Atiyah/Macdonald Commutative Algebra

Atiyah/Macdonald Commutative Algebra Linus Setiabrata ls823@cornell.edu http://pi.math.cornell.edu/ ls823 I m not sure if these arguments are correct! Most things in [square brackets] are corrections to the text, contextual definitions, representatives of an equivalence class, etc. All things in [blue font square brackets] are my own comments or occasional reminders that a hint exists. I have tried to replace the book s notation (x) for the ideal generated by x with x, but this was a change I made midway through this project so who knows. This was last updated Sep 22, 2018. I guess if you know me well, you d know who this is dedicated for. :) 1 Rings and Ideals (Mid Chapter) Exercise 1.12. [Prove that] i) a (a : b) ii) (a : b)b a iii) ((a : b) : c) = (a : bc) = ((a : c) : b) iv) ( i a i : b) = i (a i : b) v) (a : i b i) = i (a : b i ). For part i), notice that ab aa = a since a is an ideal. In particular, for any x a, we have xb a. For part ii), recall that if x (a : b), then xb a. But a is also closed under addition, so finite sums i x ib i a whenever x i (a : b) and b i b. For part iii), observe that x ((a : b) : c) means that xc (a : b). So xcb a (using additive closure of a), that is, x (a : cb) = (a : bc). Exactly the same proof works for the second equality. For part iv), we have x ( i a i : b) precisely when xb i a i, that is, xb a i for all i. This is the same as saying x i (a i : b). For part v), we have x (a : i b i) precisely when x( i b i) a. For all i, xb i x( i b i) a, so (a : i b i) i (a : b i ). On the other hand, if xb i a for all i, then since a is closed under addition we have x( i b i) a, which gives equality. (Mid Chapter) Exercise 1.13. [Prove that] i) r(a) a ii) r(r(a)) = r(a) iii) r(ab) = r(a b) = r(a) r(b) 1

iv) r(a) = 1 a = 1 v) r(a + b) = r(r(a) + r(b)) vi) if p is prime, r(p n ) = p for all n > 0. For part i), if x a then x 1 a so x r(a). For part ii), notice that r(r(a)) r(a) so we just have to show the other inclusion. The set r(r(a)) consists of x A: x n r(a), and since x n r(a) there is an m so that (x n ) m a, but then x mn a so x r(a). For part iii), recall that ab a, b so x n ab means x n a b, so r(ab) r(a b). Equivalently, x n a and x n b so x r(a), r(b), and r(ab) r(a) r(b). Conversely, if x n a, b, then x 2n ab, so if x r(a) r(b) then x r(ab). Equivalently, if x n a b, that is, x r(a b), then again x 2n ab so x r(ab). For part iv), r(a) a we have a = 1 implies r(a) = 1. Conversely, if 1 r(a), then 1 n = 1 a, so 1 a. For part v), since a + b r(a) + r(b) we have r(a + b) r(r(a) + r(b)) (if a b, then x n a means that x n b so r(a) r(b)). Conversely if x n r(a) + r(b) then x n = a + b for some a r(a), b r(b). Say that a na a and b n b b. Observe that n a ( ) n x n(na+nb) = (a + b) na+n na b + n b a+n b = a i b na+nb i + i i=0 i=n a+1 ( na + n b i ) a i b na+n b i. The first summation contains only multiples of b n b, so it is an element of b; the second summation contains only multiples of a na, so it is an element of a + b. So x n(na+n b) a + b. For part vi), first observe that r(p) = p, since if x n p then x p by definition of prime ideal. We proceed by induction: since r(p n ) = r(p n 1 ) r(p) by part 3, applying the inductive hypothesis yields r(p n ) = p p = p. (Mid Chapter) Exercise 1.18. If [A, B are rings, f : A B is a homomorphism,] a 1, a 2 are ideals of A and if b 1, b 2 are ideals of B, then (a 1 + a 2 ) e = a e 1 + a e 2, (b 1 + b 2 ) c b c 1 + b c 2, (a 1 a 2 ) e a e 1 a e 2, (b 1 b 2 ) c = b c 1 b c 2, (a 1 a 2 ) e = a e 1a e 2, (b 1 b 2 ) c b c 1b c 2, (a 1 : a 2 ) e (a e 1 : a e 2), (b 1 : b 2 ) c [=](b c 1 : b c 2), r(a) e r(a e ), r(b) c = r(b c ). The set of ideals E is closed under sum and product, and C is closed under the other three operations. [E is the set of ideals {b: b ce = b} and C is the set of ideals {a: a ec = a}.] First we have x (a 1 + a 2 ) e if and only if x = i b if(a i,1 + a i,2 ) = i b if(a i,1 ) + i b if(a i,2 ) for a i,1 a 1 and a i,2 a 2, so x a e 1 + a e 2. If x a e 1 + a e 2, then x = i b i,1f(a i,1 ) + i b i,2f(a i,2 ) for a i,1 a 1 a 1 + a 2 and a i,2 a 2 a 1 + a 2, so x (a 1 + a 2 ) e. On the other hand, if x b c 1 + b c 2 then x = a 1 +a 2 where f(a 1 ) = b 1 b 1 and f(a 2 ) = b 2 b 2. Then f(x) = b 1 +b 2 b 1 +b 2, that is, x (b 1 +b 2 ) c. If x (a 1 a 2 ) e, that is, x = i b if(a i ) for b i B, a i a 1 a 2, then in particular a i a 1 so x a e 1, and a i a 2 so x a e 2. On the other hand, x (b 1 b 2 ) c if and only if f(x) b 1 b 2, if and only if f(x) b 1 and f(x) b 2, that is, x b c 1 b c 2. We show that if x a 1 a 2 then f(x) a e 1a e 2 [wisdom/hint stolen from Michelle heh]. The point is that since a e 1a e 2 is an ideal and contains the generators of (a 1 a 2 ) e, it also contains the whole ideal. Notice that 2

