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jeffrey daniel kasik carlson: Exercises to Atiyah and Macdonald’s Introduction to Commutative Algebra
This note is intended to contain full solutions to all exercises in this venerable text,^1 as well as proofs of results omitted or left to the reader. It has none of the short text’s pith or elegance, tending rather to the other extreme, citing chapter and verse from the good book and spelling out, step-by-step, things perhaps better left unsaid. It attempts to leave no “i” undotted, no “t” uncrossed, no detail unexplained. We are rather methodical in citing results when we use them, even if they’ve likely long been assimilated by the reader. Having found some solution sets online unclear at points (sometimes due to our own shortcomings, at other times due to theirs), we in this note strive to suffer from the opposite problem. Often we miss the forest for the trees. We prefer to see this not as “pedantic,” but as “thorough.” Sometimes we have included multiple proofs if we have found them, or failed attempts at proof if their failure seems instructive. The work is our own unless explicitly specified otherwise. It is hoped that the prolix and oftentimes plodding nature of these solutions will illuminate more than it conceals.
(^1) [? ]
Theorem 1.3. Every ring A 6 = 0 has at least one maximal ideal. In order to apply Zorn’s Lemma, it is necessary to prove that if 〈a α 〉 α ∈A is a chain of ideals (meaning, recall, that for all α , β ∈ A we have a α ⊆ a β or a β ⊆ a α ) then the union a =
⋃ α ∈A a α^ is an ideal. Indeed, if a, b ∈ a, then there are α , β ∈ A such that and a ∈ a α and b ∈ a β. Without loss of generality, suppose a α ⊆ a β. Then a, b ∈ a β , so since a β is an ideal we have a − b ∈ a β ⊆ a. If x ∈ A and a ∈ a, then there is α ∈ A such that a ∈ a α. As a α is an ideal, xa ∈ a α ⊆ a. Therefore a is an ideal.
Exercise 1.12. i) a ⊆ (a : b). For each a ∈ a we have ab ⊆ ab ⊆ a, so a ∈ (a : b).
ii) (a : b)b ⊆ a. By definition, for x ∈ (a : b) we have xb ⊆ a.
iii)
(a : b) : c
= (a : bc) =
(a : c) : b
x ∈
(a : b) : c
⇐⇒ xc ⊆ (a : b) ⇐⇒ xcb ⊆ a ⇐⇒ x ∈ (a : bc); x ∈
(a : c) : b
⇐⇒ xb ⊆ (a : c) ⇐⇒ xbc ⊆ a ⇐⇒ x ∈ (a : bc).
iv)
i ai :^ b
⋂ i (ai :^ b).
x ∈
i
ai : b
⇐⇒ xb ⊆
⋂
i
ai ⇐⇒ ∀i (xb ⊆ ai) ⇐⇒ x ∈
⋂
i
(ai : b).
v)
a : (^) ∑i bi
⋂ i (a^ :^ bi).
x ∈
i
bi
⇐⇒ a ⊇ x
i
bi
i
xbi ⇐⇒ ∀i (xbi ⊆ a) ⇐⇒ x ∈
⋂
i
(a : bi).
For an A-module M and subsets N ⊆ M and E ⊆ A, define (N : E) := {m ∈ M : Em ⊆ N}; for subsets N, P ⊆ M and E ⊆ A, define (N : P) := {a ∈ A : aP ⊆ N}. Note for future use that then ii) holds equally well for subsets a, b ⊆ M, or b ⊆ A and a ⊆ M; iii) holds for a, b ⊆ M and c ⊆ A; and iv) and v) hold for modules a, ai and modules or ideals b, bi.
Exercise 1.13. −i) a ⊆ b =⇒ r(a) ⊆ r(b).
If x ∈ r(a), for some n > 0 we have xn^ ∈ a ⊆ b, so x ∈ r(b).
i) r(a) ⊇ a. For each a ∈ a we have a^1 ∈ a, so a ∈ r(a).
Chapter 1: Rings and Ideals
ii) r
r(a)
= r(a).
x ∈ r
r(a)
⇐⇒ ∃n > 0
xn^ ∈ r(a)
⇐⇒ ∃n, m > 0
(xn)m^ = xmn^ ∈ a
⇐⇒ x ∈ r(a).
iii) r(ab) = r(a ∩ b) = r(a) ∩ r(b).
For the first equality, note (a ∩ b)^2 ⊆ ab ⊆ a ∩ b, so by parts 0) and −i), r(a ∩ b) = r
(a ∩ b)^2
⊆ r(ab) ⊆
r(a ∩ b). For the second, note that if m, n > 0 are such that xm^ ∈ a and xn^ ∈ b, then xmax{m,^ n}^ ∈ a ∩ b, and conversely.
iv) r(a) = ( 1 ) ⇐⇒ a = ( 1 ).
If r(a) = ( 1 ), then 1 ∈ r(a), so for some n we have 1 = 1 n^ ∈ a, and then a = ( 1 ).
v) r(a + b) = r
r(a) + r(b)
Since a, b ⊆ a + b, by part −i) we have r(a), r(b) ⊆ r(a + b), so r(a) + r(b) ⊆ r(a + b). By parts −i), and ii), we see r
r(a) + r(b)
⊆ r
r(a + b)
= r(a + b). Conversely, by part i), we have a ⊆ r(a) and b ⊆ r(b), so adding, a + b ⊆ r(a) + r(b). By part −i), r(a + b) ⊆ r
r(a) + r(b)
vi) If p is prime, r(pn) = p for all n > 0.
By part 0), r(pn) = r(p). By (1.14), r(p) = p.
Proposition 1.17. iv*) Both extension and contraction are order-preserving with respect to containment; i.e. for ideals a 1 ⊆ a 2 of A we have ae 1 ⊆ ae 2 and for ideals b 1 ⊆ b 2 of B we have bc 1 ⊆ bc 2.^1
If a 1 ⊆ a 2 , then f(a 1 ) ⊆ f(a 2 ), so ae 1 = Bf(a 1 ) ⊆ Bf(a 2 ) = ae 2. If b 1 ⊆ b 2 , then bc 1 = f −^1 (b 1 ) ⊆ f −^1 (b 2 ) = bc 2.
