Commutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald

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1 Commutative Algebra Notes Introduction to Commutative Algebra Atiyah & Macdonald Adam Boocher 1 Rings and Ideals 1.1 Rings and Ring Homomorphisms A commutative ring A with identity is a set with two binary operations (addition and multiplication so that for all x, y, z A: 1. A is an abelian group with respect to addition (so it contains a zero element, 0, and every x A has an additive inverse x. 2. Multiplication is associative ((xy)z = x(yz)) and distributive over addition (x(y + z) = xy + xz) 3. Multiplication is commutative (xy = yx) 4. There is an identity element 1 with x1 = 1x = x The identity element is unique. Comment. If 1 = 0 then for any x A, x = x1 = x0 = 0 and we call A the zero ring denoted by 0. A ring homomorphism is a mapping f of a ring A into a ring B such that for all x, y A, f(x + y) = f(x) + f(y), f(xy) = f(x)f(y) and f(1) = 1. The usual properties of ring homomorphisms can be proven from these facts. A subset S of A is a subring of A if S is closed under addition and multiplication and contains the identity element of A. The identity inclusion map f : S A is then a ring homomorphism. The composition of two homomorphisms is a homomorphism. 1

2 1.2 Ideals, Quotient Rings A subset a of A is an ideal of A if a is closed under addition and Aa = aa = a. (Meaning that ra a for all r A, a a). The quotient group A/a is then a ring by the obvious multiplication (a + a)(b + a) = (ab + a). We call this ring the quotient ring A/a. The map π : A A/a defined by π(x) = x + a is a surjective ring homomorphism. Proposition 1.1. There is a 1-1 correspondence between the ideals b containing a and the ideals b of A/a. Proof. We prove this with careful detail. Let b be an ideal containing a. Then define φ(b) = {x + a : x b}, an ideal in A/a. Conversely let b be an ideal of A/a. Then take ψ(b) = {x A : x + a b}. This is an ideal of A and contains a. If we can check that φ ψ = id and ψ φ = id then we are done. This is easy, though. For the first, Let b be an ideal of A/a. Then ψ(b) = {x A : x+a b}. Then φ(ψ(b)) = {x + a : x ψ(b)} = {x + a : x {x A : x + a b}} which translates to saying that φ(ψ(b)) = {x+a : x+a b}. So that φ(ψ(b)) = b. The other composition is similar. If f : A B is any ring homomorphism, the kernel of f, that is is the set of all a A so that f(a) = 0 is an ideal of A and the image of f is a subring of B, and f induces a ring isomorphism A/kerf = Imf. Which you might like to call the first isomorphism theorem for rings. notation x y (mod a) means that x y a. The 1.3 Zero-Divisors, Nilpotent Elements, Units A zero-divisor in a ring A is an element x so that there exists a nonzero element y A with xy = 0. A ring with no nonzero zero-divisors is called an integral domain. An element x A is nilpotent if there exists and n > 0 so that x n = 0. A nilpotent element is a zero-divisor (unless A = 0) since x n 1 x = 0. A unit in A is an element x so that there exists an element y A with xy = 1. This y is determined uniquely and is denoted x 1. The units of A form a multiplicative group. The multiples ax of an element x A is called the principal ideal generated by x and is denoted (x). Note that x is a unit iff (x) = A = (1). A field is a ring A in which 1 0 and every non-zero element is a unit. Every field is an integral domain. (If xy = 0, then y = x 1 xy = 0). Proposition 1.2. Let A be a nonzero ring. Then the following are equivalent: 2

