pq[x] = p[x], and q[x] is in fact p[x]-primary. Tn n (iv) Let a = i=1 qi be an irredundant primary decomposition of an ideal a of A. The prime ideals (pi[x])i=1 of A[x] are distinct because they have distinct contractions to A. Furthermore, for each j we have   \ \ (qi[x]) =  qi [x] * qj[x] i6=j i6=j T because i6=j qi * qj. Therefore, n ! n \ \ a = qi [x] = qi[x] i=1 i=1 is an irredundant primary decomposition of a. (v) Suppose that p is an isolated prime ideal of a and let p0 be a prime ideal of A[x] such that a[x] ⊆ p0 ⊆ p[x]. Taking contractions yields a ⊆ (p0)c ⊆ p. Since p is minimal, (p0)c = p. On the other hand, p0 ⊇ (p0)ce = p[x]. Thus p0 = p[x]. So p[x] is an isolated prime ideal of a[x].

4.8

Let k be a field and define ideals pi = (x1, . . . , xi) of k[x1, . . . , xn] for each i = 1, . . . , n. The map k[x1, . . . , xn] → k[xi+1, . . . , xn] obtained by evaluating each xj at 0 for 1 ≤ j ≤ i is surjective with kernel pi. Since k[xi+1, . . . , xn] is a domain, pi is prime ideal. If i = n, then pi is a maximal ideal. By Proposition 4.2, the powers of pi are pi-primary in k[x1, . . . , xi]. By Exercise 4.7 (iii), the powers of pi are pi-primary in k[x1, . . . , xn] as well.

4.9

Let A be a ring and define D(A) = {p ∈ Spec(A): p ⊇ (0 : a) is a minimal for some a ∈ A}. 4 PRIMARY DECOMPOSITION 48

If x ∈ A is a non-zero, zero-divisor, then x ∈ (0 : a) for some non-zero a ∈ A. If p is a minimal prime ideal above (0 : a), then x ∈ p and p ∈ D(A). Let x be a non-zero element of A and suppose that there is a prime ideal p such that x ∈ p ∈ D(A). By the definition of D(A), there is a ∈ A such that p is a minimal prime ideal above (0 : a). Thus x is nilpotent in n (A/(0 : a))p. In other words, there is s ∈ A \ p and a positive integer n such that x s ∈ (0 : a). Since s∈ / p, we know that sa is non-zero. If n is the smallest such integer, then xn−1sa is non-zero, but x · xn−1sa = 0. Therefore, x is a zero-divisor. Let S be a multiplicatively closed subset of A. By definition, S−1p ∈ D(S−1A) if and only if S−1p is a minimal prime ideal above (0 : a/s) for some a ∈ A and s ∈ S. By Proposition 3.11 (ii), (0 : a/s) = S−1(0 : a). Therefore, by Proposition 3.11 (iv), this is equivalent to the condition that p is a minimal prime ideal above (0 : a) for some −1 a ∈ A and S ∩ p = ∅. In turn, this is holds precisely when p ∈ D(A) ∩ Spec(S A). Suppose that (0) is a decomposable ideal and p is one of its associated prime ideals. By Theorem 4.5, p = p(0 : a) for some a ∈ A. Thus p ∈ D(A). On the other hand, suppose that p ∈ D(A) is a minimal prime ideal above (0 : a) for some a ∈ A. Then p pp ∈ D(Ap). So pp is a minimal prime ideal above (0 : a)p. Since Ap is local, pp = (0 : a)p. By Proposition 4.2, (0 : a)p is pp-primary. Taking contractions yields c
c (0 : a) = 0 : (a)p ⊇ (0 : a)p ⊇ (0 : a). Thus (0 : a) is a primary ideal. Furthermore, by Proposition 4.8, (0 : a) is p-primary. So p is an associated prime ideal of (0).

4.10

For every prime ideal p of A, let Sp(0) be the kernel of the natural map A → Ap.

(i) Since (0)p ⊆ pp, Sp(0) ⊆ p by Proposition 3.11. p p
p √ (ii) If Sp(0) = p, then pp = Sp(0) p = (Sp(0))p = 0 in Ap. Therefore, Ap has only one prime ideal, and p is a minimal prime ideal of A. p If p is a minimal prime ideal of A, then pp is the nilradical of Ap. By Exercise 1.18, Sp(0) = p. 0 (iii) If p ⊇ p , then the universal property of localization gives a unique lift of A → Ap0 to Ap → Ap0 such that the following diagram commutes:

A Ap

Ap0

Therefore, Sp(0) ⊆ Sp0 (0). T (iv) Suppose x ∈ p∈D(A) Sp(0), and p is a minimal prime ideal above (0 : x). By definition, p ∈ D(A). Thus x ∈ Sp(0) by hypothesis. If x 6= 0, then there is s∈ / p such that sx = 0. In turn, this implies that s ∈ (0 : x) ⊆ p, which is a contradiction. Therefore, x = 0.

4.11

Let p be a minimal prime ideal of A and let q be p-primary. Since p is minimal, pp is the unique prime ideal ideal of Ap. Therefore, (0)p is pp-primary by Proposition 4.2, and its contraction, Sp(0), is p-primary by Proposition 4.8. Furthermore, by Exercise 1.18 and Proposition 4.8, (0) ⊆ q implies that Sp(0) ⊆ q.

Let a be the intersection of all Sp(0) for which p is a minimal prime ideal of A. Then a ⊆ Sp(0) ⊆ p for each T p minimal prime ideal. As a consequence, a ⊆ p minimal p = (0). 4 PRIMARY DECOMPOSITION 49

Suppose (0) is a decomposable ideal of A. Then A has only finitely many minimal prime ideals p1,..., pn by Corollary 4.11. Tn If a = (0), then i=1 Spi (0) is a primary decomposition of (0). Note that the prime ideals p1,..., pn are distinct. T T If Spj (0) ⊇ i6=j Spi (0) for some j, then take radicals to obtain pj ⊇ i6=j pi. By Proposition 1.11 (ii), pj ⊇ pi for some i. Since pj is a minimal prime ideal, we have equality. This contradicts the distinctness of the minimal Tn prime ideals p1,..., pn. Therefore, i=1 Spi (0) is an irredundant primary decomposition of (0). By Theorem 4.5, p1,..., pn are the associated prime ideals of (0). So every associated prime ideal of (0) is isolated. T Tn On the other hand, if every associated prime ideal of (0) is isolated, then (0) = p∈D(A) Sp(0) = i=1 Spi (0) = a by Problems 4.9 and 4.10 (iv).

4.12

Let S be a multiplicatively closed subset of the ring A and for each ideal a of A define the saturation of a, S(a), as the contraction of S−1a to A. (i) By Proposition 3.11 (v) and Exercise 1.18,

c c c c S(a) ∩ S(b) = S−1a
∩ S−1b
= S−1a ∩ S−1b
= S−1 (a ∩ b)
= S(a ∩ b).

(ii) By Proposition 3.11 (iv) and Exercise 1.18,

√ √ c √ c q S a
= S−1 a
= S−1a = (S−1a)c = pS(a).

(iii) By Proposition 1.17, S(a) = (1) if and only if S−1a = S−1A. By Proposition 3.11 (ii), this is equivalent to S ∩ a 6= ∅. (iv) Let f : A → S−1A be the natural map and note that x ∈ S(a) if and only if f(x) ∈ S−1a. This is equivalent to the existence of s ∈ S such that xs ∈ a. Therefore,

x ∈ S1(S2(a)) if and only if xs1 ∈ S2(a) for some s1 ∈ S1

if and only if xs1s2 ∈ a for some s1 ∈ S1, s2 ∈ S2

if and only if x ∈ (S1S2)(a).

Tn Tm Suppose a = i=1 qi is an irredundant primary decomposition. By Proposition 4.9, S(a) = j=1 qij is a irredundant primary decomposition for some 1 ≤ i1, . . . , im ≤ n. Since there are only finitely many subsets of {1, . . . , n}, there are only finitely many ideals S(a).

4.13

(n) n Let p be a prime ideal in a ring A, and define the nth symbolic power of p as p = Sp(p ), where Sp = A \ p. q √ −1 n −1 n −1 −1 n −1 (i) By Proposition 3.11 (v), Sp p = Sp p = Sp p. Thus Sp p is an Sp p-primary ideal by Proposition (n) −1 n 4.2. Since p is the contraction of Sp p , it is a p-primary ideal by Proposition 4.8 (ii). n Tm (ii) Suppose that p = i=1 qi is an irredundant primary decomposition and J is the set of indices j for which (n) n T p (n) √ qj ⊆ p. By Proposition 4.9, p = Sp(p ) = j∈J qj. By (i), p = p ⊆ qj for each j ∈ J. Therefore, qj is p-primary for each j ∈ J. Since the decomposition is irredundant by hypothesis, there is only one index j in J, (n) (n) n and qj = p . Thus p is the p-primary component of p . (m) (n) Tk (iii) Suppose that p p = i=1 qi is an irredundant primary decomposition and J is the set of indices j for −1 (m) (n) −1 m+n (m) (n)
(m+n) which qj ⊆ p. Since Sp p p = Sp p , Proposition 4.8 (ii) implies that Sp p p = p . On the (m) (n)
T p (m+n) √ other hand, Proposition 4.9 tells us that Sp p p = j∈J qj. By (i), p = p ⊆ qj for each j ∈ J. 4 PRIMARY DECOMPOSITION 50

Therefore, qj is p-primary for each j ∈ J. Since the decomposition is irredundant by hypothesis, there is only one (m+n) (m+n) (m) (n) index j in J, and qj = p . Therefore, p is the p-primary component of p p . (iv) If p(n) = pn, then pn is p-primary because p(n) is by (i). If pn is p-primary, then pn = p(n) by (ii).