if x = i a i,1a i,2 for a i,1 a 1 and a i,2 a 2 then f(x) = i f(a i,1)f(a i,2 ) a e 1a e 2 as desired. Conversely, if x a e 1a e 2 then ( )( ) x = b i,1 f(a i,1 ) b j,2 f(a i,2 ) = b i,1 b j,2 f(a i,1 a i,2 ) (a 1 a 2 ) e. i j i j On the other hand, if x b c 1b c 2 then x = i a i,1a i,2 for some a i,1 f 1 (b 1 ) and a i,2 f 1 (b 2 ). But then f(x) = i f(a i,1)f(a i,2 ) so x (b 1 b 2 ) c. If x (a 1 : a 2 ) then for any y = i b if(a i,2 ) a e 2 we have f(x)y = i b if(xa i,2 ), where xa i,2 a 1. So f(x)y a e 1, that is, f(x) (a e 1 : a e 2). On the other hand, if x (b 1 : b 2 ) c, that is, f(x) b 1 : b 2, then for any y b c 2 we have f(y) b 2 so f(x)f(y) b 1 and xy b c 1. Hence x (b c 1 : b c 2). In fact, conversely, if x (b c 1 : b c 2) then for all y b c 2 we have xy b c 1; equivalently, for all f(y) b 2 we have f(x)f(y) b 1, so f(x) (b 1 : b 2 ). If x r(a) then x n a. But then f(x n ) = (f(x)) n a e, so f(x) r(a e ). On the other hand, if x r(b) c if and only if (f(x)) n = f(x n ) b, if and only if x n b c, that is, x r(b c ). Finally, if b 1, b 2 E then b 1 = a e 1 and b 2 = a e 2. Thus and similarly b 1 + b 2 = a e 1 + a e 2 = (a 1 + a 2 ) e = (a 1 + a 2 ) ece = ((a 1 + a 2 ) e ) ce ) = (a e 1 + a e 2) ce = (b 1 + b 2 ) ce b 1 b 2 = a e 1a e 2 = (a 1 a 2 ) e = (a 1 a 2 ) ece = ((a 1 a 2 ) e ) ce ) = (a e 1a e 2) ce = (b 1 b 2 ) ce. Also, if a 1, a 2 C then a 1 = b c 1 and a 2 = b c 2. Thus and similarly a 1 a 2 = b c 1 b c 2 = (b 1 b 2 ) c = (b 1 b 2 ) cec = ((b 1 b 2 ) c ) ec = (b c 1 b c 2) ec = (a 1 a 2 ) ec (a 1 : a 2 ) = (b c 1 : b c 2) = (b 1 : b 2 ) c = (b 1 : b 2 ) cec = ((b 1 : b 2 ) c ) ec = (b c 1 : b c 2) ec = (a 1 : a 2 ) ec and finally r(a) = r(b c ) = r(b) c = r(b) cec = (r(b) c ) ec = r(b c ) ec = r(a) ec. Exercise 1: Let x be a nilpotent element of a ring A. Show that 1 + x is a unit of A. Deduce that the sum of a nilpotent element and a unit is a unit. Let n be such that x n = 0. We claim that (1 + x) 1 = n 1 i=0 ( x)i ; indeed, ( n 1 ) (1 + x) ( x) i = 1 + ( 1) n 1 x n = 1. i=0 Now if u is a unit and x is nilpotent, then u 1 x is also nilpotent (if x n = 0, then (u 1 x) n = u n x n = 0). Therefore, 1 + u 1 x is a unit, and u(1 + u 1 x) = u + x is also a unit (units of A form a subgroup under multiplication). Exercise 2: Let A be a ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a 0 + a 1 x + + a n x n A[x]. Prove that i) f is a unit in A[x] a 0 is a unit in A and a 1,..., a n are nilpotent. [Hint] ii) f is nilpotent a 0, a 1,..., a n are nilpotent. iii) f is a zero-divisor there exists a 0 in A such that af = 0. [Hint] 3