3 1. A is a field 2. the only ideals in A are 0 and (1); 3. every homomorphism of A into a non-zero ring B is injective. Proof. 1 = 2: Let a be a nonzero ideal with x a nonzero. Then x is a unit and thus a (x) = (1). 2 = 3. Let f : A B be a homomorphism and B nonzero. Then kerf is an ideal of A and is either (1) or 0. If it is (1) then f = 0 which is impossible since f(1) = 1. Thus kerf = 0 and f is injective. 3 = 2. Let x be a nonzero element of A. Suppose that x is not a unit so that (x) is not equal to (1). Then A/(x) is a nonzero ideal and hence the natural homomorphism π : A A/(x) is injective. But this means that (x) =ker(f) = 0 which is a contradiction. 1.4 Prime Ideals and Maximal Ideals An ideal p in A is prime if p (1) and if xy p then either x p or y p. An ideal m in A is maximal if m (1) and if there is no ideal a such that m a (1) (strict inclusions). Equivalently we have the following mini-proposition/definition: p is prime A/p is an integral domain; m is maximal A/m is a field Should we prove this? This first is quite easy, so we ll work out the second. Let m be maximal. Then since there is a correspondence between ideals of A/m and ideals containing m, the maximality of m says that A/m has no non-trivial ideals and Proposition 1.2 thus guarantees that A/m is a field. Conversely, if A/m is a field, then by the correspondence again, there are no ideals between m and A. Comment. Note that this implies that maximal ideals are prime, but not necessarily vice versa. Also note that the zero ideal is prime A is an integral domain. Proposition 1.3. If f : A B is a ring homomorphism (henceforth abbrev. HM) and b is a prime ideal in B then f 1 (b) is a prime ideal in A. Proof. This is very direct. The fact that f 1 (b) is an ideal is immediate. Let xy f 1 (b). Then f(xy) b and therefore f(x)f(y) b it follows since b is prime that either f(x) or f(y) b and thus either x or y f 1 (b). Comment. The corresponding statement about maximal ideals is not true in general, since if A is any ring that is not a field and F is any field, then 0 is maximal in F but its inverse image in A may not be maximal at all. (Just let f be injective!) 3

4 Theorem 1.1. Every ring A 0 has at least one maximal ideal. The proof will use Zorn s Lemma so remind you first what it says. Let S be a partially ordered set (sometimes called a poset), that is, one with a relation that is reflexive, transitive and anti-symmetric. A subset T of S is called a chain if any two elements of T are comparable. That is to say that if x, y T then either x y or y x. An upper bound for a T in S is an element x S such that t x for every t T. Finally, a maximal element in S is an element x S so that for all y such that x y, we have x = y. Theorem 1.2 (Zorn s Lemma). If every chain T of S has an upper bound in S then S has at least one maximal element. Zorn s Lemma is equivalent to the axiom of choice. Proof. (of Theorem) Let Σ be the set of all ideals not equal to (1) in A. Order Σ by inclusion. Σ is not empty, since 0 Σ. We must show that every chain in Σ has an upper bound in Σ. Thus we are inspired to let (a α ) be a chain of ideals in Σ, so that for each pair of indices α, β we have either a α a β or a β a α. Let a = α a α. We claim that a is an ideal. Indeed, a is clearly closed under multiplication by A, so we show closure under addition. Let x, y a. Then x a α, y a β for some α, β. Then one of these ideals contains the other, since they are elements of a chain and we therefore have x, y contained in the same ideal and thus x + y a. Note that 1 / a since 1 / a α for all α. Hence a Σ and a is an upper bound of the chain. Thus by Zorn s Lemma, Σ contains a maximal element. What is a maximal element in Σ? It is an ideal that does not contain 1 so that it there is larger ideal in Σ containing it; a maximal ideal in A. Corollary 1.1. If a (1) is an ideal of A, there exists a maximal ideal of A containing a. Proof. Note that A/a has a maximal ideal m by the above theorem. Denote by n the corresponding ideal of A containing a. We claim that n is maximal in A. The claim is justified as follows: suppose that there is an ideal p strictly between (1) and n. Then the ideal p = {x + a : x p} (1) is an ideal of A/a that strictly contains m which is a contradiction. Corollary 1.2. Every non-unit of A is contained in a maximal ideal. (Just let a = (x)). Comment. There exist rings with exactly one maximal ideal, for example fields. A ring A with exactly one maximal ideal m is called a local ring and the field k = A/m is called the residue field of A. Proposition 1.4. Let A be a ring and m (1) and ideal of A such that every x A m is a unit in A. Then A is a local ring and m its maximal ideal. 4