4.14

Let a be a decomposable ideal in a ring A and let p be a maximal element of the set of ideals of the form (a : x) where x ∈ A \ a. If yz ∈ p and y∈ / p, then z ∈ (a : xy) and p ⊆ (a : xy). Since p is maximal, z ∈ (a : xy) = p. So p is a prime ideal and p = p(a : x) is an associated prime ideal of a by Theorem 4.5.

4.15

Let a be a decomposable ideal in a ring A, let Σ be an isolated set of prime ideals associated to a, let qΣ be the intersection of the corresponding primary components. Suppose that there is f ∈ A such that f ∈ p if and only p ∈/ Σ, for each associated prime ideal p of A, and let Sf be the multiplicative monoid generated by f. Tm By Proposition 4.9, Sf (a) = i=1 qi = qΣ, where {p1,..., pm} = Σ. By Proposition 3.11 (ii), Sf (a) = S j √ k Tn j≥0(a : f ). Since f ∈ pi = qi for i = m + 1, . . . , n, there is some k such that f ∈ i=m+1 qi. Therefore, k S j Sf (a) ⊆ (a : f ) ⊆ j≥0(a : f ) = Sf (a).

4.16

Let A be a ring in which every ideal is decomposable, let S be a multiplicatively closed subset, and let b be an −1 −1 Tn ideal of S A. By Proposition 3.11 (i), b = S a for some ideal a of A. Let i=1 qi be an irredundant primary −1 Tm −1 decomposition of a. By Proposition 4.9, S a = i=1 S qi is an irredundant primary decomposition of b.

4.17

(L1) For every proper ideal a and every prime ideal p, there exists x∈ / p such that Sp(a) = (a : x).

Let A be a ring with the property (L1), and suppose a is a proper ideal of A. Let p1 be a minimal prime ideal above a. The ideal q1 = Sp1 (a) is the contraction of the p1-primary ideal Sp1 (0) ⊆ A/a to A, so q1 is p1-primary by Exercise 4.11. By (L1), q1 = (a : x) for some x∈ / p1. Clearly, a ⊆ q1 ∩ (a + (x)). If f = a + bx ∈ q1 ∩ (a + (x)), 2 then fx − ax = bx ∈ a, because f ∈ q1 = (a : x) and a ∈ a. Thus b ∈ Sp1 (a) = q1 = (a : x). So f ∈ a and we have a = q1 ∩ (a + (x)). Let a1 be an ideal maximal among those for which a = q1 ∩ a1 and x ∈ a1. Now we have a ( a1 * p1.

Given a = q1 ∩ · · · ∩ qn ∩ an where aj−1 ( aj * pj for each j = 1, . . . , n, if an is a proper ideal, then we may repeat the process of the previous paragraph to extend the strictly increasing sequence of ideals a ( a1 ( ... ( an.

4.18

(L2) For every ideal a and every descending chain S1 ⊇ S2 ⊇ S3 ⊇ ... of multiplicatively closed subsets, there exists a positive integer N such that Sn(a) = SN (a) for every n ≥ N. Tn (i)⇒(ii) Let a be an ideal of A, let p be a prime ideal, and let a = i=1 qi be an irredundant primary decomposition of a with associated prime ideals p1,..., pn. Let Σp be the set of associated prime ideals of a contained in p. Clearly, Σp is an isolated set of prime ideals. Without loss of generality, {p1,..., pm} = Σp and p + pm+1 ··· pn. Thus there is f ∈ pm+1 ··· pn \ p. By Exercise 4.15 and Proposition 4.9, k Sp(a) = qΣp = (a : f ) for some positive integer k. Therefore, A satisfies (L1). 4 PRIMARY DECOMPOSITION 51

Let S1 ⊇ S2 ⊇ ... be a descending chain of multiplicatively closed subsets of A. Without loss of generality, the first mj associated prime ideals of a meet Sj, for every j ≥ 1. Since m1, m2, m3,... is a decreasing sequence of positive integers, it has a least element m . By Proposition 4.9, if j ≥ N, then S (a) = Tn q = Tn q = N j i=mj i i=mN i SN (a). Therefore, A satisfies (L2).

(ii)⇒(i) Let a be an ideal of A. Using the property (L1), obtain the prime ideals pi and ideals ai, where we are using the notation of Exercise 4.17. Now, define Sn = Sp1 ∩ · · · ∩ Spn for each positive integer n. By (L2), there exists a positive integer N such that n ≥ N implies that Sn(a) = SN (a).

4.19

Let p be a prime ideal in a ring A, and let q be p-primary. Since (0) ⊆ q, Sp(0) ⊆ Sp(q) = q by Proposition 4.8.

Suppose that for every prime ideal p of A, Sp(0) is the intersection of all p-primary ideals of A, and let p1,..., pn be distinct non-minimal prime ideals of A. If n = 1, then a = p1 has p1 as its only associated prime ideal. Suppose that pn is maximal among the pi, and that there is a decomposable ideal b with irredundant primary decomposition Tn−1 i=1 qi, where each qi is pi-primary. Let p be a minimal prime ideal of A contained in pn. If b ⊆ Spn (0), then b ⊆ Spn (0) ⊆ Sp(0) ⊆ p by Exercise 4.10 (iii). By taking radicals, this implies that pi ⊆ p for some i. This contradicts our hypothesis that the pi’s are non-minimal prime ideals. By our assumption on A, this implies that there is a pn-primary ideal, qn, such that b * qn. Since p1,..., pn were assumed to be distinct, we have that a = p1 ∩ · · · ∩ pn is an irredundant primary decomposition.

4.20

Let N be a submodule of an A-module, M. Define the M-radical of N as √ M N = {x ∈ A : xqM ⊆ N for some q > 0}. √ It follows directly from the definition that, M N = p(N : M) = pAnn (M/N). Analogues of Exercise 1.13: Let N,N 0 be submodules of M and let a be an ideal of A. It is unclear what analogues the author is requesting. The following are what come to my mind. √ (i) M N ⊇ (N : M) (The analogy is that a is an annihilator of A/a) p √ √ (ii) M N ⊇ M N (iii) I will break this into two formulas. The first deals with intersections of submodules. The second deals with the product of an ideal and a submodule.

√ M N ∩ N 0 = p(N ∩ N 0 : M) = p(N : M) ∩ (N 0 : M) = p(N : M) ∩ p(N 0 : M) √ √ = M N ∩ M N 0

√ √ M M k k Let x ∈ aM ∩ N, then there are positive integers k1, k2 such that x 1 ∈ (aM : M) and x 2 ∈ (N : M). If k1 Pk m ∈ M, then x m = i=1 aimi for some a1, . . . , ak ∈ a and m1, . . . , mk ∈ M. Thus,

k k k1+k2 k2 X X k2 x m = x aimi = ai(x mi) ∈ aN. i=1 i=1 √ √ √ Therefore, M aM ∩ M N ⊆ M aN. 4 PRIMARY DECOMPOSITION 52

On the other hand, √ √ √ M aN = p(aN : M) ⊆ p(aM ∩ N : M) = M aM ∩ M N. Given the previous paragraph, we have established √ √ √ M aN = M aM ∩ M N. √ (iv) M N = (1) if and only if 1 ∈ Ann (M/N). √ √ k M M 0 k1 k2 0 (v) Suppose x = y + z ∈ N + N , and get k1, k2 so that y ∈ (N : M) and z ∈ (N : M). Then p √ √ √ xk(k1+k2) ∈ (N + N 0 : M). Therefore, M N + M N 0 ⊆ M N + N 0. (vi) If p is a prime ideal, then

M Mp M Mp Mp ppnM = p (pn−1M) = ppM ∩ pn−1M = pn−1M √ √ √ for every positive integer n. By induction, p ⊆ M pM = M pnM. Now suppose that M/pM and M pM are finitely √ r M
  generated A-modules. By Propositions 3.11 (v) and 3.14, we have pM p = Ann (M/pM)p = (0) because √ (M/pM) is a κ -vector space. By Exercise 3.1, there is s∈ / p such that s · M pM = (0) ⊆ p. Since p is prime p √p √ ideal, we have M pnM = M pM ⊆ p. √ In particular, if A is Noetherian, and M/pM is a finitely generated A-module, then M pnM = p.

4.21

A proper submodule Q of M is primary if and only if every zero-divisor of M/Q is nilpotent. Suppose that Q is a primary submodule of M and that xy ∈ (Q : M), but y∈ / (Q : M). Let m ∈ M such that ym∈ / Q. Then x is a zero-divisor of M/Q. Since Q is primary, xn ∈ (Q : M) for some positive integer n. Therefore, (Q : M) is a primary ideal of A. Tn Analogue of Lemma 4.3: Let Q1,...,Qn be p-primary submodules of M and let Q = i=1 Qi. By Lemma 4.3, Tn (Q : M) = i=1(Qi : M) is a p-primary ideal. Therefore, Q is a p-primary submodule of M. Analogue of Lemma 4.4: Let Q be a p-primary submodule of M. (i) If m ∈ Q, then (Q : m) = (1) because Q is an A-module. (ii) If m∈ / Q and xy ∈ (Q : m) with y∈ / (Q : m), then xym ∈ Q and ym∈ / Q. Since Q is primary, there is a positive integer n such that xn ∈ (Q : M) ⊆ (Q : m). Letting y = 1, we see that (Q : M) ⊆ (Q : m) ⊆ p(Q : M) = p. Therefore, (Q : m) is p-primary. (iii) If x∈ / p and m ∈ (Q : x), then xm ∈ Q, but x is not nilpotent in M/Q. Therefore, m ∈ Q, and we see that (Q : x) = Q.