iv) f is said to be primitive if (a 0, a 1,..., a n ) = 1. Prove that if f, g A[x], then fg is primitive f and g are primitive. For part i), we first prove the backward direction with induction on n. If n = 0, then the claim holds. Now let f = a 0 + a 1 x + + a n x n with a 0 a unit and a 1,..., a n nilpotent. and define ˆf = f a n x n ; by induction, it has an inverse ĝ = b 0 + b 1 x + + b m x m, that is, ˆfĝ = 1. Then fĝ = 1 + an x n ĝ. But a n is nilpotent in A, so a n x n ĝ is nilpotent in A[x]. By Exercise 1, fĝ is a unit and has an inverse ĥ. So fĝĥ = 1, and ĝĥ is an inverse for f. For the forward direction, we follow the hint in the book. Fix a polynomial f = a 0 + a 1 x + + a n x n is a unit with n 1 (for n = 0 this is trivial). If g = b 0 + b 1 x + + b m x m is its inverse, then we claim a r+1 n b m r = 0 for all 0 r n. Indeed, the coefficient of x n+m in fg is a n b m = 0. Now for 0 r m, consider the coefficient of x n+m r. On one hand, it s 0, because n 1, and on the other hand, it is equal to a n b m r + a n 1 b m r+1 + + a n r b m, so multiplying by a r n doesn t change the fact that the above expression is 0. But this is a r+1 n b m r + a r na n 1 b m r+1 + + a r na n r b m = 0, but every term except for the first one is zero by induction. So a r+1 n b m r = 0 too. For r = m, this says that a m+1 n b 0 = 0, but b 0 = a 1 0 is a unit and is not a zero-divisor. So a m+1 n = 0 and a n is nilpotent in A. Hence a n x n is nilpotent in A[x], and by Exercise 1, g := f(x) + ( a n x n ) is the sum of a nilpotent and a unit, and is hence a unit. But g is a unit with leading coefficient a n 1. We can repeat this process whenever deg g 1 to reduce to the case where g is constant. For part ii), if f is nilpotent then 1 + f is a unit by Exercise 1, so a 0 + 1 is a unit and a 1,..., a n are nilpotent. Furthermore a 0 is nilpotent because if f n = 0 then the constant term is a n 0 = 0. Conversely, if a 0,..., a n are nilpotent, then there are integers m 0,..., m n so that a mi i = 0. Let N = i m i, and observe that ( n ) N ( ) f N = a i x i N = a c0 0 a 0,..., a ac1 1... acn n x c1+2c2+ +ncn. n i=0 c 0+ +c n=n But if c i = m i then for some i we have c i m i, and so a ci i summation has a coefficient of 0, so f N = 0. = 0, that is, each term in the above For part iii), we induct on the degree of f (the forward direction is trivial). If f is constant then this is also trivial. Say that f = a 0 + a 1 x + + a n x n is a degree n polynomial which is a zero divisor, so there is g = b 0 + b 1 x + + b m x m such that fg = 0. Then in particular the coefficient of x n+m in fg is a n b m = 0. But then b m f has degree at most n 1, so there is a 0 A so that a(b m f) = 0. For part iv), given f A[x] denote by c(f) to be the ideal generated by the coefficients of f. Notice that c(fg) c(f)c(g), because any coefficient of fg is in the ideal c(f)c(g). If either c(f) or c(g) are strictly smaller than A then so is c(f g), which proves the forward direction. For the backward direction, say that f = a 0 + a 1 x + + a n x n and g = b 0 + b 1 x + + b m x m, and fg = c 0 + c 1 x + + c m+n x m+n. If c 1,..., c m+n m but some a i, b j m. Let a I, b J be the minimal indices so that a I, b J m. Then c I+J = I+J i=0 a ib I+J i m, and each a i b I+J i m for i I; we have c I+J I+J i=0,i I a i b I+J i = a I b J m, but since m is prime it follows that either a I or b J m, contrary to assumption. Exercise 3: Generalize the results of Exercise 2 to a polynomial ring A[x 1,..., x r ] in several indeterminates. 4

For part i), first observe that A[x 1,..., x r ] = (A[x 1,..., x r 1 ])[x r ], that is, f A[x 1,..., x r ] can always be written has f = f 0 (x 1,..., x r 1 ) + f 1 (x 1,..., x r 1 )x r + + f n (x 1,..., x r 1 )x n r for some f i A[x 1,..., x r 1 ]. So Exercise 2 part i) guarantees that f is a unit if and only if f 0 is a unit in A[x 1,..., x r 1 ] and f i is nilpotent for all other i. Thus induction and repeated use of Exercise 2 part i) and ii) gives us that f is a unit if and only if a 0 (the constant term) is a unit in A and all other coefficients are nilpotents in A. For part ii), induction and Exercise 2 part ii) yields that f = f 0 (x 1,..., x r 1 ) + f 1 (x 1,..., x r 1 )x r + + f n (x 1,..., x r 1 )x n r is nilpotent if and only if each coefficient is nilpotent. [Part iii) not done] [Part iv) not done] Exercise 4: [Prove that] In the ring A[x], the Jacobson radical is equal to the nilradical. If f is in the nilradical n, that is, f is in every prime ideal, then it is also in every maximal ideal (because maximal implies prime), so f is in the Jacobson J. On the other hand, if f = a 0 +a 1 x+ +a n x n J then 1 fg is a unit for all g A[x] by Proposition 1.9, but for g(x) = 1 this says that 1 + f is a unit, so by Exercise 2.1 we have 1 + a 0 is a unit and a 1,..., a n are nilpotent. Furthermore for g(x) = x this says 1 + fx is a unit, which implies that 1 is a unit and a 0 is nilpotent. Hence f is nilpotent, and is in n. Exercise 5: Let A be a ring and let A[[x]] be the ring of formal power series f = n=0 a nx n with coefficients in A. Show that i) f is a unit in A[[x]] a 0 is a unit in A. ii) If f is nilpotent, then a n is nilpotent for all n 0. Is the converse true? iii) f belongs to the Jacobson radical of A[[x]] a 0 belongs to the Jacobson radical of A. iv) The contraction of a maximal ideal m of A[[x]] is a maximal ideal of A, and m is generated by m c and x. v) Every prime ideal of A is the contraction of a prime ideal of A[[x]]. For part i), first notice that the forward direction is trivial: if fg = 1 with g = n=0 b nx n then a 0 b 0 = 1 and a 0 is a unit. Conversely, if a 0 is a unit then define b 0 = a 1 0 and for n 1 iteratively define Then n=0 b nx n =: g is the desired inverse. b n := a 1 0 (a 1b n 1 + a 2 b n 2 + + a n b 0 ). For part ii), we induct on the a i in the following way: obviously a 0 is nilpotent (if f n = 0, the constant term in f n is a n 0 = 0), and we claim that if f is nilpotent, and additionally a 0,..., a k is nilpotent, then a k+1 is also nilpotent. Indeed, define f k (x) = a 0 + a 1 x + + a k x k, and note that f k is nilpotent (by Exercise 2 part ii), which we can apply here because of the ring homomorphism id: A[x] A[[x]]). Now if f and f k are nilpotent then so is f f k since the nilradical is an ideal and in particular is closed under addition (alternatively, if f n = 0 and f n k k = 0 then (f f k ) n+n k = 0). But the coefficient of x (k+1)n in (f f k ) n is a n k+1, so a k+1 is nilpotent because f f k is nilpotent. Note that the converse is not true: let A = Z[ε 0, ε 1,... ]/(ε 2 0 = 0, ε 2 1 = 0,... ) be the ring of (countably many) infinitesimals. Then, consider f = i ε ix i : each coefficient is by definition nilpotent, but f itself is not because the coefficient of x (n 2) in f n has a nonzero multiple of ε 0 ε 1... ε n 1, and so f n is not the zero polynomial. 5