5 Proof. Note that m is clearly maximal in A since if an ideal contained m and any other element, it would contain a unit. Also, all ideals (1) consist of nonunits. Note that m contains all non-units so that m contains all ideals (1), and is therefore the only maximal ideal. Proposition 1.5. Let A be a ring with m a maximal ideal of A, such that every element of 1 + m is a unit in A. Then A is a local ring. Proof. Let x A m. Then the ideal generated by m and x is (1) since m is maximal. It follows then that there exist m m and t A so that m + xt = 1. This implies that xt = 1 m 1 + m and thus xt is a unit. This also implies that x is a unit. (Do you see why?) Thus we can apply the previous proposition and conclude that A is a local ring. A ring with only finitely many maximal ideals is called semi-local. Example 1.1. See Atiyah & MacDonald for examples. Proposition 1.6. A principal ideal domain (PID) is an integral domain A where every ideal is principle. In such a domain, every nonzero prime ideal is maximal. Proof. Suppose that (x) is a prime ideal, but not maximal. Let (y) (x). Then since (x) sits inside of (y), we see that x = yt for some t A. Thus yt (x) and y / (x). Thus it follows that t (x) since (x) is prime and hence t = xw for some w A. Substituting this, we see that x = yt = yxw = yw = 1 and y is a unit so that (y) = (1) and (x) is maximal. 1.5 Nilradical and Jacobson Radical Proposition 1.7. The set R of all nilpotent elements in a ring A is an ideal and A/R has no nilpotent elements 0. Proof. If x is nilpotent then so is ax for all a A. Now let x, y be nilpotent elements, say x n, y m = 0. Then (x + y) n+m 1 = 0 since each term of the expansion must contain either a power of x greater than n 1 or a power of y greater than m 1. To see that A/R has no nilpotent elements, note that x + R A/R is nilpotent if and only if x n + R = 0 in A/R which is equivalent to saying that x n R which would imply that x R. The ideal R defined above is called the nilradical of A. The following proposition gives another definition of R. 5

6 Proposition 1.8. The nilradical of A is the intersection of all the prime ideals. Proof. Let R denote the intersection all prime ideals of A. Then if f is nilpotent, then f n = 0 for some n > 0. Since 0 p for all ideals and p is prime, we have that f p for all prime ideals p and hence f R. Conversely, we will show that if f is not nilpotent, then it is not in the intersection of all prime ideals. Suppose that f is not nilpotent. Then let Σ be the set of all ideals a such that no power of f is in a. Ordering Σ by inclusion we can apply Zorn s Lemma to conclude that it has a maximal element, p. We shall show that p is prime by showing that x, y / p implies xy / p. Indeed, if x, y / p then p + (x) and p + (y) properly contain p and thus are not elements of Σ by the maximality of Σ. Thus it follows that there exist some n, m so that f n p + (x) f m p + (y) which clearly imply that f n+m p + (xy) which implies that xy / p. Thus p is prime and does not contain f as required. The Jacobson radical R of A is the intersection of all the maximal ideals of A. It can be characterized as follows: Proposition 1.9. x R 1 xy is a unit in A for all y A. Proof. : Suppose 1 xy is not a unit for some y A. Then by Corollary xy is contained in some maximal ideal m of A. But since x R m we have 1 xy m = 1 m which is absurd. : Suppose x / m for some maximal ideal m. Since x and m generate A we have m + xy = 1 for some elements m m and y A. Thus 1 xy m contradicting the fact that 1 xy is a unit. 2 Operations of Ideals We define the sum of two ideals a and b to be the the set of all x+y where x a and y b. We denote the sum by a + b. It is the smallest ideal containing a and b. In general we define the sum i I a i of any family (possibly infinite) of ideals a i of A; its elements are all sums x i, where x i a i for all i I and all but finitely many of the x i are zero. It is the smallest ideal of A which contains all of the ideals a i. The intersection of any family of ideals is an ideal. (Proof is easy) The product of two ideals a, b in A is the ideal generated by all products xy where x a and y b. It is the set of all finite sums x i y i where x i a and y i b. We can similarly define the product of any finite family of ideals. In particular, the powers a n of an ideal are defined. We also make the convention that a 0 = (1). Example