4.22

Tn Analogue of Theorem 4.5: Let N = i=1 Qi be an irredundant primary decomposition. For every m ∈ M, n p \ p \ (N : m) = (Qi : m) = pj.

i=1 m/∈Qj p If (N : m) is prime ideal, then it equals pj for some Qj not containing m, by Proposition 1.11 (ii). On the T other hand, since the decomposition is irredundant, we may choose mj ∈ i6=j Qi \ Qj for each index j. Thus p (N : mj) = pj for each j. By the Lattice Isomorphism theorem, the natural map M → M/N yields an order preserving bijection between the submodules of M containing N and the submodules of M/N. In particular, sums, intersections, inclusions, and quotients are preserved. Thus the associated prime ideals of N in M are the associated prime ideals of 0 in M/N. 4 PRIMARY DECOMPOSITION 53

4.23

By the Lattice Isomorphism Theorem, we may assume that N is the zero submodule. Tn Analogue of Proposition 4.6: Suppose 0 is a decomposable submodule of M, let 0 = i=1 Qi be an irredundant Tn primary decomposition, and let p be a prime ideal of A. If p ⊇ Ann(M) = i=1(Qi : M), then taking radicals Tn shows that p ⊇ i=1 pi. By Proposition 1.11 (ii), p contains an associated prime ideal of 0. Therefore, every minimal prime ideal of A/ Ann(M) is an isolated prime ideal of 0 ≤ M. Analogue of Proposition 4.7: Suppose 0 is a decomposable submodule of M, and let D be the set of zero- divisors of M. Then x ∈ D if and only if there exists a non-zero m ∈ M such that xm = 0. In particular, S p p D = 06=m∈M (0 : m). If p is an associated prime ideal of 0, then Exercise 4.22 implies that p = (0 : m) for some m ∈ M. Therefore, p ⊆ D. On the other hand, for every non-zero m ∈ M, there is some associated prime ideal, p, of 0 such that p \ q \ (0 : m) = (Qj : m) = pj ⊂ p.

m/∈Qj m/∈Qj Sn Thus D ⊆ i=1 pi, and we see that the set of zero-divisors of M is the union of the associated prime ideals of 0 in M. Analogue of Proposition 4.8: Let S be a multiplicatively closed subset of A and let Q be a p-primary submodule of M. (i) If s ∈ S ∩ p, then sn ∈ S ∩ (Q : M) for some positive integer n. Therefore, S−1M/S−1Q = S−1(M/Q) = 0. (ii) Analogue of Proposition 4.9: Analogue of Theorem 4.10: Analogue of Corollary 4.11: 5 INTEGRAL DEPENDENCE AND VALUATIONS 54

5 Integral Dependence and Valuations

5.1

Let f : A → B be an integral map of rings and let b be an ideal of B. By Exercise 1.21 (iii), f ∗(V (b)) ⊆ V (f −1(b)). Let p ∈ V (f −1(b)). By Proposition 5.6 (i), B/b is integral over A/f −1(b). By Proposition 5.10, there is q ∈ Spec(B/b) = V (b) such that f ∗(q) = p. Therefore, f ∗(V (b)) = V (f −1(b)), and f ∗ is a closed map.

5.2

Let B/A be an integral extension of rings, let Ω be an algebraically closed field, and let f : A → Ω be a ring map. Let m = ker(f). By Proposition 5.10, there is a prime ideal n of B over m. Since Ω is a field, m is maximal. By Corollary 5.8, n is maximal. By Proposition 5.6 (i), κn/κm is an integral, hence algebraic, extension of fields. Therefore, f : κm → Ω lifts to some g : κn → Ω. Precompose g with the natural map B → κn to get g : B → Ω. Our construction yields the following diagram:

f A κm Ω id g B κn Ω, where the vertical maps are the extension maps A → B, κm → κn, and the identity map. The horizontal composites are f and g, respectively, and the right square commutes by construction. On the other hand, the left square commutes since the central map is the quotient of the map A → B. Therefore the entire diagram commutes and we see that g is an extension of f.

5.3

Let f : B → B0 be a map of A-algebras and let C be an A-algebra. Suppose f is integral. If x ∈ B0, then n−1 i B[x] = ⊕i=0 Bx as a B-module, for some non-negative integer n. Therefore,

(B ⊗A C)[x ⊗ 1] = B[x] ⊗A C n−1 i = ⊕i=0 (Bx ⊗A C) n−1 i = ⊕i=0 (B ⊗A C)(x ⊗ 1)

0 is a finitely generated B ⊗A C-module. Thus x ⊗ 1 is integral over B ⊗A C by Proposition 5.1. Since B ⊗A C is the B ⊗A C-span of {x ⊗ 1}x∈B0 , f ⊗ idC is an integral map of A-algebras.

5.4

Let A = k[y],B = k[x], where k is a field and x is a root of t2 − (y + 1). Note that t2 − (y + 1) ∈ A[t] is irreducible by Eisenstein’s criterion. Let n = (x − 1) and m = (y) be maximals ideal so B and A respectively, and 1 note that n ∩ A = m. Additionally, let z = x+1 ∈ Bn and let K = A(0). 2 z 2 2 1 2 Since (x + 1)z = 1, we have yz + 2z − 1 = 0. Therefore, mK (t) = t + y t − y . Let g(t) = yt + 2t − 1, and n suppose z is integral over Am. Then f(z) = 0 for some f(t) = c0t + ... + cn ∈ A[t] with c0 ∈/ m. Without loss z of generality, f is primitive of minimum positive degree. Since f ∈ K[t] and f(z) = 0, f(t) = mK (t)q(t) for some q(t) ∈ K[t]. Thus yf(t) = g(t)q(t). By Gauss’ Lemma, q(t) ∈ A[t] and y divides q(t). Therefore, f(t) = g(t)h(t) for some h(t) ∈ A[t]. However, this implies that y divides c0, contrary to hypothesis. 5 INTEGRAL DEPENDENCE AND VALUATIONS 55

5.5

Let B/A be an integral extension of rings. −1 −n n−1 −1 n−1 (i) Suppose x ∈ A and x ∈ B. Then x +a1x +...+an = 0 implies x = −(a1 +a2x+...+anx ) ∈ A. (ii) By Corollary 5.8, every maximal ideal of B contracts to a maximal ideal of A. By Theorem 5.10, every maximal ideal of A is the contraction of a prime ideal of B, which is maximal by Corollary 5.8. Thus the contraction of the Jacobson radical of B is the Jacobson radical of A.

5.6

Qn Let B1,...,Bn be integral A-algebras and let B = i=1 Bi. Then B is clearly an A-algebra by the diagonal map A → B. Let x = (x1, . . . , xn) ∈ B and let mi(t) be the minimal polynomial of xi over A for each i. Define Qn p(t) = i=1 mi(t) ∈ A[t]. By construction, p(t) is monic and

p(x) = (p(x1), . . . , p(xn)) = 0.

Thus x is integral over A.

5.7

Let A be a subring of B such that B \ A is closed under multiplication, and let x ∈ B be integral over A with n n−1 minimal polynomial t + a1t + ... + an. If x∈ / A, then

n−1 n−2 x(x + a1x + ... + an−1) = −an ∈ A

n−1 n−2 0 implies that x + a1x + ... + an−1 = a ∈ A. Thus

n−1 n−2 0 x + a1x + ... + an−1 − a = 0.

Since n was assumed to be minimal, we must have n = 1 and the minimal polynomial of x is t + a1. So x ∈ A, contrary to hypothesis. Therefore, x ∈ A.

5.8

Let A be a subring of B, and let C be the integral closure of A in B. (i) Suppose B is a domain, let f, g ∈ B[x] be monic such that fg ∈ C[x], and let K be a splitting field of f, g Qn Qm over B(0). Then we may factor as f(x) = i=1(x − ξi) and g(x) = j=1(x − ηj). Since fg ∈ C[x], the ξ’s and η’s are integral over C. Therefore coefficients of f, g ∈ B[x] are polynomials in elements of K, integral over C. Since C is integrally closed in B, we have f, g ∈ C[x]. (ii) Let h ∈ B[x] be monic, let B0 = B[x]/hB[x], let i : B → B[x] be the inclusion, and let q : B[x] → B0 be the quotient map. We claim that q ◦ i is injective. Note that a ∈ hB[x] if and only if there exists some b ∈ B[x] such that a = bh. Since h is monic, a has positive degree. Since the domain of q ◦ i contains no elements of B[x] of positive degree, q ◦ i is injective. Therefore, B0 is a finite B-algebra with an injection B → B0. Furthermore, 0 0 h(t) = (t − x)h1(t) over B , for some h1(t) ∈ B [t] (it is here that we are using the fact that a non-unique form of Euclidean division lets us divide by a monic polynomial over an arbitrary ring). Let f, g ∈ B[x] be monic such that fg ∈ C[x]. Repeat the preceding construction with f and g to obtain a finite B-algebra B0 such that f and g split over B0. The same argument as in (i) concerning polynomials in the roots of f and g establishes that f, g ∈ C[x]. 5 INTEGRAL DEPENDENCE AND VALUATIONS 56

5.9

Let A be a subring of B, and let C be the integral closure of A in B. Let f ∈ B[x] be integral over A[x] n n−1 with minimal polynomial p(t) = t + a1t + ... + an, where a1, . . . , an ∈ A[x]. Let r be an integer greater than r deg a1,..., deg an, deg f, and n. If f0 = f − x , then

r n n−1 r 0 = p(f) = p(f0 + x ) = f0 + h1f0 + ... + hn−1f0 + p(x ).