For part iii), notice that f is in the Jacobson radical if and only if for all g = n=0 b nx n A[[x]] we have 1 fg a unit. By part i) 1 fg is a unit if and only if 1 a 0 b 0 is a unit, so 1 fg is a unit for all g if and only if 1 a 0 b 0 is a unit for all b 0 (of course, this happens if and only if a 0 is in the Jacobson radical of A). [Part iv) not done] [Part v) not done] Exercise 6: A ring A is such that every ideal not contained in the nilradical contains a nonzero idempotent (that is, an element e such that e 2 = e 0). Prove that the nilradical and Jacobson radical of A are equal. Let J denote the Jacobson radical and n the nilradical. We have J n (maximal implies prime), so we have to show J\n is empty. If there was an element x J\n, then consider the ideal xa (which is not contained in the nilradical). Thus there is a nonzero idempotent xa xa. But then (xa) 2 xa = 0, so xa(xa 1) = 0, that is, xa 1 is a zero divisor. But then it cannot be a unit, which contradicts Proposition 1.9 (since x J by assumption). Exercise 7: Let A be a ring in which every element x satisfies x n = x for some n > 1 (depending on x). Show that every prime ideal in A is maximal. Let p be a prime ideal, and let x p. We will show that p, x = 1 : this implies that p is maximal (if it is strictly contained in a maximal ideal m, pick an x m\p and observe that p, x m 1 ). Indeed, 0 = x n x = x(x n 1 1) p, so since p is prime and x p we have x n 1 1 p. So x n 1 1 p, x. On the other hand, since x p, x we also have x n 1 p, x, so 1 p, x. Exercise 8: Let A be a ring 0. Show that the set of prime ideals of A has minimal elements with respect to inclusion. The set of prime ideals of A form a poset under reverse inclusion (ie. p q if p q). Let {p i } i I be a chain. We claim that p := i I p i is an upper bound. In particular, first observe that p is a prime ideal: define the two sets I x := {i: x p i } I y := {i: y p i } and observe that I x I y = I, since each p i is prime and hence contains either x or y. We further claim that either I x = I or I y = I: if not, then we can find i x I x \I y and i y I y \I x. This leads to a contradiction, because either p ix p iy or p ix p iy ; in the former case we have x p iy so i y I x, and in the latter case we have y p ix, so i x I y. Thus p is prime, and is contained in every p i so it is an upper bound. So there is a maximal element by Zorn s Lemma. Exercise 9: Let a be an ideal (1) in a ring A. Show that a = r(a) a is an intersection of prime ideals. The forwards direction is just Proposition 1.14: we have r(a) = p a p is an intersection of prime ideals, so if a = r(a) then a, too, is an intersection of prime ideals. The backwards direction follows from (Mid Chapter) Exercise 1.13.iii) and 1.13.vi): if a = α p α then ( ) r(a) = r p α = r(p α ) = p α = a. α α α Exercise 10: Let A be a ring, R its nilradical. Show that the following are equivalent: i) A has exactly one prime ideal; ii) every element of A is either a unit or nilpotent; 6

iii) A/R is a field. We first show that i) implies ii). Recall that A has at least one maximal ideal (Theorem 1.3), and maximal ideals are prime: it follows that the unique prime ideal of A, say p, is maximal. Furthermore, p = R since the nilradical is the intersection of all prime ideals. So p consists of only nilpotent elements. On the other hand Corollary 1.5 states that every non-unit of A is contained in a maximal ideal, so A\p consists only of units. Then, ii) implies iii) because any nonzero [x] A/R is invertible (since x 1 A exists, hence so does [x 1 ] = [x] 1 ). Finally, iii) implies i) because R is an ideal so if A/R is a field then R is a maximal ideal. But it is contained in all the prime ideals: if there were at least two distinct prime ideals, then R would be strictly contained in one of these ideals, and wouldn t be maximal. Exercise 11: A ring A is Boolean if x 2 = x for all x A. In a Boolean ring A, show that i) 2x = 0 for all x A; ii) every prime ideal p is maximal, and A/p is a field with two elements; iii) every finitely generated ideal in A is principal. For part i), we compute 4x = 4x 2 = (2x) 2 = 2x for all x A, and subtracting 2x from both sides yields 2x = 0. For part ii), exercise 7 guarantees that every prime ideal is maximal. Notice that A/p is a field with [x] 2 = [x 2 ] = [x] for every element [x] A/p. But in a field, this can only happen to [0] and [1] (since [x] is now invertible if [x] [0] and [x] 2 = [x] implies [x] = [1]). For part iii), first notice that x 1 (x 2 x 1 x 2 ) = 0, so that x 1 (x 1 +x 2 x 1 x 2 ) = x 1 and x 2 (x 1 +x 2 x 1 x 2 ) = x 2. It follows that every ideal with two generators is principal: (x 1, x 2 ) (x 1 +x 2 x 1 x 2 ) (the reverse inclusion is obvious). Now we can induct on the number of generators, since (x 1, x 2,..., x r ) = (x 1 +x 2 x 1 x 2, x 3,..., x r ). [The choice of x 1 + x 2 + x 1 x 2 is actually motivated by the hint in Exercise 24] Exercise 12: [Prove that] A local ring contains no idempotent 0, 1. Let x be an idempotent that is not 0 or 1. Since x 2 x = x(1 x) = 0, neither x nor 1 x are units since they are zero-divisors (notice that 1 x is not 0 or 1, either, so they are both nontrivial zero-divisors). But Corollary 1.5 says that these are both contained in some maximal ideal; since there is only one maximal ideal m we have x, 1 x m so their sum is in m, a contradiction. [I can t quite prove it [probably because it s false], but this proof takes me tantalizingly close to a local ring contains no x 0, 1 satisfying x n = x...] Exercise 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 x f, one for each f Σ. Let a be the ideal of A generated by the polynomials f(x f ) for all f Σ. Show that a (1). 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 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=1k n. 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. [not done] Exercise 14: In a ring A, let Σ be the set of ideals in which every element is a zero-divisor. Show that the set Σ has maximal elements [wrt inclusion?] 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. 7