7 1. If A = Z, a = (m), b = (n) then a + b is the ideal generated by gcd(m, n). This follows since the gcd of any two numbers can always be represented as an integer combination of the two numbers; a b is the ideal generated by their lcm (proof easy); and ab = (mn) (proof easy) Thus it follows that ab = a b m, n are coprime. 2. A = k[x 1,..., x n ], a = (x 1,..., x n ) = ideal generated by (x 1,..., x n ). Then a m is the set of all polynomials with no terms of degree < m. The above operations are all commutative and associative. We also have a distributive law for products. a(b + c) = ab + ac In general, and + are not distributive over each other. The best we can do is the modular law a (b + c) = a b + a c if a b or a c. Indeed, if a b and x a (b + c) then x = b + c for some b b a and c c. Since x a we have b + c a and thus that c a. The rest is easy to do. Finally, (a + b)(a b) ab. (since (a + b)(a b) = (a(a b) + b(a b) ab). Also we clearly have that ab a b, hence a b = ab provided a + b = (1). We say that two ideals a, b are coprime if a + b = (1). Thus for coprime ideals we have a b = ab. Two ideals are coprime iff there exist x a and y b such that x + y = 1. Let A 1,..., A n be rings. Their direct product A = n i=1 is the set of all sequences x = (x 1,..., x n ) with x i A i with componentwise addition and multiplication. A is a commutative ring with identity (1,..., 1). We have projections p i : A A i defined by p i (x) = x i ; they are ring homomorphisms. Let A be a ring and a 1,..., a n ideals of A. Define a homomorphism φ : A by the rule φ(x) = (x + a 1,..., x + a n ). A i n (A/a i ) Proposition 2.1. i. If a i, a j are coprime whenever i j, then Πa i = a i. i=1 ii. φ is surjective a i, a j are coprime whenever i j. 7

8 iii. φ is injective a i = (0). Proof. i). We will induct on n, the number of ideals. When n = 2 the argument is handled above. Suppose we have a 1,..., a n and the result is true for any set of n 1 ideals. Let b = n 1 i=1 a i = n 1 i=1 a i (by assumption). Since a i + a n = 1 for each i we know that we have equations Note that x i + y i = 1 with x i a i, y i a n n 1 i=1 n 1 x i = (1 y i ) 1(mod a n ). i=1 Hence a n + b = (1) so they are coprime. Thus n a i = ba n = b a n = i=1 n a n. ii). : We will show, for example, that a 1 and a 2 are coprime. We accomplish this by asserting noting that since φ is surjective, there is an x so that φ(x) = (1 + a 1, 0,..., 0). This means that x 1 mod a 1 and x 0 mod a 2. This means that i=1 1 = (1 x) + x a 1 + a 2 so that a 1 and a 2 are coprime. : It is enough to show that we can obtain (1 + a 1, 0,..., 0) for example. Since a 1, a i are coprime for i > 1 we have a set of equations c i + y i = 1 where c i a 1, y i a i. Let x = n i=2 y i. Then x is 0 mod a i when i > 2. On the other hand, x = n (1 c i ) which is 1 mod a 1. Thus φ(x) = (1 + a 1, 0,..., 0). iii). Note that a i = ker(φ) which obviously implies the result. i=2 In general, the union of two ideals is not an ideal. The reason for this is that the sum of two elements need not be contained in the ideal. For an example, consider, Z with the union of the ideals (3), (5). This is the set of all integers which are multiples of either 3 or 5. It is not closed under addition since 3+5 = 8 is not in the union. We can do better than that, however, with the following Prime Avoidance Lemma. 8