Exercise 1.18. Let f : A → B be a ring homomorphism, and let a, aj be ideals of A and b, bj ideals of B. ( ∑ aj
)e = ∑ aej.
Because the homomorphism f distributes over finite sums and multiplication of ideals distributes over addition,
(
)e = Bf
(a 1 a 2 )e^ = ae 1 ae 2.
Because 1 ∈ B, we have B = BB; as f is a homomorphism and ideal multiplication commutes,
(a 1 a 2 )e^ = Bf(a 1 a 2 ) = BBf(a 1 ) f(a 2 ) = Bf(a 1 )Bf(a 2 ) = ae 1 ae 2. ( ∑ bj
)c ⊇ (^) ∑ bcj.
Given any finitely many nonzero aj ∈ f −^1 (bj), we have f
∑j aj
= (^) ∑j f(aj) ∈ (^) ∑j bj.
(b 1 b 2 )c^ ⊇ bc 1 b 2 c.
If f(aj) ∈ bj for j = 1, 2, then f(a 1 a 2 ) = f(a 1 ) f(a 2 ) ∈ b 1 b 2.
( (^) ⋂ j aj
)e ⊆
⋂ j a e j.^ ( (^) ⋂ aj
)e = Bf
aj
⋂ f(aj) =
⋂ Bf(aj) =
⋂ aej.
( (^) ⋂ bj
⋂ bcj. ( (^) ⋂ bj
)c = f −^1
bj
⋂ f −^1 (bj) =
⋂ bcj.
(^1) This is trivial, but the book never seems to explicitly state that it is the case, so here is a place to cite when we use it.
Chapter 1: Rings and Ideals Ex. 1.
Multiplying by arn, we get
0 = ar n+ 1 bm−r + arnbm−r+ 1 ︸ ︷︷ ︸ 0
(an− 1 an) + · · · + anbm ︸ ︷︷ ︸ 0
(an−r ar n− 1 ),
so 0 = ar n+ 1 bm−r, completing the induction. When we get to r = m we see b 0 am n +^1 = 0, so since b 0 is a unit, an is nilpotent. Hence an xn^ is nilpotent, and by [1.1], f − an xn^ is a unit. This has degree < n, so by induction, a 1 ,... , an− 1 are nilpotent.
ii) f is nilpotent ⇐⇒ a 0 , a 1 ,... , an are nilpotent.
⇐=: Since N
A[x]
is an ideal by (1.7), if all aj ∈ N, then all aj xj^ ∈ N, so f = ∑ aj xj^ ∈ N. =⇒: On the other hand, for each prime p A, we have p[x] A[x] prime since it is the kernel of the surjection A[x] (A/p)[x], whose image is an integral domain by [1.2.iii]: if aj xj^ ∈ A[x] is a zero-divisor,
there exists a nonzero c ∈ A with c · (^) ∑ aj xj^ = 0, so each c · aj = 0, and hence aj = 0 as A/p is an integral domain. Thus
N
A[x]
p[x]
p
[x] = N(A)[x].^3
iii) f is a zero-divisor ⇐⇒ there exists a 6 = 0 in A such that af = 0.
The “if” direction is trivial; the “only if” we prove by induction. We prove something slightly more specific: if a nonzero g = b 0 + b 1 x + · · · + bm xm^ ∈ (0 : f ) is of least possible degree m, then bm f = 0.
For the base case, if f = a 0 has degree zero, then of course bm a 0 = bm f = 0. Suppose inductively that for all zero-divisors f ′^ of degree n − 1 we know there is some b ∈ A such that bf ′^ = 0. Let deg f = n and g and m be as before. Since fg = 0, also anbm = 0, so deg(an g) < m. As angf = 0 as well, by minimality of m, we know ang = 0. Now 0 = fg − an xn^ g = ( f − an xn)g. Since f ′^ = f − an xn^ is of degree < n, by the inductive assumption bm f ′^ = 0, so bm f = bm an xn^ + bm f ′^ = 0, completing the induction.
iv) f is said to be primitive if (a 0 , a 1 ,... , an) = ( 1 ). Prove that if f, g ∈ A[x], then fg is primitive ⇐⇒ f and g are primitive.
We can actually let A[x] := A[x 1 ,... , xr ] be a polynomial ring in several indeterminates, writing x for the sequence x 1 ,... , xr, in the proof below.
Note that a polynomial is primitive just if no maximal ideal contains all its coefficients. Let m A be maximal. Since A/m is a field, A[x]/m[x] ∼= (A/m)[x] is an integral domain. Thus
f, g ∈/ m[x] ⇐⇒ f, ¯¯g 6 = 0 in (A/m)[x] ⇐⇒ fg 6 = 0 in (A/m)[x] ⇐⇒ fg ∈/ m[x].
Therefore no maximal ideal contains all the coefficients of fg just if the same holds for f and g.^4
This result is called Gauß’s Lemma and was originally proven in his Disquisitiones Arithmeticae for A = Z. Cf. also [9.2].
(^3) Alternate proof: suppose f m (^) = 0 and j is minimal with aj 6 = 0. Then the lowest term amj xjm (^) of f is 0, so aj is nilpotent and f − aj xj^ is nilpotent. Repeatedly lopping off lowest terms, we see each aj ∈ N(A). (^4) We also have the following generalization of the classical proof. Suppose f is not primitive, so that for some maximal m A we have f ∈ m[x]. Then fg ∈ m[x], so fg is not primitive. The same holds if g is not primitive. Now suppose fg is not primitive; we show one of f and g is also not. There is a maximal ideal m containing all of the coefficients cr = ∑j aj br−j of fg; we suppose that neither all of the aj nor all of the bk lie in m and obtain a contradiction. There are a least J and a least K such that aJ , bK ∈/ m. Now m contains the coefficient cJ+K = (^) ∑j<J aj bJ+K−j + aJ bK + (^) ∑k<K aJ+K−k bk, and each of the sums is in m by assumption, so aJ bK ∈ m as well. Since m is prime, we have aJ or bK in m, a contradiction.