Therefore, n−1 n−2 f0, f0 + h1f0 + ... + hn−1 ∈ B[x] and n−1 n−2 r f0(f0 + h1f0 + ... + hn−1) = −p(x ) ∈ C[x].

By Exercise 5.8, f0 ∈ C[x].

5.10

Let f ∗ : Spec B → Spec A be the map associated to a ring homomorphism f : A → B. ∗ (i) (a)⇒(b) Suppose that f is closed, p1 ⊆ p2 are prime ideals of A, and q1 is a prime ideal of B over p1. Then

∗ ∗ f (V (q1)) = V (f (q1)) = V (p1)

∗ and p2 ∈ V (p1). Thus there is q2 ∈ V (q1) such that f (q2) = p2. ∗ (b)⇔(c) The map f : Spec(B/q) → Spec(A/p) is surjective if and only if every prime ideal of A containing p has a lift to a prime ideal of B over q. ∗ (ii) (a)⇒(b) Suppose that f is open, p2 ⊆ p1 are prime ideals of A, and q1 is a prime ideal of B over p1. Let ∗ ∗ U be an open neighborhood of q1 in Spec B. Since f (q1) = p1 and f (U) is open, there is s ∈ A \ p1 such that ∗ ∗ Xs is an open neighborhood of p1 in f (U). Since p2 ⊆ p1, we also have p2 ∈ Xs ⊆ f (U). By Exercise 3.26, ∗ \ ∗ \ ∗ f (Spec Bq1 ) = f (Spec Bt) = f (Yt).

t/∈q1 t/∈q1 ∗ Since p2 ∈ f (Yt) for all t∈ / q1, there is some q2 ⊆ q1 over p2. ∗ (b)⇔(c) The map fb : Spec Bq → Spec Ap is surjective if and only if every prime ideal of A contained in p has a lift to a prime ideal of B contained in q.

5.11

∗ Let f : A → B be a flat ring homomorphism. By Exercise 3.18, f : Spec Bq → Spec Ap is surjective whenever q is a prime ideal of B and p = qc. By Exercise 5.10 (ii), f has the going-down property.

5.12

Let G be a finite group of automorphisms of a ring A, let AG be the ring of G-invariants, and let x ∈ A. Q Define p(t) = σ∈G(t − σ(x)) ∈ A[t]. Since the coefficients of p(t) are symmetric polynomials in the G-orbit of x, p(t) ∈ AG[t]. Furthermore, p(x) = 0. So x is integral over AG. Let S be a G-stable multiplicatively closed subset of A and let SG = S ∩ AG. For any σ ∈ G, the universal σ −1 −1 −1 a
σ(a) property of localization lifts the map A −→ A → S A uniquely to a map σ : S A → S A such that σ s = σ(s) . This is the desired group action. 5 INTEGRAL DEPENDENCE AND VALUATIONS 57

Since AG ⊆ A and SG ⊆ S, the universal property of localization yields a unique map f :(SG)−1AG → S−1A G a
a a G −1 G extending the inclusion A → A such that f s = s . Given s ∈ (S ) A and σ ∈ G,  a a σ(a) a a σ f = σ = = = f . s s σ(s) s s

So the image of f is in (S−1A)G. a −1 G 0 Q G Let s ∈ (S A) , and define s = σ∈G σ(s) ∈ S .

5.13

G Let p be a prime ideal of A and let P be the set of prime ideals of A over p. Suppose that p1, p2 ∈ P and x ∈ p1. Then we have Y G σ(x) ∈ p1 ∩ A = p ⊆ p2. σ∈G S Since p2 is prime ideal, there is σ ∈ G such that x ∈ σ(p2). Therefore, p1 ⊆ σ∈G σ(p2). By the prime ideal G G G Avoidance Lemma, p1 ⊆ p2 for some σ. Since A is integral over A and p1 ∩ A = p2 ∩ A , we know that p1 = σ(p2).

5.14

Let A be an integrally closed domain with K = A(0), L/K a finite, separable, normal field extension, and let B be the integral closure of A in L, and let G = Gal(L/K). Suppose b ∈ B has minimal polynomial n n−1 p(x) = x + a1x + ... + an. Given σ ∈ G, 0 = σ(0) = σ(p(b)) = p(σ(b)), because G fixes A. Since B is the integral closure of A in L, B is G-stable. Furthermore, we have shown that σ permutes the roots p(x). Therefore, there is b0 ∈ B such that σ(b0) = b. If b ∈ BG, then b ∈ K because it is fixed, and b is integral over A because it is in B. Since A is integrally closed, b ∈ A. Therefore, BG ⊆ A.

5.15

Let A be an integrally closed domain with K = A(0), L/K a finite extension of fields, and let B be the integral closure of A in L.

5.16

Let k be an infinite field and let A be a finitely generated k-algebra. Then A = k[x1, . . . , xn]/a for some ideal a of k[x1, . . . , xn]. If a = (0) or n = 0, then the lemma holds trivially. Suppose that a 6= (0), n > 0, and that the lemma holds for any number of generators less than n. Let f ∈ a be non-zero. Without loss of generality, xn is one of the variables appearing in f. Let F ∈ k[x1, . . . , xn] be the homogeneous component of f of maximum degree, d. Since k is infinite, there are scalars λ1, . . . , λn−1 such that F (λ1, . . . , λn−1, 1) 6= 0. Define the variables, 0 xi = xi − λixn for 1 ≤ i < n. For an arbitrary monomial, we have

n n−1 n−1 ! Y νi νn Y 0 νi Y νi d d−1 xn = xn (xi + λixn) = λi xn + a1xn + ... + ad, i=1 i=1 i=1

0 0 where a1, . . . , ad ∈ k[x1, . . . , xn−1]. Therefore, by change of coordinates, we may write

d d−1 f = F (λ1, . . . , λn−1, 1)xn + b1xn + ... + bd, 5 INTEGRAL DEPENDENCE AND VALUATIONS 58

0 0 0 0 0 0 with b1, . . . , bd ∈ k[x1, . . . , xn−1]. Let a be the kernel of the natural map k[x1, . . . , xn−1] → A, and let A = 0 0 0 0 0 k[x1, . . . , xn−1]/a . Note that the map A → A is injective. By induction, there y1, . . . , yr ∈ A that are alge- 0 braically independent over k, and k[y1, . . . , yr] → A is finite, injective, and y1, . . . , yr are k-linear combinations of 0 0 x1, . . . , xn−1. By Proposition 2.16, k[y1, . . . , yr] → A is finite, injective, and y1, . . . , yr are k-linear combinations of x1, . . . , xn. (Practically, the process of proof can be repeated until a = 0. At that point we have obtained the desired algebraically independent set of generators.)

5.17

Let k be an algebraically closed field and let X ⊆ kn be an affine algebraic set with ideal I(X) 6= (1). Let P (X) = k[t1, . . . , tn]/I(X) be the coordinate ring of X. Since I(X) 6= (1), we know that P (X) is not the zero ring. By Exercise 1.27, we know that any maximal ideal m of P (X) has the form (t1 − a1, . . . , tn − an) for some a1, . . . , an ∈ k. Let f ∈ I(X). Since m is a maximal ideal of P (X), we know that f ∈ m. Therefore, f(a1, . . . , an) = 0. Since this is true for every f ∈ I(X), the point (a1, . . . , an) is in X.

5.18

Let k be a field and let B be a finitely generated k-algebra that is also a field. Let B be generated by x1, . . . , xn over k. If n = 1, then 1 n n−1 = x1 + a1x1 + ... + an x1 for some a1, . . . , an ∈ k shows that x1 is integral over k. Therefore, B/k is a finite field extension. Therefore, we may assume that n > 1.

Let A = k[x1] and let K = A(0). By the inductive hypothesis, B/K is a finite field extension. Therefore, x2, . . . , xn are integral over K, hence algebraic over A. Let f be the product of the lead coefficients of the minimal polynomials of x2, . . . , xn over A. Then x2, . . . , xn are integral over Af . By the generation assumption on B, this implies that B, hence K is integral over Af . If x1 is transcendental over k, then A is a unique factorization domain, hence integrally closed. By Proposition 5.12, Af is integrally closed. Since K/Af is integral, K = Af .

5.19 Undone

5.20

Let B be a finitely generated faithful A-algebra that is a domain, and let S = A \ (0). Then S−1B is a finitely generated K-algebra, where K = S−1A is the field of fractions of A. By Noether’s Normalization Lemma, −1 −1 there are x1, . . . , xn ∈ S B, algebraically independent over K, such that S B is integral over K[x1, . . . , xn]. d ai1 d −1 aidi Let z1, . . . , zm generate B as an A-algebra, and let mi(t) = t i + t i + ... + be the minimal polynomial si1 sidi of zi over K[x1, . . . , xn] for each 1 ≤ i ≤ m. For each 1 ≤ k ≤ n, let sk ∈ S be such that skxk ∈ B. Define Q  Q s = i,j sij ( k sk), define yk = sxk ∈ B, and let mb i(t) ∈ A[x1, . . . , xn][t] be monic polynomials such that mb i(st) = smi(t) for each i. Then y1, . . . , yn are algebraically independent over K, hence over A. 0 Let B = A[y1, . . . , yn]. For each i, polynomial mb i(t) gives an equation of integrality for szi over A[x1, . . . , xn]. xi 0 0 Since yi = s , zi ∈ Bs is integral over Bs. Since Bs is generated by z1, . . . , zn over Bs, the ring extension is integral.