[not done] Exercise 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)). ii) V (0) = X, V (1) =. iii) if (E i ) i I is any family of subsets of A, then ( ) V E i = V (E i ). i I iv) V (a b) = V (ab) = V (a) V (b) for any ideals a, b of A. i I 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). For part i), notice that E a so if p a then p E. So for the first equality it remains to show that if p E then p a. But this follows from the fact that p is an ideal [michelle wisdom again!]: if y a then y = i a ix i for x i E, a i A, and since each x i p we have y p. The second equality is similar: since a r(a), if p r(a) then p a. For the other direction, notice that if p a, then if y r(a), that is, y n a for some n > 0, we have y n p and since p is a prime ideal then y p. For part ii), just observe that every proper ideal contains 0 and does not contain 1. For part iii), just observe that p V ( i E i ) precisely when p i E i, which happens if and only if p E i for each i, that is, p i V (E i ). For part iv), we have that a b ab, so if p a b then p ab. On the other hand, if p ab, then if x a b then x 2 ab, and since p is prime if x 2 p then x p, that is, p ab. Additionally, if p a or p b, then p a b because a b a and a b b. Conversely, if p a b, then Proposition 1.11 says that p a or p b. Exercise 16: Draw pictures of Spec(Z), Spec(R), Spec(C[x]), Spec(R[x]), Spec(Z[x]). I refuse to draw these pictures, so I ll just compute the spectra. An ideal n Z is prime if and only if n is prime; a closed set V (I) = V ( n ) consists of the finite set { p } p n of prime ideals, and conversely, any finite set of prime ideals { p 1,..., p n } all contain p 1... p n (and are the only ideals to do so). There are countably many primes, so Spec(Z) is the cofinite topology on Z. Observe that R is a field, so the only prime ideal is {0}. Hence Spec(R) is a single point with the only topology it could have. Observe that C is algebraically closed, so that if (z α 1 )... (z α n ) p for some prime ideal p, we have that z α i p for some i. It follows that p contains some monomials. Observe that p actually only has exactly one monomial; if z z 1, z z 2 p, then z 1 z 2 p and since z 1 z 2 C it is invertible and p = 1. Thus, just like in the Z case, Spec(C[x]) is the complex plane endowed with the cofinite topology: for an ideal I, define X I = {z C: f(z) = 0 f I}, and observe that X I is nonempty (if it was empty, take f, g with no common roots and notice that 1 = gcd(f, g) I), so say X I = {z 1,..., z n } (it s finite because any polynomial has at most finitely many roots) and observe that {(z z 1 ),..., (z z n )} are precisely the prime ideals that contain I. Conversely, any finite set of prime ideals {(z z 1 ),..., (z z n )} all contain the ideal generated by (z z 1 )... (z z n ) [Spec(R[x]), Spec(Z[x])] 8

Exercise 17: For each f A, let X f denote the complement of V (f) in X = Spec(A). The sets X f are open. Show that they form a basis of open sets for the Zariski topology, and that i) X f X g = X fg ; ii) X f = f is nilpotent; iii) X f = X f is a unit; iv) X f = X g r( f ) = r( g ); v) X is quasi-compact (that is, every open covering of X has a finite subcovering). [Hint] vi) More generally, each X f is quasi-compact. vii) An open subset of X is quasi-compact if and only if it is a finite union of sets X f. The sets X f are called basic open sets of X = Spec(A). We first show that the X f form a basis of open sets for the Zariski topology. Indeed, let U be an open set, so that U c = V (E) for some subset E of A. In particular, ( ) U c = V f = V (f) so ( U = f E f E ) c V (f) = f E f E V (f) c = X f. For part i), note that X f consists of the prime ideals p that do not contain f. Thus p is in X f X g if and only if f, g p. Notice that f, g p if and only if fg p (the negation of the former is f or g p and that of the latter is fg p; the forward direction is the definition of ideal and the backward direction is the definition of prime ). For part ii), recall the nilradical is the intersection of all the prime ideals (Proposition 1.8), that is, f is nilpotent if and only if it is in all prime ideals. Also, X f is the set of prime ideals that do not contain f. For part iii), the backwards direction follows from the fact that no (proper) ideal contains any unit. Conversely, every nonunit is contained in a maximal ideal (Corollary 1.5), which are in particular prime. The contrapositive of this gives the forward direction. For part iv), recall that for an ideal a, Proposition 1.14 says that r(a) is the intersection of the prime ideals containing a. Then, for the forward direction, let p be an ideal containing f, so that in particular it contains f. Since X f = X g we have that p contains g, too, and so it contains g. Dually if p g then p f, so p = p p f since the intersection is taken over the same set of prime ideals. For the backwards direction, say that f p. Then f p, so we have p p = r( f ) = r( g ) {g}, p f p g so g p, and dually if g p then f p. So f p if and only if g p. [Part v) not done] [Part vi) not done] [Part vii) not done] f E 9