Ex. 1.3 Chapter 1: Rings and Ideals
3. Generalize the results of Exercise 2 to a polynomial ring A[x 1 ,... , xr ] in several indeterminates.
We start with an assumption about the ring B = A[x 1 ,... , xr ] and prove the corresponding statement about the ring B[y] = A[x 1 ,... , xr, y] in one more indeterminate. For a multi-index α = 〈j 1 ,... , jr 〉 we write a α , k := aj 1 , ..., jr , k and x α^ := xj 11 · · · xj rr. If f ∈ B[y], we can write it as f = (^) ∑ α , k a α , k x α yk^ = (^) ∑k hkyk, where hk = (^) ∑ α a α , k x α^ ∈ B.
i) f is a unit in B[y] ⇐⇒ a0, 0 is a unit in A and all other a α , k are nilpotent.
f ∈ B[y]×^ [1.2.i] ⇐⇒ B[y]/B
h 0 ∈ B×^ and other hk ∈ N(B) [1.3.i], [1.3.ii] ⇐⇒ B/A
a0, 0 ∈ A×^ and other a α , k ∈ N(A).
Use of [1.3.ii] is permissible because it is independent, but we could also perform the induction in both exercises at the same time.
ii) f is nilpotent ⇐⇒ all a α , k are nilpotent.
f ∈ N
B[y]
) (^) [1.2.ii] ⇐⇒ B[y]/B
all hk ∈ N(B)
[1.3.ii] ⇐⇒ B/A
all a α , k ∈ N(A).
iii) f is a zero-divisor ⇐⇒ there exists a 6 = 0 in A such that af = 0. The inductive assumption will be that if g ∈ B is a zero-divisor, and b ∈ B is of minimal multidegree α (in the reverse lexicographic order) such that bg = 0, then if a is the coefficient of the leading term of b, we have ag = 0. The “if” is again trivial. For the “only if,” suppose f is a zero-divisor in B[y]. By [1.2], there exists a nonzero b ∈ B such that bf = 0. Thus bhk = 0 for each k. By the inductive assumption the highest coefficient a ∈ A of b is such that ahk = 0 for each k. Then af = 0.
iv) Prove for f, g ∈ A[x 1 ,... , xr ] that fg is primitive over A ⇐⇒ f and g are primitive over A. The proof in [1.2.iv] goes through equally well in this case.
4. In the ring A[x], the Jacobson radical is equal to the nilradical.
We know N ⊆ R by (1.8), since maximal ideals are prime, so it remains to show all elements of R are nilpotent. Let f = (^) ∑ aj xj^ ∈ R, where aj ∈ A. By (1.9), 1 − xf is a unit. By [1.2.i], then, all aj ∈ N, so by [1.2.ii], f ∈ N.
5. Let A be a ring and let A[[x]] be the ring of formal power series f = (^) ∑∞ n= 0 an xn^ with coefficients in A. Show that i) f is a unit in A[[x]] ⇐⇒ a 0 is a unit in A.
⇐=: Supposing a 0 is a unit, we construct an inverse g = (^) ∑m bm xm^ to f. Let b 0 = a 0 − 1. We want fg = ∑j cj xj^ = 1, so for j ≥ 1 we want cj = ∑ j n= 0 anbj−n^ =^ 0. Now suppose we have found satisfactory coefficients bj for j ≤ k. We need ck+ 1 = a 0 bk+ 1 + ∑k n+=^11 anbk+ 1 −n = 0; but we can solve this to find the solution bk+ 1 = −a− 01
∑k n+=^11 anbk+ 1 −n
. Since we can do this for all k, we have constructed an inverse to f. =⇒ : If g = (^) ∑m bm xm^ is an inverse of f, then fg = 0 implies a 0 b 0 = 1 so that a 0 is a unit.
ii) If f is nilpotent, then an is nilpotent for all n ≥ 0. Is the converse true? The two proofs of “ =⇒ ” in [1.2.ii] both hold, mutatis mutandis, here. The converse is untrue. Let B = C[y 1 , y 2 ,.. .] be a polynomial ring in countably many indeterminates over an integral domain C, and let b = (y 1 , y^22 , y^33 ,.. .) be the smallest ideal containing each ynn for n ≥ 1. Then writing zn for the image of yn in A = B/b, we have znn = 0 and zn n− 16 = 0. Thus an element of A is equal to zero just if, written as a polynomial in the zn over C, each term is divisible by some znn. Now let f = (^) ∑∞ n= 0 zn xn^ ∈ A[[x]]. By construction, each coefficient is nilpotent. However, for each n, one term of the coefficient in A of xn(n+^1 )^ in f n^ is znn+ 1 , which is nonzero and cannot be cancelled, so f n^6 = 0.
iii) f belongs to the Jacobson radical of A[[x]] ⇐⇒ a 0 belongs to the Jacobson radical of A.
Ex. 1.11 Chapter 1: Rings and Ideals
11. A ring A is Boolean if x^2 = x for all x ∈ A. In a Boolean ring A, show that i) 2 x = 0 for all x ∈ A;
x + 1 = (x + 1 )^2 = x^2 + 12 + 2 x = (x + 1 ) + 2 x, so subtracting x + 1 from both sides, 0 = 2 x.
ii) every prime ideal p is maximal, and A/p is a field with two elements;
By [1.7], every prime ideal is maximal. x^2 − x = x(x − 1 ) = 0 holds for each x ∈ A, so ¯x( x¯ − 1 ) = 0 holds for each x¯ ∈ A/p; as A/p is an integral domain, this means each element is either 0 or 1, so A/p ∼= F 2.
iii) every finitely generated ideal in A is principal.
We induct on the number of generators. The one-generator case is trivial. Suppose every ideal gener- ated by n elements is principal, and a = (x 1 ,... , xn, y). Let x generate (x 1 ,... , xn), and let z = x + y − xy. Then xz = x^2 + xy − x^2 y = x and yz = y, so a = (x, y) = (z).