5.21

Let B be a finitely generated faithful A-algebra that is a domain. By Exercise 5.20, there exist y1, . . . , yn ∈ B that are algebraically independent over A and a non-zero s ∈ A such that Bs is integral over A[y1, . . . , yn]s. Let Ω be an algebraically closed field and let f : A → Ω be a ring homomorphism for which f(s) 6= 0. Since the y1, . . . , yn are algebraically independent over A, f has an extension to g : A[y1, . . . , yn] → Ω given by mapping 5 INTEGRAL DEPENDENCE AND VALUATIONS 59

each yi to 0. Furthermore, f(s) 6= 0 implies that g(s) is a unit. By the universal property of localization, g has a lift to a map gb : A[y1, . . . , yn]s → Ω. Since Bs is finitely generated and integral over A[y1, . . . , yn]s, it is a finite A[y1, . . . , yn]s-module. Let z1, . . . , zm be a list of generators. We can successively extend gb by mapping each zi to a root of its minimal polynomial in Ω. Since there are only finitely many zi’s, this process terminates in a map g : B → Ω.

5.22 Undone

Let B be a finitely generated faithful A-algebra that is a domain.

5.23

Let A be a ring. (i)⇒(ii) By the First Isomorphism Theorem, homomorphic images of A may be identified with quotients. Let a be an ideal of A. The nilradical of A/a is the intersection of all prime ideals of A/a. By the Lattice Isomorphism Theorem we may identify ideals of A containing a with ideals of A/a in such a way that containment, primality, maximality, and intersections are preserved. If p is a prime ideal of A containing a, then there is a T family of maximal ideals {mi}i∈Ip such that p = i∈Ip mi. By the lattice isomorphism theorem, this intersection is preserved on passing to the quotient. Therefore the nilradical of A/a is an intersection of maximal ideals. Since the Jacobson radical is the intersection of all maximal ideals, the Jacobson radical is contained in the nilradical of A/a. The reverse containment holds in general, so the Jacobson radical and nilradical of A/a are the same. (ii)⇒(iii) Let p be a non-maximal prime ideal of A. We apply the Lattice Isomorphism Theorem. The nilradical of A/p corresponds to the ideal p. The Jacobson radical of A/p corresponds to the intersection of all maximal ideals containing p. By hypothesis, these are equal. If q is a prime ideal of A strictly containing p, then q is contained in a maximal ideal. Therefore, the intersection of all prime ideals strictly containing p is an ideal containing the nilradical of A/p and contained in the Jacobson radical of A/p. Since these are equal, the intersection of all prime ideals strictly containing p is p. (iii)⇒(i) Suppose A has a prime ideal ideal which is not an intersection of maximal ideals. By passing to the quotient we may assume that A is a domain whose Jacobson radical is non-zero. Let f be a non-zero element of the Jacobson radical of A. Since A is a domain, Af 6= 0. Let p be the contraction of a maximal ideal of Af to A. Then p is a prime ideal of A not containing f and is maximal with respect to this property. Therefore, the intersection of all prime ideals strictly containing p contains f, and is not equal to p.

5.24

Let A be a Jacobson ring and let B be an A-algebra. √ (i) Suppose that B is integral over A. Let b be an ideal of B. We want√ to show that b is the intersection b b b a b of maximal ideals of B containing . Without loss of generality, let =√ , and let be the contraction of to A. Since taking the radical commutes with taking the contraction, a = a as well. By Proposition 5.6 (i), B/b is integral over A/a. Let J be the Jacobson radical of B/b. By Exercise 5.5 (ii), J c is the Jacobson radical of A/a. Since A is a Jacobson ring, J c = (0). n n−1 Suppose there is f ∈ J that is not nilpotent and f has minimal polynomial t + a1t + ... + an over A/a. n n−1 c Since f + a1f + ... + an = 0, an ∈ J, hence an ∈ J = (0). Since f is not nilpotent, f∈ / q for some prime ideal q of B/b, then n−1 n−2 f · (f + a1f + ... + an−1) = 0 ∈ q n−1 n−2 implies that f +a1f +...+an−1 ∈ q. Let p be the contraction of q to A/a. Passing to the quotient gives B/q integral over A/p, f is in the Jacobson radical of B/q with minimal polynomial of degree n − 1, the contraction of the J to A/p is the Jacobson radical of A/p and is (0), and f is again not nilpotent. As long as n > 1 we may repeat this process until we arrive at the case that f has minimal polynomial t + a1. Since f is in the Jacobson 5 INTEGRAL DEPENDENCE AND VALUATIONS 60

radical of B/q and f = −a1 ∈ A/p, f is in the Jacobson radical of A/p. Therefore, f ∈ p ⊆ q. This contradicts the hypothesis on q. Therefore, the Jacobson radical of B/b is equal to the nilradical, and B is a Jacobson ring. (ii) Suppose that B is finitely generated over A. Let q be a non-maximal prime ideal of B and let p be the contraction of q to A. By passing to the quotient we may assume that B is a finitely generated faithful A-algebra which is also a domain and that the Jacobson radical of A is (0). By Exercise 5.22, we know that the Jacobson radical of B is (0). Since every non-trivial prime ideal of B is contained in a maximal ideal, the intersection of all non-trivial prime ideals is contained in the Jacobson radical of B, and is therefore (0). This establishes criterion (iii) of Exercise 5.23.

Since Z is a Jacobson ring and every field is a Jacobson ring, finitely generated rings and finitely generated k-algebras are Jacobson rings.

5.25

Let A be a ring. (i)⇒(ii) Let A be a Jacobson ring and let B be a field which is a finitely generated A-algebra. Since B is a field, the kernel of A → B is a prime ideal ideal. Passing to the quotient, we may assume that A is a domain whose Jacobson radical is (0), and B is a faithful A-algebra. Let s ∈ A be as in Exercise 5.21. Since s 6= 0, there is a maximal ideal m of A not containing s. Let k = A/m and let Ω be the algebraic closure of k. Compose the natural maps A → k and k → Ω to obtain a map f : A → Ω such that f(s) 6= 0. By Exercise 5.21, f has an extension to some g : B → Ω. Since B is a field, g is injective. Therefore, B is an algebraic extension of k. Since B is finitely n n−1 generated over k, B/k is a finite field extension. Given x ∈ B, let m(t) = t + a1t + ... + an ∈ (A/m)[t] be the minimal polynomial of x over k, where a1, . . . , an ∈ A. Then m(t) lifts to some monic m(t) ∈ A[x] such that m(x) = a0 ∈ m. Thus x satisfies the polynomial m(t)−a0 and is integral over A. Therefore, B is a finite A-algebra. (ii)⇒(i) Let p be a non-maximal prime ideal of A and let B = A/p. Let f be a non-zero element of B. Then Bf is a finitely generated A-algebra. Suppose that Bf is a field. By hypothesis, it is a finite A-algebra. Thus it is a finite B-algebra, hence integral over B. By Proposition 5.7, B is a field, in contradiction to our hypothesis on p. Therefore, Bf is not a field. Contract a maximal ideal of Bf to B to obtain a prime ideal q not containing f and maximal with this property. Therefore, for every f ∈ \p, there is a prime ideal strictly containing p but not f. So p is the intersection of all prime ideals strictly containing it.

5.26

Let X be a topological space. Suppose U open,Cclosed ⊆ X. We claim that U ∩ C is open in its closure. Since U and C are closed, and U ∩ C ⊆ U ∩ C, we have U ∩ C ⊆ U ∩ C. Thus

c c U ∩ C ∩ C ⊆ U ∩ C ∩ C = ∅. Therefore,

U ∩ C \ (U ∩ C) = U ∩ C ∩ (U c ∪ Cc) = (U ∩ C ∩ U c) ∪ (U ∩ C ∩ Cc) = (U ∩ C ∩ U c), which is closed. Thus U ∩ C is open in U ∩ C. On the other hand, suppose that S ⊆ X is open in S. Then there is U open ⊆ X such that S = U ∩ S, and we see that S is the intersection of an open and a closed set.

Let X0 be a subset of X. closed (1)⇒(2) Suppose that every non-empty locally closed subset of X meets X0, and let E ⊆ X. Let x ∈ E and let U be a neighborhood of x. Since U ∩ E is locally closed, U ∩ E ∩ Xo 6= ∅. Therefore, x is in the closure of E ∩ X0. The reverse containment holds in general. 5 INTEGRAL DEPENDENCE AND VALUATIONS 61

closed (2)⇒(3) Suppose that E ∩ X0 = E for every E ⊆ X, and let U, V be open subsets of X such that c c U ∩ X0 = V ∩ X0. Taking X0-complements yields U ∩ X0 = V ∩ X0. Now take closures in X to obtain

c c c c U = U ∩ X0 = V ∩ X0 = V .

Thus U = V and the correspondence is injective. On the other hand, the correspondence is surjective by the definition of the subspace topology.

(3)⇒(1) Suppose that the correspondence U 7→ U ∩X0 from the topology of X to that of X0 is bijective, and let S be a non-empty locally closed subset of X. By definition, there are U open,Cclosed ⊆ X such that S = U ∩C. Since ∅ 7→ ∅ and the correspondence is bijective, U ∩ X0 6= ∅. Since S is non-empty, X 7→ X0, and the correspondence c is bijective, C ∩ X0 6= X0. Therefore, S ∩ X0 6= ∅.

Let A be a ring, let X = Spec A, and let X0 be the set of maximal ideals of A.