9 Proposition 2.2. (Prime Avoidance Lemma) Let p 1,..., p n be prime ideals and let a be an ideal contained in n i=1 p i. Then a p i for some i. Proof. This is a somewhat complicated argument so we first walk through the steps. We would like to show that a n p i = a p i for some i. i=1 We will accomplish this by proving the contrapositive: a p i n for all i = a p i. i=1 We will do this by induction in fact! The statement is certainly true when n is 1. We now prove it for n > 0. Assume that the result is true for n 1. Suppose a p i for all i = 1... n. Then for each i the remaining n 1 ideals satisfy the induction hypothesis so we can say a n p j \p i. j=1 This is the same as saying that there is an element x i a so that x i is not in any of the p j when j i. If for any i we have that x i / p i then we will have succeeded in showing the statement for n and are through. Thus suppose that x i p i for all i. Consider the element y = n x 1 x 2 x i 1 x i+1 x 1+2 x n. i=1 Then y a (clearly) and y / p i (1 i n) (since y is the sum of n 1 terms in p i and one term not contained in p i. In fact we need the primeness of p i to assert this latter part). Thus a n i=1 p i. Proposition 2.3. Let a 1,..., a n be ideals and let p be a prime ideal containing their intersection. Then p a i for some i. If p = a i, then p = a i for some i. Proof. This proof is straightforward. Suppose the statement is false. Then for each i there exists and x i a i such that x i / p. Then x i a i a i p. But since p is prime, the product cannot be an element of p. Thus we have a contradiction and p a i for some a i. If p = a i then p a i for all i and hence p = a i for some i. If a, b are two ideals in a ring A, their ideal quotient is (a : b) = {x A : xb a} 9

10 This is an ideal: (This is clear) In particular, (0 : b) is called the annihilator of b and is also denoted Ann(b): it is the set of all x A such that xb = 0. In this notation, the set of all zero-divisors in A is D = x 0 Ann(x) where (x) is the principal ideal generated by x. In fact, if b = (x) is a principal ideal, we will just use (a : x) instead of the fancier (a : (x)). Example 2.2. Let A = Z, a = (m), b = (n), where m = 2 µ2 3 µ3 5 µ5, n = 2 ν2 3 ν3 5 ν5 Then (a : b) is the set of all numbers which when multiplied with n give a multiple of m. It is clear that (a : b) = (q) where q = 2 γ2 3 γ3 5 γ5 and γ p = max(µ p ν p, 0). In other words, q = m/ gcd(m, n). Exercise 1.12 i) a (a : b). (Clear) ii) (a : b)b a. If x (a : b) then xb a which is what this sentence is saying. iii) ((a : b) : c) = (a : bc) = ((a : c) : b). This argument is done by noticing that x ((a : b) : c) = xc (a : b) = xcb a = x(bc) a = x (a : bc) All the other directions are the same. iv)( i a i : b) = i (a i : b). These are both easy to see. v)(a : i b i) = i (a : b i). Suppose that x( i b i) a. Then in particular, xb i a for each i. Thus x i (a : b i). Conversely, if x i (a : b i), then x( i b i) a as required. If a is any ideal of A, the radical of a is r(a) = {x A : x n a for some n > 0} If φ : A A/a is the standard homomorphism, then r(a) = φ 1 (R A/a ) (the nilradical of the quotient). Thus since it is the preimage of an ideal, it is an ideal. Exercise 1.13 i) r(a) a. (Clear) ii) r(r(a)) = r(a). Suppose x r(r(a)). Then for some n, x n r(a) and thus for some m, (x n ) m = x nm a. The other inclusion follows from i). iii) r(ab) = r(a b) = r(a) r(b). We prove the first inequality. Suppose that x r(ab). Then for some n, x n ab a b so x r(a b). Conversely, suppose x r(a b). Then for some n, x n a, x n b which implies that x 2n ab so x r(ab). 10