12. A local ring contains no idempotent 6 = 0, 1.
Let A be a ring. For any idempotent unit e, we have e = e−^1 e^2 = e−^1 e = 1. Suppose e^2 = e 6 = 0, 1 in A. Then e is not a unit, and by (1.5) is contained in some maximal ideal m. Similarly ( 1 − e)^2 = 1 − 2 e + e^2 = 1 − e is another idempotent 6 = 0, 1, hence not a unit. But were A local, e would be in m = R, so 1 − e would be a unit by (1.9).^7
13. Let K be a field and let Σ be the set of all irreducible monic polynomials f in one indeterminate with coefficients in K. Let A be the polynomial ring over K generated by indeterminates xf, one for each f ∈ Σ. Let a be the ideal of A generated by the polynomials f (xf ) for all f ∈ Σ. Show that a 6 = ( 1 ).
If a = ( 1 ), there exist finitely many af ∈ A such that 1 = ∑ af f(xf ). The set I of xg occurring in this expression (not only those in the f (xf ), but also those occurring in the af ) is finite. We may enumerate I as x 1 ,... , xi,... , xn, corresponding to irreducible polynomials fi, and suppose n is minimal such that such an equation holds. Write B = K[x 1 ,... , xn− 1 ], C = B[xn], and b :=
f 1 (x 1 ),... , fn− 1 (xn− 1 )
By minimality of n, the ideal b is proper, so be^ = b[xn] C is as well, while by the equation above, be^ +
fn(xn)
= C. Since b 6 = B, we know B/b 6 = 0. Let g be the image of fn(xn) in (B/b)[xn]. Since fn is irreducible in K[xn], its degree degxn fn ≥ 1 and also degxn g ≥ 1. Then
be^ +
fn(xn)
b[xn]
(g)
B[xn]/
b[xn]
(g)
∼= (B/b)[xn] (g)
which is a contradiction.^8
Let m be a maximal ideal of A containing a and let K 1 = A/m. Then K 1 is an extension field of K in which each f ∈ Σ has a root. Repeat the construction with K 1 in place of K, obtaining a field K 2 , and so on. Let L =
⋃∞ n= 1 Kn. Then L is a field in which each f ∈ Σ splits completely into linear factors. Let K be the set of all elements of L which are algebraic over K. Then K is an algebraic closure of K.
We should probably show that K is closed under subtraction and multiplication. Let a, b ∈ K have conjugates ai, bj over K. Then ∏i, j(x − ai + bj) is symmetric in the ai and the bj, and so has coefficients in K, so a − b ∈ K. Similarly (^) ∏i, j(x − aibj) is symmetric, so ab ∈ K.
(^7) Cf. the implication iii) =⇒ ii) in [1.22] for another proof: by (1.4), each of (e) and ( 1 − e) is contained in a maximal ideal, and we can show the two are coprime, so no maximal ideal can contain both. Actually, there is even an isomorphism A −→∼ A/(e) × A/( 1 − e), so A obviously has more than one maximal ideal. (^8) This proof is taken from [? ].
Chapter 1: Rings and Ideals Ex. 1.
14. In a ring A, let Σ be the set of all ideals in which every element is a zero-divisor. Show that the set Σ has maximal elements and that every maximal element of Σ is a prime ideal. Hence the set of zero-divisors in A is a union of prime ideals. Order Σ by inclusion; to show it has maximal elements, it suffices by Zorn’s Lemma to show every chain 〈a α 〉 α ∈A has an upper bound in Σ. Let a =
⋃ α a α. It contains only zero-divisors, since if^ x^ ∈^ a, then there is α such that x ∈ a α , and then by definition x is a zero-divisor. Let p be a maximal element of Σ; we must show it to be prime. Suppose x, y /∈ p. Then there are non-zero-divisors in (x) + p and (y) + p, and their product is an element of (xy) + p that is again a non- zero-divisor. Thus xy ∈/ p, lest there be something in p other than a zero-divisor. This shows p is prime. Thus Σ has maximal elements and every element of Σ is contained in one; considering principal ideals, this shows every zero-divisor is in a maximal element of Σ. The last statement follows.
The prime spectrum of a ring
15. Let A be a ring and let X be the set of all prime ideals of A. For each subset E of A, let V(E) denote the set of all prime ideals of A which contain E. Prove that i) if a is the ideal generated by E, then V(E) = V(a) = V(r(a)). Let p ∈ X. We have E ⊆ a, so if a ⊆ p, then E ⊆ p. On the other hand, if E ⊆ p, then a = AE ⊆ Ap = p. Thus V(E) = V(a). By (1.14), r(a) =
⋂ V(a), so p ⊇ r(a) ⇐⇒ p ⊇ a and V(a) = V
r(a)
ii) V( 0 ) = X, V( 1 ) = ∅. For every prime ideal p we have 0 ∈ p and 1 /∈ p.
iii) if 〈Ei〉i∈I is any family of subsets of A, then V
i∈I
Ei
⋂
i∈I
V(Ei).
p ∈ V
i
Ei
⋃
i
Ei ⊆ p ⇐⇒ ∀i ∈ I (Ei ⊆ p) ⇐⇒ ∀i ∈ I (p ∈ V(Ei)) ⇐⇒ p ∈
⋂
i
V(Ei).
Note also for future use that
⋃ Ei ⊆ p ⇐⇒
⋃ AEi ⊆ Ap = p ⇐⇒ (^) ∑ AEi ⊆ p, so in particular for ideals ai we have V
ai
∑ ai
iv) V(a ∩ b) = V(ab) = V(a) ∪ V(b) for any ideals a, b of A. Suppose ab ⊆ p and b 6 ⊆ p. Then there is b ∈ b\p, and ab ∈ p for all a ∈ a, so the primality of p gives a ∈ p and thus a ⊆ p. Thus if p ∈ V(ab), we have shown a ⊆ p or b ⊆ p, so p ∈ V(a) ∪ V(b). On the other hand, if p contains a or contains b, then it must contain the subset ab. Thus V(ab) = V(a) ∪ V(b). Now ab ⊆ a ∩ b, so if a ∩ b ⊆ p, then ab ⊆ p. On the other hand, if ab ⊆ p, then we have shown either a ⊆ p or b ⊆ p, so since a ∩ b is a subset of both of these we have a ∩ b ⊆ p. Thus V(a ∩ b) = V(ab).