(i)⇒(ii) Suppose that A is a Jacobson ring. Let a be an ideal of A and let V (b) = V (a) ∩ X0. By construction, b a a b a a V ( ) ⊆ V ( ). On the other hand, V ( ) ∩ X0 ⊆ V ( ) is the set of maximal ideals√ containing√ . Therefore, and b are contained in exactly the same maximal ideals. Since A is a Jacobson ring, a, b are each the intersection of the maximal ideals containing them, and are thus equal. In summary, V (a) ∩ X0 = V (a), and V (a) was an arbitrary closed set. So X0 is very dense in X.

(ii)⇒(iii) Suppose that X0 is very dense in X and suppose that {x} is locally closed. By criterion (1), {x}∩X0 is non-empty. Therefore, x is a maximal ideal of A, and {x} is closed. (iii)⇒(i) Suppose that every locally closed subset of X consisting of single point is closed. Let p be a non- maximal prime ideal of A and let f ∈ A \ p. Since Xf ∩ V (p) is locally closed and {p} is not closed, Xf ∩ V (p) contains a point q other than p, and f∈ / q. Since f was arbitrary, p is equal to the intersection of prime ideal ideals strictly containing it, and A is a Jacobson ring.

5.27

Let K be a field and let Σ be the set of all local subrings of K. Since K ∈ Σ, Σ is non-empty. Order Σ by S S domination and let A1 ≤ A2 ≤ ... be a chain. Let A = i≥1 Ai and let m = i≥1 mi, where mi is the maximal ideal of Ai for each i. Then A is a ring and m is an ideal. Suppose f ∈ A \ m. Then f ∈ Ai \ mi for some i. Since −1 Ai is a local ring, f ∈ Ai ⊆ A. Therefore, A is a local ring and m is its maximal ideal. By Zorn’s Lemma, Σ has maximal elements. Let A, m be a maximal element of Σ and let f ∈ K \ A. Then A is a subring of A[f]. If mA[f] is a proper ideal, then it is contained in some maximal ideal n of A[f]. Since we are working inside of an ambient field, A is a subring of the local ring (A[f])n and m = n ∩ A. Since A is maximal, A = (A[f])n, contrary to our assumption on f. Therefore, 1 ∈ mA[f], f −1 ∈ m ⊆ A, and we see that A is a valuation ring of K. Suppose that A is a valuation ring of K. By Proposition 5.18 (i), A, m is a local ring. Suppose that B, n is a local subring of K dominating A, m. If f ∈ B \ A, then f −1 ∈ A is a non-unit. So f −1 ∈ m ⊆ n. Thus, 1 ∈ n, contrary to hypothesis. Therefore, A, m is a maximal element of Σ.

5.28

Let A be a domain with fraction field K. (i)⇒(ii) Suppose that A is a valuation ring of K with ideals a, b, let f ∈ b \ a, and let 0 6= g ∈ a. Since A is f g f f g a valuation ring, g ∈ A or f ∈ A. If g ∈ A, then f = g · g ∈ a, contrary to hypothesis. Therefore, f ∈ A and g g = f · f ∈ b. Thus a ⊆ b. f (ii)⇒(i) Suppose that the ideals of A are totally ordered by containment and let g ∈ K with f, g ∈ A. Then f either (f) ⊆ (g) or (f) ⊇ (g). So f = ag for some a ∈ A or bf = g for some b ∈ A. Therefore, g = a ∈ A or g f = b ∈ A, and A is a valuation ring. 5 INTEGRAL DEPENDENCE AND VALUATIONS 62

If A is a valuation ring with prime ideal p, then the ideals of A are totally ordered by containment. Since the respective lattices of ideals of Ap and A/p are sublattices of the lattice of ideals of A, they are also totally ordered by containment. Therefore, Ap and A/p are valuation rings of their fields of fractions.

5.29

Let A be a valuation ring of a field K and let B be a subring of K containing A. Since A is a valuation ring of K, B is also a valuation ring of K by Proposition 5.18 (ii). By Proposition 5.188 (i), B is a local ring with maximal ideal n. Let p = n ∩ A. If s ∈ A \ p, then s ∈ B \ n. So s is a unit in B. By the universal property of localization, the inclusion A → B lifts uniquely to a map Ap → B. Furthermore, this map is injective because A is a domain. If f ∈ B \ A, then f −1 ∈ A because A is a valuation ring. Since f ∈ B, f −1 is not in n, hence not in −1 −1 p. Therefore, f = (f ) ∈ Ap. Therefore, the localization map is surjective as well.

5.30

Let A be a valuation ring of a field K, let U ≤ K∗ be the group of units of A, and let Γ = K∗/U. For ξ, η ∈ Γ define ξ ≥ η by xy−1 ∈ A for some coset representatives x, y ∈ K∗ of ξ, η ∈ Γ. We claim that this is a well-defined total ordering on Γ that is compatible with the group structure. (Well-Defined) Let a, b ∈ U and let x, y ∈ K∗ such that xy−1 ∈ A. Then

(ax)(by)−1 = ab−1xy−1 ∈ A.

So the relation is well-defined on elements of Γ. (Total) Since A is a valuation ring, xy−1 ∈ A or yx−1 ∈ A for every x, y ∈ K∗. (Reflexive) Since xx−1 = 1 ∈ A, the relation is reflexive. (Anti-Symmetric) If xy−1, yx−1 ∈ A, then xy−1 ∈ U. Therefore, ξη−1 = 1, and ξ = η. (Transitive) Let ζ ∈ Γ with coset representative z ∈ K∗, and suppose that xy−1, yz−1 ∈ A. Then xz−1 = xy−1 · yz−1 ∈ A. (Compatible) If xy−1 ∈ A, then (xz)(yz)−1 = xy−1 ∈ A. Let v : K∗ → Γ be the quotient map. Since A is a valuation ring, xy−1 ∈ A or yx−1 ∈ A. Without loss of generality, xy−1 ∈ A. So min(v(x), v(y)) = v(y). Therefore,

(x + y)y−1 = xy−1 + 1 ∈ A, and we see that v(x + y) ≥ min(v(x), v(y)).

5.31

Let Γ be a totally ordered abelian group (written additively) and let K be a field. A valuation of K with values in Γ is a map v : K∗ → Γ such that (i) v(xy) = v(x) + v(y) and (ii) v(x + y) ≥ min(v(x), v(y)). Let A ⊆ K be the set of x ∈ K such that x = 0 or v(x) ≥ 0. By (i) and (ii), A is closed under addition and multiplication. By (i), v(1) = v(1) + v(1), which implies that v(1) = 0. Therefore, 1 ∈ A. On the other hand, 0 = v(1) = v(−1) + v(−1) implies that v(−1) = 0 (even if K has characteristic 2). Thus v(−x) = v(x) for all x ∈ A by (i). So A is closed under the taking of additive inverses, and A is a ring. 5 INTEGRAL DEPENDENCE AND VALUATIONS 63

5.32 Undone

5.33 Undone

5.34 Undone

5.35 Undone 6 CHAIN CONDITIONS 64

6 Chain Conditions

6.1

Let M be an A-module and let u : M → M be A-linear. (i) Suppose that u is surjective and M is Noetherian. Then the ascending chain, 0 ≤ ker u ≤ ker u2 ≤ ..., stabilizes at some index n ≥ 1. Since u is surjective, un is as well. If x ∈ ker u and x = un(y) for some y ∈ M, then 0 = u(x) = un+1(y). Thus y ∈ ker un+1 = ker un. So x = un(y) = 0. Therefore, u is injective. (ii) Suppose that u is injective and M is Artinian. Then the descending chain, 0 ≥ im u ≥ im u2 ≥ ..., stabilizes at some index n ≥ 1. Since u is injective, it lifts to an injective map coker ui → coker ui+1 for each i. If x ∈ coker u, then un(x) ∈ coker un+1 = coker un. Thus un(x) = 0. Therefore, x = 0, and u is surjective.

6.2

Let M be an A-module such that every non-empty set of finitely generated submodules has a maximal element, let N be a submodule of M, and let Σ be the set of finitely generated submodules of N. Since 0 ∈ Σ, it is non- empty. Let N 0 be a maximal element of Σ. Since N 0 ∈ Σ, N 0 is finitely generated. If x ∈ N, then Ax + N 0 is a finitely generated submodule of N containing N 0. Since N 0 is maximal, N 0 = Ax + N 0. Since x was arbitrary, N 0 = N. We have proven that every submodule of M is finitely generated. Therefore, M is Noetherian.

6.3

Let N1,N2 be submodules of an A-module M.

(Noetherian) Suppose M/N1, M/N2 are Noetherian, and let

M1 ≤ M2 ≤ M3 ≤ ... be an ascending chain of submodules of M containing N1 ∩ N2. For j = 1, 2, adjoin Nj and pass to the quotient to obtain an ascending chain, (M1 + Nj)/Nj ≤ (M2 + Nj)/Nj ≤ ..., of submodules of M/Nj. Since M/Nj is Noetherian, the ascending chain terminates at some index nj. Let n = max(n1, n2). Then we have the two identities,

(Mn + N1)/N1 = (Mn+1 + N1)/N1 and (Mn + N2)/N2 = (Mn+1 + N2)/N2.