Exercise 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 was a point of X = Spec(A). When thinking of x as a prime ideal of A, we denote it by p x (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) p x is maximal; ii) {x} = V (p x ); iii) y {x} p x p y ; 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). For part i), recall that {x} = V (I) for some I if and only if p x is the unique ideal containing I. By definition, I is maximal, and in fact I = p x. For part ii), first notice that if I 1 I 2, then V (I 1 ) V (I 2 ). Now observe that any closed set E = V (I) has x E if and only if I p x, so E V (p x ). On the other hand, x V (p x ). For part iii), just use part ii), that is, {x} = V (p x ), so that y V (p x ) is by definition p y p x. For part iv), notice that if every open set contains either both or none of x, y then in particular all the open sets X f = V (f) either contain both or none of x, y. In other words, f p x precisely when f p y, so p x = p y. Exercise 19: A topological space X is said to be irreducible if X 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. Recall that the sets X f = V (f) form a basis for the open sets in X, and observe that X f is empty if and only if f is nilpotent (because the nilradical is the intersection of all the prime ideals; its elements are precisely those contained in every prime ideal). It is enough to show that whenever V f and V g are nonempty, V f V g is nonempty too (given two general nonempty open sets U 1, U 2, we can find nonempty V f U 1 and V g U 2 since the V f form a basis of open sets). Indeed, by Exercise 17 part i), V f V g = V fg is nonempty whenever fg is not nilpotent. So X is irreducible if and only if f, g non-nilpotent implies fg non-nilpotent (contrapositively, f g nilpotent implies f or g nilpotent). Exercise 20: Let X be a topological space. i) If Y is an irreducible subspace of X, then [prove that] the closure Y of Y in X is irreducible. ii) [Prove that] every irreducible subspace is contained in a maximal irreducible subspace. iii) [Prove that] the maximal irreducible subspaces are closed and cover X. They are called the irreducible components of X. What are the irreducible components of a Hausdorff space? iv) If A is a ring and X = Spec(A), then [prove that] the irreducible components of X are the closed sets V (p), where p is a minimal prime ideal of A. For part i), let U 1, U 2 be nonempty open subsets of Y ; we have U 1 Y and U 2 Y nonempty open subsets of Y, and since Y is irreducible these intersect nontrivially. [Part ii) not done] For part iii), recall that the closure of an irreducible subspace is again irreducible (part i)), so maximal irreducible subspaces are closed. Furthermore, they cover X because each singleton set {x} X is an irreducible subspace, and is hence contained in a maximal irreducible subspace. The irreducible components of a Hausdorff space are precisely the singleton sets; any subspace Y X containing two points y 1, y 2 contains 10

two disjoint open sets U 1, U 2 (with y 1 U 1, y 2 U 2, though this is not necessary), and cannot be irreducible. [Part iv) not done] Exercise 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 (X f ) = Y ϕ(f), and hence that ϕ is continuous. ii) If a is an ideal of A, then (ϕ ) 1 (V (a)) = V (a e ). iii) If b is an ideal of B, then ϕ (V (b)) = V (b c ). 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/R) (where R is the nilradical of A) are naturally homeomorphic.) v) If ϕ is injective, then ϕ (Y ) is dense in X. More precisely, ϕ (Y ) is dense in X ker(ϕ) R. vi) Let ψ : B C be another ring homomorphism. Then [prove that] (ψ ϕ) = ϕ ψ. vii) Let A be an integral domain with just one non-zero prime ideal p, and let K be the field of fractions of A. Let B = (A/p) K. Define ϕ: A B by ϕ(x) = ( x, x), where x is the image of x in A/p. Show that ϕ is bijective but not a homeomorphism. For part i), observe that (ϕ ) 1 (X f ) consists precisely of the y Y so that ϕ (y) X f, that is, the ideals p y B such that f ϕ 1 (p y ). This is true if and only if ϕ(f) p y, that is, y Y ϕ(f). For part ii), again observe that (ϕ ) 1 (V (a)) consists precisely of the y Y so that ϕ (y) V (a), that is, the ideals p y B such that ϕ 1 (p y ) a. But then p y ϕ(ϕ 1 (p y )) ϕ(a). But since p y is an ideal we in fact have p y (ϕ(a)) = a e, so indeed y V (a e ). Conversely, if p y a e then ϕ 1 (p y ) a ec a. For part iii), we first notice that ϕ (V (b)) V (b c ), since ϕ (V (b)) consists of ideals {ϕ 1 (p y ): p y b} which all contain ϕ 1 (b) = b c. Conversely, if all ϕ 1 (p y ) contain some ideal I, then the ideal p y b ϕ 1 (p y ) = ϕ 1 ( p y b p y ) = ϕ 1 (r(b)) also contains I. In particular, I (r(b)) c = r(b c ) (recall the last statement of Mid Chapter Exercise 1.18), so V (I) V (r(b c )) = V (b c ) (recall Exercise 15). For part iv), first notice that ϕ really is a map Y V (ker(ϕ)), since ϕ 1 (p y ) ϕ 1 (0) = ker(ϕ)). To see that it is a homeomorphism, first recall that it is continuous by part i), and that ϕ(ϕ 1 (I)) = I because ϕ is surjective. So ϕ is injective because if ϕ 1 (p) = ϕ 1 (q) then their images under ϕ are equal, so p = q. Furthermore ϕ is surjective because [idk why lol], and so the inverse defined by [idk lol] is also continuous because [idk why lol]. In particular, the natural homomorphism ϕ: A A/R gives a homeomorphism Spec(A/R) V (R) Spec(A). Recall that R = r(0), so that V (R) = V (0) = Spec(A). For part v), recall that Y = V (0) so by part iii) ϕ (V (0)) is dense in V (ker(ϕ)) = V (0) = X. More precisely, if ker(ϕ) R, then if p R is a prime ideal then p ker(ϕ). This implies that X = V (R) V (ker(ϕ)). Then part iii) asserts that ϕ (Y ) = ϕ (V (0)) is dense in V (ker(ϕ)) = X. Conversely, if ϕ (Y ) is dense in X, then X ϕ (V (0)) = V (ker(ϕ)). If V (ker(ϕ)) = X, this means that every prime ideal contains ker(ϕ). In particular, the intersection of all prime ideals (that is, R) contains ker(ϕ). For part vi), it is enough to recall that ϕ 1 ψ 1 = (ψ ϕ) 1. For part vii), notice that both Spec(A) and Spec(B) are both two point spaces [observation is due to michelle]: of course, {0} A is a prime ideal since A is an integral domain, and p is the unique nonzero 11