These results show that the sets V(E) satisfy the axioms for closed sets in a topological space. The resulting topology is called the Zariski topology. The topological space X is called the prime spectrum of A, and is written Spec(A).
16. Draw pictures of Spec( Z ), Spec( R ), Spec( C [x]), Spec( R [x]), Spec( Z [x]).
There is only one point, ( 0 ), in Spec( R ). In Spec( Z ), the elements are ( 0 ) and (p) for each positive prime p ∈ N , and the closed sets are X, ∅, and all finite sets containing ( 0 ). In C [x], all polynomials split into linear factors, so the irreducible polynomials are x − α for α ∈ C. Since C is a field, this means the only primes are ( 0 ) and (x − α ) for α ∈ C. The closed sets are again X, ∅, and all finite sets containing ( 0 ). As a point set, it makes sense to think of X as the complex plane plus one additional dense point. In R [x], all polynomials split into linear factors and polynomials of the form (x − α )(x + α ) for α ∈ C with Im ( α ) > 0. Thus the primes of R [x] correspond to points of R , points of the upper half plane, and the dense point ( 0 ) again. In Z [x], there are irreducible polynomials f of all degrees ≥ 1, giving rise to prime ideals ( f ), there are ideals (p) for all positive primes p ∈ N , and there are ideals (p, f ) = (p) + ( f ), which are maximal.
Chapter 1: Rings and Ideals Ex. 1.
Each Xf is compact, and a union of a finite collection of compact sets is compact^12 , so a finite union of basic open sets Xf is compact. On the other hand, suppose a set is open and compact. Since it is open, we can write it as a union of some basic open sets X (^) f α ; since it is compact, we can take a finite subcover, showing it is a union of finitely many basic open sets.
18. For psychological reasons it is sometimes convenient to denote a prime ideal of A by a letter such as x or y when thinking of it as a point of X = Spec(A). When thinking of x as a prime ideal of A, we denote it by px (logically, of course, it is the same thing). Show that i) The set {x} is closed (we say that x is a “closed point”) in Spec(A) ⇐⇒ px is maximal;
⋂^ Let^ Y^ ⊆^ X, and let^ V(a)^ ⊆^ X^ be a closed set. Recall our Galois correspondence:^ Y^ ⊆^ V(a)^ ⇐⇒^ a^ ⊆ y∈Y py.
V(a) : Y ⊆ V(a)
V(a) : a ⊆
⋂
y∈Y
py
a : a ⊆
⋂
y∈Y
py
y∈Y
py
so {x} is closed just if {x} = V(px ), or in other words iff no other prime contains px.
ii) {x} = V(px );
{x}
??
{px }
= V(px ).
iii) y ∈ {x} ⇐⇒ px ⊆ py;
y ∈ {x} ii) = V(px ) ⇐⇒ px ⊆ py.
iv) X is a T 0 -space (this means that if x, y are distinct points of X, then either there is a neighborhood of x which does not contain y, or else there is a neighborhood of y which does not contain x).
If every neighborhood of x contains y and vice versa, then y ∈ {x} and x ∈ {y}, so by part iii); px ⊆ py ⊆ px and x = y.
19. A topological space X is said to be irreducible if X 6 = ∅^ and if every pair of non-empty open sets in X intersect, or equivalently if every non-empty open set is dense in X. Show that Spec(A) is irreducible if and only if the nilradical of A is a prime ideal. ∅ (^) is not an intersection of two nonempty open sets just if X is not a union of two proper closed sets V(a), V(b). By the proof of [1.17.iv],
X = V( 0 ) 6 = V(a) ⇐⇒ r(a) 6 ⊆ r( 0 )
(1.8) = N
(1.13) ⇐⇒ a 6 ⊆ N,
and by [1.15.iv], V(a) ∪ V(b) = V(ab), so X is irreducible just if for all ideals a, b 6 ⊆ N we also have ab 6 ⊆ N; by contraposition, this is that ab ⊆ N =⇒ a ⊆ N or b ⊆ N. But this condition is just a rephrasing of primality: if N is prime, then as in [1.15.iv], ab ⊆ N =⇒ a ⊆ N or b ⊆ N; conversely, if the condition holds, then ab ∈ N =⇒ (a)(b) ⊆ N =⇒ (a) or (b) ⊆ N =⇒ a or b ∈ N, so N is prime.
20. Let X be a topological space. i) If Y is an irreducible (Exercise 19) subspace of X, then the closure Y of Y in X is irreducible.
Let open U, V ⊆ Y be non-empty. Then U ∩ Y and V ∩ Y are non-empty by the definition of closure. Since Y is irreducible, U ∩ V ∩ Y 6 = ∅, and a fortiori U ∩ V 6 = ∅.
ii) Every irreducible subspace of X is contained in a maximal irreducible subspace. (^12) To see this, let a cover V of the finite union K = ⋃nj= 1 of compact sets Kj be given. For each Kj take a finite subcollection Uj ⊆ V covering Kj; then ⋃ j Uj^ ⊆ V^ is a finite collection covering^ K.
Ex. 1.21 Chapter 1: Rings and Ideals
We apply Zorn’s Lemma. Order the irreducible subspaces Σ of X by inclusion, and let 〈Y α 〉 be a linearly ordered chain. Set Z =
⋃ α Y α ; we will be done if we can show^ Z^ ∈^ Σ. Let^ U,^ V^ ⊆^ Z^ be open and non- empty. By definition, U meets some Y α and V meets some Y β. Without loss of generality, suppose α ≤ β. Then as Y α ⊆ Y β , we have both U ∩ Y β and V ∩ Y β non-empty and open in Y β , so by irreducibility of Y β we have U ∩ V ∩ Y β 6 = ∅ and U ∩ V 6 = ∅, showing Z is irreducible.
iii) The maximal irreducible subspaces of X are closed and cover X. They are called the irreducible components of X. What are the irreducible components of a Hausdorff space? To see that a maximal irreducible subspace is closed, note that its closure is irreducible by part i) and contains it, and so equals it by maximality. To see the maximal irreducible subspaces cover X, note that each singleton is irreducible and contained in some maximal irreducible subspace. Every subspace of a Hausdorff space is Hausdorff. If a Hausdorff space has two distinct points, they have two disjoint neighborhoods by definition, so the space is not irreducible. Thus the irreducible com- ponents of a Hausdorff space are the singletons.
iv) If A is a ring and X = Spec(A), then the irreducible components of X are the closed sets V(p), where p is a minimal prime ideal of A (Exercise 8). Any closed subset of X is of the form V(a) for some ideal a A, and is homeomorphic to Spec(A/a) by [1.21.iv]. By [1.19], V(a) = V
r(a)
is irreducible if and only if N(A/a) is prime, or equivalently if r(a) is prime in A; so the irreducible closed subspaces of X are V(p) for p ∈ X. Such a V(p) is maximal just if there is no q ∈ X with V(p) ( V(q), or equivalently, by the proof of [1.17.iv], there is no prime q ( p.