Let f ∈ Mn+1. By the first identity, there is f1 ∈ N1 such that f + f1 ∈ Mn. By the second identity, there is f2 ∈ N2 such that f + f1 = f + f2. Therefore, f1 = f2 ∈ N1 ∩ N2, and f ∈ Mn + N1 ∩ N2. Since f was arbitrary, Mn+1 + N1 ∩ N2 = Mn + N1 ∩ N2. Therefore, the chain,

M1/(N1 ∩ N2) ≤ M2/(N1 ∩ N2) ≤ M3/(N1 ∩ N2) ≤ ..., of submodules of M/(N1 ∩ N2) terminates. So M/(N1 ∩ N2) is Noetherian.

(Artinian) If M/N1, M/N2 are Artinian, the same proof works with the reverse ordering. 6 CHAIN CONDITIONS 65

6.4

Let M be a Noetherian A-module, let a = Ann(M), and suppose that a1 ≤ a2 ≤ ... is an ascending chain of ideals of A containing a. Since M is Noetherian, it is finitely generated by some x1, . . . , xk. Again, since M is Noetherian, for each i the ascending chain of submodules, a1xi ≤ a2xi ≤ ..., stabilizes for some index ni. Let n = max(n1, . . . , nk). Then anxi = an+1xi for each i. Hence, anM = an+1M. Let a ∈ an+1. Since anM = an+1M, the multiplication by a map M/anM → M/anM is the zero map. Since a ⊆ an, the annihilator of M/anM is an. Therefore, a ∈ an. Since a was arbitrary, an+1 ⊆ an, and the chain stabilizes. Therefore, A/a is a Noetherian ring.

Let M = Z2/Z as a Z-module (that is M consists of fractions whose denominators are powers of 2, modulo 1). n Let N be a submodule of M. If 2k ∈ N is in lowest terms, then there are a, b ∈ Z such that

n an + b2k 1 a · = = ∈ N. 2k 2k 2k

−k Therefore, (2 Z)/Z ⊆ N. If N contains elements of arbitrarily large denominator, then N = M. Otherwise, −k N = (2 Z)/Z for some k ≥ 0. Now, having classified all of the Z-submodules of M, we can see that M is Artinian. k ∼ If n ∈ Z annihilates M, then 2 |n for all k ≥ 0. Thus Ann(M) = (0). However, Z/(0) = Z is not Artinian because (2) ⊇ (4) ⊇ (8) ⊇ ... is a non-terminating descending chain of ideals.

6.5

Let X be a Noetherian topological space, let Y be a subspace, and let U1 ⊆ U2 ⊆ ... be an ascending chain of open subsets of Y . By the definition of the subspace topology, there is some open subset V1 of X such that 0 V1 ∩ Y = U1. Given open subsets V1 ⊆ V2 ⊆ ... ⊆ Vn−1 of X such that Vi ∩ Y = Ui for each i, let Vn be an open 0 0 subset of X such that Vn ∩ Y = Un, and define Vn = Vn−1 ∪ Vn. Then we have Vn−1 ⊆ Vn and

0 Vn ∩ Y = (Vn−1 ∩ Y ) ∪ (Vn ∩ Y ) = Un−1 ∪ Un = Un, so the inductive selection of the family of V ’s may continue. Since X is a Noetherian topological space, there is an index, N, at which the chain of V ’s stabilizes. Therefore,

UN = VN ∩ Y = Vn ∩ Y = Un, for all n ≥ N. So Y is a Noetherian topological space.

Let {Wi}i∈I be an open cover of X. Without loss of generality, I is an ordinal. For each i ∈ ω ∩ I, define S Yi = j

6.6

Let X be a topological space. (i)⇒(iii) If X is Noetherian, and Y is a subspace of X, then Y is Noetherian by Exercise 6.5. Furthermore, Exercise 6.5 implies that Y is quasi-compact. (iii)⇒(ii) This is obvious.

(ii)⇒(i) Suppose that every open subspace of X is quasi-compact, let U1 ⊆ U2 ⊆ ... be an ascending chain of S open subsets of X, and define U = i≥1 Ui. Since U is a union of open subsets it is open, hence quasi-compact, and {Ui}i≥1 is an open cover of U. Therefore, there is some n such that U1,...,Un cover U, and the chain stabilizes. 6 CHAIN CONDITIONS 66

6.7

6.8

Let A be a Noetherian ring, let X = Spec A, and let V1 ⊇ V2 ⊇ ... be a descending chain of closed subsets of X. By the definition of the Zariski topology, there are radical ideals ai of A such that Vi = V (ai) for each i and a1 ⊆ a2 ⊆ ... is an ascending chain of ideals. Since A is Noetherian, this chain terminates at some index, n. Therefore, the descending chain of closed subspaces of X terminates at n as well. Therefore, X is Noetherian. By our method of proof one can see that if X is Noetherian, then A satisfies the ACC on the set of radical ideals.

6.9 Undone

6.10 Undone

6.11 Undone

6.12 Undone 7 NOETHERIAN RINGS 67

7 Noetherian Rings

7.1 8 ARTIN RINGS 68

8 Artin Rings 9 DISCRETE VALUATION RINGS AND DEDEKIND DOMAINS 69

9 Discrete Valuation Rings and Dedekind Domains 10 COMPLETIONS 70

10 Completions 11 DIMENSION THEORY 71

11 Dimension Theory 12 CONVENTIONS 72

12 Conventions

1. Don’t use “∈”.

2. Don’t use “$”.

3. Direct proofs when possible.

4. Give constructive and non-constructive proof when possible.

5. Colon for definitions.

6. Reader should not need to look anything up. 13 REMARKS 73

13 Remarks

Remark 1. I would like to prove an analogue of Exercise 1.2 (iii) by the following induction: If f is a zero-divisor, then by Exercise 1.2 (iii), there is an element a of Ar−1 such that af = 0. Then claim that the lead coefficient of a as a polynomial in xr−1 kills f as well. However, the argument in Exercise 1.2 (iii) made use of the hypothesis that g was of minimum degree, and we have no obvious such control on the degree of the lead coefficient of a with respect to a single variable. If you find an example showing that this method of proof will not work, or if you find a way of making it work, please send me an email. Remark 2. The following proof is for the case that A is reduced. I would like to extend this approach to the non-reduced case, but I am skeptical that this is possible. In particular, a regular function on a reduced scheme is uniquely determined by its values. This is no longer true for non-reduced schemes. I suspect that this feature is at the heart of the matter. Remark 3. The following proof uses machinery from Chapter 3. The merit of the proof is that it has a geometric interpretation: Z/mZ corresponds to a module supported on the finite closed subset V (m) of Spec Z. If m, n are coprime, then V (m) ∩ V (n) is empty. Therefore the tensor product module has empty support. Remark 4. There are two other proofs of this fact that are somewhat straightforward, albeit tedious. You can prove L that M ⊗A N satisfies the universal property of the direct sum or you can prove that i∈I (Mi ⊗A N) satisfies the universal property of the tensor product. Since universal objects are unique up to unique isomorphism, in either case you have the desired result. Remark 5. More generally, suppose M is an A-module with endomorphism φ such that φ2 = φ. Then

M = (ker φ) ⊕ (im φ) .

This can be seen by applying φ and idM −φ to the decomposition m = (m−φ(m))+φ(m). Such an endomorphism is called a linear projection. f g Suppose that 0 → K −→ M −→ N → 0 is a short exact sequence of A-modules. If there is σ : N → M such that g ◦ σ = idN , then σ ◦ g is a projection on M. If there is τ : M → K such that τ ◦ f = idK , then f ◦ τ is a projection on M. In either case, we obtain an identification of M with K ⊕ N.

Remark 6. Here we give an alternative construction of the direct limit of the directed system (Mi, µij : Mi → Mj). ` Let X = i∈I Mi be the disjoint union of the sets (Mi)i∈I . Define the relation ∼ on X by (x, i) ∼ (y, j) if there is k ≥ i, j such that µik(x) = µjk(y). We claim that ∼ is an equivalence relation on X. Reflexivity and symmetry follow immediately from those properties of equality. Suppose that (x, i) ∼ (y, j) and (y, j) ∼ (z, k). Then there are m, n ∈ I such that m ≥ i, j; n ≥ j, k; µim(x) = µjm(y); and µjn(y) = µkn(z). Let p ∈ I such that p ≥ m, n. Then we have

µip(x) = µmp(µim(x)) = µmp(µjm(y)) = µjp(y) = µnp(µjn(y)) = µnp(µkn(z)) = µkp(z).

That is, (x, i) ∼ (z, k). Let M = X/ ∼. We will put the structure of an A-module on M. Let [(x, i)], [(y, j)] ∈ M. Define

[(x, i)] + [(y, j)] := [(µik(x) + µjk(y), k)], where k ∈ I is such that k ≥ i, j. Let us show that addition is well-defined. Suppose that l ≥ i, j. Then there is m ≥ k, l such that µkm(µik(x) + µjk(y)) = µim(x) + µjm(y) = µlm(µil(x) + µjl(y)).

So the equivalence class [(µik(x) + µjk(y), k)] does not depend on the choice of k.

Suppose (u, p) ∼ (x, i) and (v, q) ∼ (y, j). Pick k ≥ i, j, p, q such that µpk(u) = µik(x) and µqk(v) = µjk(y). Then µik(x) + µjk(y) = µpk(u) + µqk(v). Therefore, addition does not depend on the choice of class representatives. 13 REMARKS 74

Let a ∈ A and define a · [(x, i)] := [(ax, i)]. If (x, i) ∼ (y, j), let k ≥ i, j. Then

µik(ax) = aµik(x) = aµjk(y) = µjk(ay).

So scalar multiplication is well-defined.

The axioms for M to be an A-module follow from the maps µij being A-linear and from each Mi being an A-module. The can be tediously verified.