prime ideal by assumption. There are also two prime ideals in B: observe that if (x, y) I, where I is an ideal of B, and x, y 0, then x 1 and y 1 exist since p is in fact maximal so A/p is a field. Hence (x, y) (x 1, y 1 ) = (1, 1) I so A/p K I. So a prime ideal of B contains only points of the form (x, 0) or (0, y). In the former case, if x 0 then again (x, 0) (x 1, 1) = (1, 0) I and hence A/p {0} I, and if y 0 then {0} K I. The only other ideal of B is {0} {0}. Notice that {0} {0} is not a prime ideal, since (0, 1) (1, 0) {0} {0}, and that the other two ideals A/p {0} and {0} K are prime, since both A/p and K are fields (and in particular integral domains). We now show that ϕ is a bijection and not a homeomorphism. Indeed, the Zariski topology on Spec(A) consists of the closed sets, {0}, {0, p}, so the open sets are, {p}, {0, p}. On the other hand, the Zariski topology on Spec(B) is the discrete topology. Since ϕ 1 (A/p {0}) = {(0, 0)} and ϕ 1 ({0} K) = p this really is a bijection; it is not a homeomorphism because ϕ is not an open mapping (it sends the open set {0} K} to the singleton set {p}, which is not open in Spec(A)). Exercise 22: Let A = n i=1 A i be the direct product of rings A i. Show that Spec(A) is the disjoint union of open (and closed) subspaces X i, where X i is canonically homeomorphic with Spec(A i ). Conversely, let A be any ring. Show that the following statements are equivalent: i) X = Spec(A) is disconnected. ii) A = A 1 A 2 where neither of the rings A 1, A 2 is the zero ring. iii) A contains an idempotent 0, 1. In particular, the spectrum of a local ring is always connected. Let p be a prime ideal of A. We claim that p = A 1 p i A n for some prime ideal p i of A i. To shorten notation, for this problem we will denote p i := A 1 p i A n. Indeed, say that there were two distinct indices j, k so that there existed (0,..., y j,..., 0) p and (0,..., y k,..., 0) p, then p couldn t be prime (because their product is (0,..., 0) p). Thus, if π i : A A i is the restriction onto the ith entry, we must have π i (p) = A i for all but at most one i (if it was a strict subset of A i for two indices j, k, then take elements in the complement y j A j \π j (p) and y k A k \π k (p) and observe that (0,..., y j,..., 0) p, and (0,..., y k,..., 0) p). Since p A, there must exist a unique index i so that p = p i for some ideal p i ; we should prove that p i is prime. Indeed, let (0,..., x i y i,..., 0) = (0,..., x i,..., 0) (0,..., y i,..., 0) p. Since p is prime either x i π i (p) or y i π i (p); it follows that π i (p) is a prime ideal of A i. It follows that p = A 1 p i A n. Thus as a set Spec(A) = X i, where X i is the set of prime ideals p i. Endowed with the Zariski topology, X i is closed because 0 i is contained by precisely X i, and X i is open because 0 A i 0 is contained by precisely X 1 ˆX i X n ; we also have that X i = Spec(Ai ) via the homeomorphism p i p i. For the converse, we observe that ii) is equivalent to iii): if A = A 1 A 2 then (0, 1) 2 = (0, 1) is idempotent, and conversely if x 2 = x then (1 x) 2 = 1 2x + x 2 = 1 x. It is enough to observe that x 1 x = 0 since if xa = (1 x)b then multiplying both sides by x yields xa = x(1 x)b = 0, and x + 1 x contains x+(1 x) = 1. Thus A = x 1 x, because if I and J are ideals of A such that I J = 0 and I +J = A, the map ϕ: I J A given by (i, j) i + j is an isomorphism: it is certainly a homomorphism since ϕ(i 1, j 1 ) ϕ(i 2, j 2 ) = (i 1 + j 1 )(i 2 + j 2 ) = i 1 i 2 + i 1 j 2 + j 1 i }{{} 2 +j 1 j 2 = i 1 i 2 + j 1 j 2 = ϕ(i 1 i 2, j 1, j 2 ) =0+0 because i 1 j 2, j 1 i 2 I J, so ϕ(i 1, j 1 ) ϕ(i 2, j 2 ) = ϕ(i 1 i 2, j 1, j 2 ); it is surjective because I + J = A and it is injective because ϕ(i, j) = 0 means that i = j, which implies i, j I J and i = j = 0. [i) and ii) not done] Exercise 23: Let A be a Boolean ring, and let X = Spec(A). 12