21. Let φ : A → B be a ring homomorphism. Let X = Spec(A) and Y = Spec(B). If q ∈ Y, then φ −^1 (q) is a prime ideal of A, i.e., a point of X. Hence φ induces a mapping φ ∗^ : Y → X. Show that i) If f ∈ A then φ ∗−^1 (Xf ) = Y φ ( f ), and hence that φ ∗^ is continuous.
q ∈ Y φ ( f ) ⇐⇒ φ ( f ) ∈/ q ⇐⇒ f /∈ φ −^1 (q) ⇐⇒ φ ∗(q) = φ −^1 (q) ∈ Xf ⇐⇒ q ∈ ( φ ∗)−^1 (Xf ).
ii) If a is an ideal of A, then φ ∗−^1
V(a)
= V(ae). First, note that (1.17.i) implies extension and contraction of ideals form an isotone Galois correspon- dence:^13 a ⊆ bc^ ⇐⇒ ae^ ⊆ b. Indeed, if a ⊆ bc, then extending, ae^ ⊆ bce^ ⊆ b, and if ae^ ⊆ b, then contracting, a ⊆ aec^ ⊆ bc. Now for q ∈ Spec(B),
q ∈ ( φ ∗)−^1
V(a)
⇐⇒ φ ∗(q) ∈ V(a) ⇐⇒ a ⊆ qc^ ⇐⇒ ae^ ⊆ q ⇐⇒ q ∈ V(ae).
iii) If b is an ideal of B, then φ ∗(V(b)) = V(bc).
By Eq. ?? from [1.18.i], φ ∗(V(b)) is the set of prime ideals containing
⋂ φ ∗
V(b)
, which ideal equals
⋂ {qc^ : b ⊆ q ∈ Y}
b⊆q∈Y
q
(1.14)
r(b)c^
(1.18) = r(bc).
But V
r(bc)
= V(bc).
iv) If φ is surjective, then φ ∗^ is a homeomorphism of Y onto the closed subset V
ker( φ )
of X. (In particular, Spec(A) and Spec(A/N) (where N is the nilradical of A) are naturally homeomorphic.) If φ : A → A/a is surjective, (1.1) gives an containment-preserving and -reflecting bijection between the set of ideals b A containing a and the ideals b/a A/a. Since this relation preserves and reflects primes (p. 9), for any ideal b/a of A/a,
φ ∗(V
b/a)
p ∈ Spec(A) : b/a ⊆ p/a ∈ Spec(A/a)
= V(b), (^13) [? ]
Ex. 1.23 Chapter 1: Rings and Ideals
23. Let A be a Boolean ring (Exercise 11), and let X = Spec(A). i) For each f ∈ A, the set Xf (Exercise 17) is both open and closed in X.
Xf is open because it is the complement of V( f ). It is open because V( f ) = X 1 − f. Indeed, X = Xf ∪ X 1 − f , since any ideal containing both f and 1 − f contains 1; and Xf ∩ X 1 − f = Xf ( 1 − f ) = X 0 = ∅ by [1.17.i] and [1.17.ii].
ii) Let f 1 ,... , fn ∈ A. Show that X (^) f 1 ∪ · · · ∪ X (^) fn = Xf for some f ∈ A. ⋃ Xfi =
X\V( fi)
⋂ V( fi) = X\V
∑ (^ fi)
by [1.15]. By [1.11.iii], (^) ∑ ( fi) = ( f ) for some f ∈ A, so
⋃ Xfi = X\V( f ) = X (^) f.
iii) The sets Xf are the only open subsets of X which are both open and closed.
Let U be both open and closed. Since it is closed and X is compact, U is compact. By [1.17.vii], U is a union of finitely many X (^) fj. By part ii), it is an Xf for some f ∈ A.
iv) X is a compact Hausdorff space.
The compactness of X is [1.17.v]. To show X is Hausdorff, let x, y ∈ X; we will show that if they do not have disjoint neighborhoods Xf and X 1 − f , then x = y. Now Xf 3 x and X 1 − f 3 y just if f ∈/ px and 1 − f ∈/ py. By part i), this is the same as saying f /∈ px and f ∈ py. If no such f exists, we have py ⊆ px, and since [1.11.ii] showed py is maximal, px = py.
24. Let L be a lattice, in which the sup and inf of two elements a, b are denoted by a ∨ b and a ∧ b respectively. L is a Boolean lattice (or Boolean algebra) if i) L has a least element and a greatest element (denoted by 0, 1 respectively); ii) each of ∨, ∧ is distributive over the other; iii) Each a ∈ L has a unique “complement” a′^ ∈ L such that a ∨ a′^ = 1 and a ∧ a′^ = 0. For example, the set of all subsets of a set, ordered by inclusion, is a Boolean lattice. Let L be a Boolean lattice. Define addition and multiplication in L by the rules
a + b = (a ∧ b′) ∨ (a′^ ∧ b), ab = a ∧ b.
Verify that in this way L becomes a Boolean ring, say A(L).
To verify the ring axioms, we first require some lemmas about Boolean algebra.
Chapter 1: Rings and Ideals Ex. 1.