Each map µi : Mi → M is defined as the inclusion Mi → X followed by the quotient X → M. It is easy to check that these maps are A-linear. Let i ≤ j and let x ∈ Mi. Then µj(µij(x)) = [(µij(x), j)] and µi(x) = [(x, i)]. Take k = j in the definition of ∼ to see that [(x, i)] = [(µij(x), j)]. Remark 7. I have a suspicion that there is a problem with the given proof. Remark 8. A useful consequence of Exercise 2.16 is that the construction of the direct limit given in the text of Exercise 2.14 and that in Remark6 are canonically isomorphic.

Remark 9. Let A be a ring and let (Mi)i∈I , (µij)i,j∈I be a directed system of A-modules. Let i0 ∈ I and let Ji0 be the set of indices j ∈ I such that j ≥ i0. Then lim Mj is canonically isomorphic to lim Mi as an A-module. −→j∈Ji0 −→i∈I Below is a proof. 0 First, we need to define the structure maps µi : Mi → lim Mi to see lim Mi as a direct limit over I. −→j∈Ji0 −→j∈Ji0 Let (µ ) be the structure maps of lim M from Exercise 2.14. If i ∈ I, then there is j ∈ J such that i ≤ j. i i∈Ji0 i i0 −→j∈Ji0 0 0 Define µi := µj ◦ µij. The identity stated in the text at the end of Exercise 2.14 shows that µi is well-defined.

We will use the characterization of lim Mi given in Exercise 2.16. Let N be an A-module and let (αi : Mi → −→ 0 N)i∈I be a family of A-linear maps such that αi = αj ◦ µij for all i ≤ j in I. Let µi(x) ∈ lim Mi. Define −→j∈Ji0 0 α(µi(x)) := αi(x). 0 0 Let us show that α is well-defined. Let µi(x) = µj(y) and let k ≥ i, j. Then

0 0 0 0 µk(µik(x)) = µi(x) = µj(y) = µk(µjk(y)).

0 So µk(µik(x) − µjk(y)) = 0. By Exercise 2.15 there is l ≥ k such that µil(x) = µjl(y). Therefore,

αi(x) = αl(µil(x)) = αl(µjl(y)) = αj(y),

0 and α is well-defined. On the other hand, the definition of α was forced by the condition α ◦ µi = αi. So α is unique. Remark 10. Here is a characterization of the Tor functor due to J. P. May (we have adapted it to commutative rings):

For every n ∈ Z there is a functor

A Torn : A − Mod × A − Mod → A − Mod such that for every A-module M and short exact sequence of A-modules

0 → N 0 → N → N 00 → 0

There are connecting homomorphisms

A 00 A 0 A 00 A 0 ∂ : Torn (N ,M) → Torn−1(N ,M) and ∂ : Torn (M,N ) → Torn−1(M,N ), natural in M and in the choice of short exact sequence that satisfy the following properties:

A 1. Torn (M,N) = 0 for all n < 0. 13 REMARKS 75

A 2. Tor0 (M,N) is naturally isomorphic to M ⊗A N. A 3. Torn (M,N) = 0 for n > 0 if either M or N is projective. 4. The following sequences are exact:

A 00 A 0 A A 00 A 0 · · · → Torn+1(N ,M) → Torn (N ,M) → Torn (N,M) → Torn (N ,M) → Torn−1(N ,M) → · · · A 00 A 0 A A 00 A 0 · · · → Torn+1(M,N ) → Torn (M,N ) → Torn (M,N) → Torn (M,N ) → Torn−1(M,N ) → · · ·

A A Furthermore, for each fixed M, the functors Torn (M, ·) and Torn (·,M) with their associated connecting homo- morphisms are uniquely determined up to isomorphism by properties 1-4. Here are some further properties from Weibel’s An Introduction to Homological Algebra:

• (Weibel 3.2.8) Let M,N be A-modules and let F∗ → M be a flat resolution of M. Then A ∼ A ∼ Tor∗ (M,N) = H∗(F∗ ⊗A N) and Tor∗ (N,M) = H∗(N ⊗A F∗).

Note that free ⇒ projective ⇒ flat.

•

Remark 11. Let a be an ideal of A and let M be an A-module. Tensor the short exact sequence

0 → a → A → A/a → 0 with M to get the following commutative diagram with exact rows:

A A Tor1 (A, M) Tor1 (A/a,M) a ⊗A M A ⊗A M (A/a) ⊗A M 0

0 0 aM M M/aM 0

Taking the kernel complex yields the exact sequence

A A 0 = Tor1 (A, M) → Tor1 (A/a,M) → ker (a ⊗A M → aM) → 0.

As a consequence, we have the following chain of implications: “M is flat”

⇒ “a ⊗A M = aM for all ideals a of A”

⇒ “a ⊗A M = aM for all finitely generated ideals of A” ⇒ “M is flat.” Remark 12. A consequence of Exercise 2.27: Every finitely generated ideal of an absolutely flat ring is projective. Remark 13. Here I will develop an alternative construction of the localization of a ring A with respect to a multiplicative subsemigroup S.

Construction of As: Let s ∈ A. Define As := A[x]/(sx − 1) and let f : A → As be the composition of the −1 canoncial maps A → A[x] and A[x] → As. In the ring As, we will denote the class of x as s . −n Representation of Elements: Every element of As has a representative of the form as for some n ≥ 0 and a ∈ A/ n n−1 Given any polynomial anx + an−1x + ... + a0 in A[x] we have the following congruence mod (sx − 1):

n n−1 n n n n n n anx + an−1x + ... + a0 ≡ anx + an−1sx + . . . a0s x = (an + an−1s + . . . a0s )x . 13 REMARKS 76

Universal Property of As: If B is an A-algebra, then # homA−alg(As,B) ≤ 1, with equality if and only if s is invertible in B.

Let g ∈ homA−alg(As,B). Then 1 = g(1) = g(ss−1) = sg(s−1) implies that s is invertible in B. Furthermore, let f : A → As, h : A → B be the structure maps. Then

g(f(a)f(s)−1) = h(a)h(s)−1 shows that g is uniquely determined by the structure map of B. Therefore, # homA−alg(As,B) ≤ 1. Conversely, suppose B is an A algebra such that s is invertible in B. By the Universal Property of Polynomial Rings, there is a unique A-algebra homomorphism A[x] → B sending x to s−1. Clearly, sx − 1 is in the kernel. So −1 −1 this A-algebra homomorphism lifts uniquely to an A-algebra homomorphism As → B such that s 7→ s . p Characterization of ker(A → As): a ∈ ker(A → As) if and only if s ∈ AnnA(a).

Suppose a ∈ A maps to 0 in As. Then there is

m m−1 q(x) = bmx + bm−1x + . . . b0 ∈ A[x] such that a = (sx − 1)q(x) in A[x]. (4)

Looking at the degree m + 1 term on the right hand side of (4) shows that sbm = 0. Therefore,

m−1 sa = (sx − 1)(sbm−1x + ... + sb0).

Repeating this same reasoning a total of m times yields

m m m+1 m s a = (sx − 1)s b0 = s b0x − s b0.

By comparing degrees, we get m m m+1 s a = −s b0 and s b0 = 0. Therefore, m+1 m+1 s a = −s b0 = 0.

Conversely, suppose that a ∈ A and sma = 0 for some m ≥ 1. Then

a ≡ smaxm mod (sx − 1) = 0.

Construction of S−1A: We will construct S−1A as a direct limit of A-algebras. 0 Let S ⊆ A be a multiplicatively closed set containing 1. Let S be the category with objects {As}s∈S and A-algebra homomorphisms as morphisms. By the “Universal Property 1” each hom-set has cardinality at most 1. Let S be a subcategory of S0 obtained by selecting a single representative of each isomorphism class of S0. Then S is a partial order. If s ∈ S, then As is uniquely isomorphic to a unique object As0 of S and we may identify these A-algebras without further comment. By abuse of notation, we will refer to an object As of S as s. In particular, s1 ≤ s2 if and only if there is an A-algebra homomorphism σs1s2 : As1 → As2 . −1
−1
Suppose that s1, s2 ∈ S. In A(s1s2) we have the identities s1 · s2(s1s2) = 1 and s2 · s1(s2s1) = 1. Therefore, s1, s2 ≤ s1s2. So S is a directed set. Properties (1) and (2) of Exercise 2.14 follow immediately from the cardinality of hom-sets. Therefore, {As}s∈S , {σs1s2 : As1 → As2 }s1≤s2 is a directed system of A-algebras. We define S−1A := lim A . −→ s s∈S If we picked a different subcategory S of S0, then our previous remarks would show that we obtain an isomorphic partially ordered set and all objects are uniquely isomorphic as A-algebras. Therefore, S−1A has been constructed up to unique isomorphism. 13 REMARKS 77

−1 −1 Universal Property of S A: (Proposition 3.1) If B is an A-algebra, then # homA−alg(S A, B) ≤ 1, with equality if and only if every element of S is invertible in B.

By the Universal Property of As and the identification,

hom (S−1A, B) = lim hom (A ,B), A−alg ←− A−alg s s∈S

−1 we can see that the set homA−alg(S A, B) has at most one element and is non-empty precisely when every element of S is invertible in B.

A Third Construction: Let A be a ring with multiplicatively closed subset S. Let B := A [{xs}s∈S] be the polynomial ring over A with an indeterminate for each element of S. Let I be the ideal of B generated −1 by the family {sxs − 1}s∈S, and define S A := B/I. It is straightforward to show that the ring B/I satisfies the universal property of S−1A by applying the Universal Property of Polynomial Rings and using the First Isomorphism Theorem.