i) [Prove that] for each f A, the set X f is both open and closed in X. ii) Let f 1,..., f n A. Show that X f1 X fn = X f for some f A. iii) [Prove that] The sets X f are the only subsets of X which are both open and closed. [Hint] iv) [Prove that] X is a compact Hausdorff space. Exercise 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. 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). Conversely, starting from a Boolean ring A, define an order on A as follows: a b means that a = ab. Show that, with respect to this ordering, A is a Boolean lattice. [Hint] In this way we obtain a one-toone correspondence between (isomorphism classes of) Boolean rings and (isomorphism classes of) Boolean lattices. Exercise 25: From the last two exercises deduce Stone s theorem, that every Boolean lattice is isomorphic to the lattice of open-and-closed subsets of some compact Hausdorff topological space. Exercise 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), 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 m x be the set of all f C(X) such that f(x) = 0. The ideal m x is maximal, because it is the kernel of the (surjective) homomorphism C(X) R which takes f to f(x). If X denotes Max(C(X)), we have therefore defined a mapping µ: X X, namely x m x. 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 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 f x m such that f x (x) 0. Since f x is continuous, there is an open neighborhood U x of x in X on which f x does not vanish. By compactness a finite number of the neighborhoods, say U x1,..., U xn, cover X. Let f = f 2 x 1 + + f 2 x n. 13

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 m x, hence m = m x 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 y = m x m y, and therefore µ is injective. iii) Let f C(X); let U f = {x X : f(x) 0} and let Ũ f = {m X : f m}[.] Show that µ(u f ) = Ũf. The open sets U f (resp. Ũ f ) form a basis of the topology of X (resp. therefore µ is a homeomorphism. Thus X can be reconstructed from the ring of functions C(X). X) and Exercise 27: Let k be an algebraically closed field and let f α (t 1,..., t n ) = 0 be a set of polynomial equations in n variables with coefficients in k. The set X of all points x = (x 1,..., x n ) k n which satisfy these equations is an affine algebraic variety. Consider the set of all polynomials g k[t 1,..., t n ] with the property that g(x) = 0 for all x X. The 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,..., t n ]/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 t i 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 X let m x 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(P (X)), we have defined a mapping µ: X X, namely x mx. It is easy to show that µ is injective: if x y, we must have x i y i for some i (1 i n), and hence ξ i x i is in m x but not in m y, so that m x m y. What is less obvious (but still true) is that µ is surjective. This is one form of Hilbert s Nullstellensatz. Exercise 28: Let f,..., f m be elements of k[t 1,..., t n ]. They determine a polynomial mapping ϕ: k n k m : if x k n, the coordinates of ϕ(x) are f 1 (x),..., f m (x). Let X, Y be affine algebraic varieties in k n, k m respectively. A mapping ϕ: X Y is said to be regular if ϕ is the restriction to X of a polynomial mapping from k n to k m. If η is a polynomial function on Y, the η ϕ is a polynomial function on X. Hence ϕ induces a k- algebra homomorphism P (Y ) P (X), namely η η ϕ. Show that in this way we obtain a one-to-one correspondence between the regular mappings X Y and the k-algebra homomorphisms P (Y ) P (X). 14

2 Modules (Mid Chapter) Exercise 2.2. [Prove that] i) Ann(M + N) = Ann(M) Ann(N). ii) (N : P ) = Ann((N + P )/N) For part i), if x(m + n) = 0 for all m M, n N, then in particular xm = 0 for all m M and xn = 0 for all n N. Conversely, if for all m M and n N we have xm = xn = 0, linearity of the A-module M + N guarantees that x(m + n) = xm + xn = 0. For part ii), if xp N then x(n + P ) N, too, since N is an abelian group. But then x annihilates (N + P )/N. Conversely if x annihilates (N + P )/N then x(n + P ) N. But then xp x(n + P ) N. (Mid Chapter) Exercise 2.15. Let A, B be rings, let M be an A-module, P a B-module and N an (A, B)-bimodule (that is, N is simultaneously an A-module and B-module and the two structures are compatible in the sense that a(xb) = (ax)b for all a A, b B, x N). Then [prove that] M A N is naturally a B-module, N B P an A-module, and we have (M A N) B P = M A (N B P ). (Mid Chapter) Exercise 2.20. If f : A B is a ring homomorphism and M is a flat A-module, then M B = B A M is a flat B-module. [Hint] Exercise 1: Show that (Z/mZ) Z (Z/nZ) = 0 if m, n are coprime. Exercise 2: Let A be a ring, a an ideal, M an A-module. Show that (A/a) A M is isomorphic to M/aM. [Hint] Exercise 3: Let A be a local ring, M and N be finitely generated A-modules. Prove that if M N = 0, then M = 0 or N = 0. [Hint] Exercise 4: Let M i (i I) be any family of A-modules, and let M be their direct sum. Prove that M is flat if and only if each M i is flat. Exercise 5: Let A[x] be the ring of polynomials in one indeterminate over a ring A. Prove that A[x] is a flat A-algebra. [Hint] Exercise 6: For any A-module, let M[x] denote the set of all polynomials in x with coefficients in M, that is to say expressions of the form m 0 + m 1 x + + m r x r (m i M). Defining the product of an element of A[x] and anelement of M[x] in the obvious way, show that M[x] is an A[x]-module. Show that M[x] = A[x] A M. 15