(x′^ ∧ y′) ∧ (x ∨ y) = (x′^ ∧ y′^ ∧ x) ∨ (x′^ ∧ y′^ ∧ y) = ( 0 ∧ y′) ∨ (x′^ ∧ 0 ) = 0 ∨ 0 = 0, (x′^ ∧ y′) ∨ (x ∨ y) = (x′^ ∨ x ∨ y) ∧ (y′^ ∨ x ∨ y) = ( 1 ∨ y) ∧ (x ∨ 1 ) = 1 ∧ 1 = 1;
since (x ∨ y)′^ is postulated to be unique with these properties, we have (x ∨ y)′^ = x′^ ∧ y′. The other law (x ∧ y)′^ = x′^ ∨ y′^ is dual; the proof is the same, exchanging ∧ for ∨ and vice versa everywhere.
From now on write · for ∧. We prove a few more miscellaneous facts.
(a ∨ b)(a′^ ∨ b′)
= (a ∨ b)′^ ∨ (a′^ ∨ b′)′^ = a′b′^ ∨ ab.
Now we can prove the ring axioms for A(L).
[a + b]′^ ∨ c′
= (ab′^ ∨ a′b ∨ c)(a′b′^ ∨ ab ∨ c′) = ab′a′b′^ ∨ ab′ab ∨ ab′c′^ ∨ a′ba′b′^ ∨ a′bab ∨ a′bc′^ ∨ ca′b′^ ∨ cab ∨ cc′^ (1.2) = 0 ∨ 0 ∨ ab′c′^ ∨ 0 ∨ 0 ∨ a′bc′^ ∨ ca′b′^ ∨ cab ∨ 0. = ab′c′^ ∨ a′bc′^ ∨ a′b′c ∨ abc.
Write x+^ = x and x−^ = x′. Then (a + b) + c is the supremum of the four possible terms a±b±c± with an odd number of +’s. This is invariant under permutations of a, b, c, so by commutativity, a + (b + c) = (b + c) + a = (a + b) + c.
[bc′] ∨ [b′c]
= abc′^ ∨ ab′c, while
ab + ac = (ab ∨ ac)
[ab]′^ ∨ [ac]′
= ab[ab]′^ ∨ ac[ab]′^ ∨ ab[ac]′^ ∨ ac[ac]′ = 0 ∨ ac[a′^ ∨ b′] ∨ ab[a′^ ∨ c′] ∨ 0 = aca′^ ∨ acb′^ ∨ aba′^ ∨ abc′ = 0 ∨ abc′^ ∨ 0 ∨ ab′c = abc′^ ∨ ab′c.
Conversely, starting from a Boolean ring A, define an ordering on A as follows: a ≤ b means that a = ab. Show that, with respect to this ordering, A is a Boolean lattice. In this way we obtain a one-to-one-correspondence between (isomorphism classes of) Boolean rings and (isomorphism classes of) Boolean lattices.
Write L = L(A) for ordered set. We verify the partial order axioms, the lattice axioms, and the Boolean algebra axioms.
Chapter 1: Rings and Ideals Ex. 1.
each distribute over the other. The complement V of an open and closed set U is open and closed, and is the unique V ⊆ P (X) with U ∪ V = X and U ∩ V = ∅. Thus B is a Boolean algebra. By [1.23.i] and [1.23.iii], B is the set of Xf for f ∈ A, so the correspondence φ : L → B taking f 7 → Xf is surjective. Since no nonzero element of A is nilpotent, by [1.17.ii], φ is also injective. To show φ is an order isomorphism (hence a Boolean algebra isomorphism), it remains to show that f ≤ g ⇐⇒ Xf ⊆ Xg. Now f ≤ g ⇐⇒ f = fg =⇒ f ∈ (g), and conversely if f = ag ∈ (g), then fg = (ag)g = ag = f. Thus f ≤ g ⇐⇒ f ∈ (g). Now f ∈ (g) ⇐⇒ ∃n ≥ 1
f = f n^ ∈ (g)
⇐⇒ f ∈ r
(g)
⇐⇒ r
( f )
⊆ r
(g)
and from the proof of [1.17.iv], r
( f )
⊆ r
(g)
⇐⇒ Xf ⊆ Xg. Therefore f ≤ g ⇐⇒ Xf ⊆ Xg, so φ is an order isomorphism.
26. Let A be a ring. The subspace of Spec(A) consisting of the maximal ideals of A, with the induced topology, is called the maximal spectrum of A and is denoted by Max(A). For arbitrary commutative rings it does not have the nice functorial properties of Spec(A) (see Exercise 21), because the inverse image of a maximal ideal under a ring homomorphism need not be maximal. Let X be a compact Hausdorff space and let C(X) denote the ring of all real-valued continuous functions on X (add and multiply functions by adding and multiplying their values). For each x ∈ X, let mx be the set of all f ∈ C(X) such that f(x) = 0. The ideal mx is maximal, because it is the kernel of the (surjective) homomorphism C(X) → R which takes f to f(x). If X denotes˜ Max
, we have therefore defined a mapping μ : X → X,˜ namely x 7 → mx. We shall show that μ is a homeomorphism of X onto X.˜ i) Let m be any maximal ideal of C(X), and let V = V(m) be the set of common zeros of the functions in m: that is, V = {x ∈ X : f(x) = 0 for all f ∈ m}. Suppose that V is empty. Then for each x ∈ X there exists fx ∈ m such that fx (x) 6 = 0. Since fx is continuous, there is an open neighborhood Ux of x in X on which fx does not vanish. By compactness a finite number of the neighborhoods, say Ux 1 ,... , Uxn , cover X. Let
f = f (^) x^21 + · · · + f (^) x^2 n. Then f does not vanish at any point of X, hence is a unit in C(X). But this contradicts f ∈ m, hence V is not empty. Let x be a point of V. Then m ⊆ mx, hence m = mx because m is maximal. Hence μ is surjective. ii) By Urysohn’s lemma (this is the only non-trivial fact required in the argument) the continuous functions separate the points of X. Hence x 6 = y =⇒ mx 6 = my, and therefore μ is injective. iii) Let f ∈ C(X); let Uf = {x ∈ X : f(x) 6 = 0 } and let U˜f = {m ∈ X˜ : f ∈/ m}. Show that μ (Uf ) = U˜f. The open sets Uf (resp. U˜f) form a basis of the topology of X (resp. X) and therefore˜ μ is a homeomorphism. Thus X can be reconstructed from the ring of functions C(X). For each f ∈ C(X), we have x ∈ Uf ⇐⇒ f(x) 6 = 0 ⇐⇒ f ∈/ mx ⇐⇒ mx ∈ U˜f, so μ restricts to a bijection Uf ↔ U˜f. It remains to show these sets form bases. The Uf will form a basis for X if whenever x ∈ W ⊆ X with W open, there is an f ∈ C(X) such that x ∈ Uf ⊆ W. But as X is compact Hausdorff, it is normal, and so the Urysohn lemma applies to show closed sets can be separated by continuous functions. Thus there is f ∈ C(X) such that f(X\W) = { 0 } and f(x) = 1, and then evidently x ∈ Uf ⊆ W. Each U˜f is open the subspace topology inherited from Spec
, being the intersection of X˜ with the open set Spec
f of [1.17]. As these sets form a basis for Spec
(see [1.17]), the U˜f form a basis for X˜.
Affine algebraic varieties
Ex. 1.27 Chapter 1: Rings and Ideals
27. Let k be an algebraically closed field and let
f α (t 1 ,... , tn) = 0 be a set of polynomial equations in n variables with coefficients in k. The set X of all points x = 〈x 1 ,... , xn〉 ∈ kn which satisfy these equations is an affine algebraic variety. Consider the set of all polynomials g ∈ k[t 1 ,... , tn] with the property that g(x) = 0 for all x ∈ X. This set is an ideal I(X) in the polynomial ring, and is called the ideal of the variety X. The quotient ring
P(X) = k[t 1 ,... , tn]/I(X)
is the ring of polynomial functions on X, because two polynomials g, h define the same polynomial function on X if and only if g − h vanishes at every point of X, that is, if and only if g − h ∈ I(X). Let ξ i be the image of ti in P(X). The ξ i ( 1 ≤ i ≤ n) are the coordinate functions on X: if x ∈ X, then ξ i(x) is the ith coordinate of x. P(X) is generated as a k-algebra by the coordinate functions, and is called the coordinate ring (or affine algebra) of X. As in Exercise 26, for each x ∈ Xlet mx be the ideal of all f ∈ P(X) such that f(x) = 0 ; it is a maximal ideal of P(X). Hence, if X˜ = Max
, we have defined a mapping μ : X → X, namely x˜ 7 → mx. It is easy to show that μ is injective: if x 6 = y, we must have xi 6 = yi for some i ( 1 ≤ i ≤ n), and hence ξ i − xi is in mx but not in my, so that mx 6 = my. What is less obvious (but still true) is that μ is surjective. This is one form of Hilbert’s Nullstellensatz (see Chapter 7). Abbreviate k[t] := k[t 1 ,... , tn]. As in [1.26], mx is maximal because it is the kernel of the surjective homomorphism f 7 → f(x) : P(X) → k. To be more explicit about what mx looks like, note that if x = 〈x 1 ,... , xn〉, then the polynomial function ξ i − xi ∈ P(X) vanishes at x, so that ξ i − xi ∈ mx. On the other hand, since (t 1 − x 1 ,... , tn − xn) is the kernel of the surjective homomorphism k[t] k taking 1 7 → 1 and ti 7 → xi, it is a maximal ideal of k[t], so by the correspondence (1.1) we have ( ξ 1 − x 1 ,... , ξ n − xn) ⊆ mx maximal, which shows the the containment must in fact be an equality. We then want to show the images of these mx are the only maximal ideals of P(X). By (1.1), it will suffice to do this for X = kn^ and P(X) = k[t] and then prove that x ∈ X ⇐⇒ I(X) ⊆ mx, which will be item ?? in a list of remarks that follows. First we show all maximal ideals of P(X) come from points. One way is to use an equivalent result traditionally called the weak Nullstellensatz; see [5.17] for a statement and proof. Another is to use Zariski’s Lemma ((5.24), [5.18], (7.9)) that any field L finitely generated as an algebra over a field K is a finite algebraic extension of K, which implies both. This is done in [5.19]. Here is a more elementary proof^16 avoiding the technology of Chapter 5, but it too runs through Zariski’s Lemma.
Lemma 1.27.1.* If an integral domain A is algebraic over a field k, then A is a field.
Proof. Let 0 6 = a ∈ A. Since A is a domain and a is algebraic over k, the kernel of the projection k[x] k[a] is a nonzero prime ideal. But k[x] is a PID, so this kernel is maximal, k[a] is a field, and a is a unit.
Proposition 1.27.2.* If k ⊆ L are fields in some integral domain B finitely generated as a k-algebra, L is algebraic over k.
Proof. Imagine there were a transcendental element a ∈ L, and include a in a finite set T of generators for B as a k-algebra. Select a maximal k-algebraically independent set S ⊆ T containing a, so that the field of fractions L′^ of B is a finite extension of k(S).^17 Picking an k(S)-basis of L′^ gives us a representation φ of multiplication on L′^ by square matrices over k(S). Writing the entries of the matrices φ (t) for t ∈ T as fractions in k[S], let g be the product of all the denominators, so that φ (B) has entries in k[S, g−^1 ]. If p ∈ k[a] is any irreducible element, then p−^1 ∈ L since L is a field, so φ (p−^1 ) is a diagonal matrix with entries p−^1. Then this entry lies in k[S, g−^1 ], so there is some positive power gm^ such that gm^ p−^1 ∈ k[S], meaning p|gm. Since p is irreducible and k[a] and k[S], being isomorphic to polynomial rings, are UFDs, it follows p is a scalar multiple of one of those finitely many irreducible factors qi ∈ k[S] of g which also (^16) [? , Prop. 1.5] (^17) This proof can be modified to use concepts from later on: pick a finite basis for L′ (^) over k(S); each basis element satisfies a monic polynomial with coefficients in k(S). Let g ∈ k[S] be a common multiple of the denominators of these coefficients. Then L′^ is integral over k[S]g by (5.3), so k[S]g is a field by (5.7) or [5.5.i]. But this is a contradiction, for the polynomial ring k[S] cannot be a Goldman domain by (5.18.1*).