11 Now we prove the second equality. Let x r(a b). Then for some n, x n a b r(a) r(b) by i). Thus x n r(a) and x n r(b), giving us that in fact x r(a) r(b). Conversely, if x r(a) r(b) then for some m, n, x m a, x n b and thus x mn a b. iv) r(a) = (1) a = (1). Suppose that r(a) = (1). Then 1 r(a) so that 1 = 1 n a, so a = (1). The other direction is clear. v)r(a + b) = r(r(a) + r(b)). Let x r(a + b). Then for some n, x n a + b r(a) + r(b) so that x r(r(a)+r(b)). Conversely, if x r(r(a)+r(b)) then for some n, x n r(a)+r(b). This means x n = p + q where p m a and q k b for some m, k. Then for a large enough power M, x M a + b and x r(a + b). vi) if p is prime, r(p n ) = p for all n > 0. First note that p is clearly contained in r(p n ). Now suppose x r(p n ). Then for some m, x m p n p. Thus since p is prime, x p. Proposition 2.4. The radical of an ideal a is the intersection of the prime ideals which contain a. Proof. Proposition 1.8 applied to A/a tells us that the nilradical of A/a is the intersection of all prime ideals of A/a which is in correspondence with the set of all prime ideals containing a. More generally, we may define the radical r(e) of any subset E of A in the same way. It is not an ideal in general. We have r( α E α ) = α r(e α ) for any family of subsets E α of A. (This is seemingly vacuous) Proposition 2.5. D = set of zero-divisors of A = x 0 r(ann(x)). Proof. D = r(d) = r( x 0 Ann(x)) = x 0 r(ann(x)). Example 2.3. If A = Z, a = (m), let p i (1 i r) be the distinct prime divisors of m. Then r(a) = (p 1 p r ) = r i=1 (p i). Proposition 2.6. Let a, b be ideals in a ring A such that r(a), r(b) are coprime. Then a, b are coprime. Proof. Suppose that r(a) + r(b) = (1). Then by v) above, and by iv) we have that a + b = (1). r(a + b) = r(r(a) + r(b)) = r(1) = (1) 11

12 3 Extension and Contraction Let f : A B be a ring homomorphism. If a is an ideal in A, in general, f(a) is not an ideal in B. (take for example, the imbedding of Z into Q and take a to be any nonzero ideal in Z. We define the extension a e of a to be the ideal Bf(a) generated by f(a) in B: explicitly, a e is the set of all sums yi f(x i ) x i a, y i B. If b is an ideal in B, then f 1 (b) is always an ideal of A called the contraction b c of b. If b is prime, then b c is prime. (This is an easy proof). If a is prime in A then a e is not necessarily prime in B. (Just inject Z into Q). We can factorize this process as A p f(a) j B where p is surjective and j is an injection. For p the situation is very simple. There is a bijective correspondence between ideals of f(a) and the ideals of A which contain ker(f). Let s see why this is true. We know that f(a) = A/ ker f(a). Now just apply (1.1). Prime ideals correspond to prime ideals. For j, the situation is much more dicey. (Insert example from algebraic number theory which is presented without proof in AM). Moving on... Proposition 3.1. Let f : A B be a ring homomorphism and a and b as before. Then i.) a a ec, b b ce ; ii.) b c = b cec, a e = a ece ; iii.) If C is the set of contracted ideals in A and if E is the set of extended ideals in B, then C = {a a ec = a}, E = {b b ce = b}, and a a e is a bijective map of C onto E, whose inverse is b b c. Proof. i.) Let x a. Then x f 1 (a e ) so x a ec. Next let y b ce. Then y is a linear combination of terms of the form b i f(x i ) with b i B and x i b c. This means that f(x i ) b and thus y b. ii.) These both really do follow from i). Indeed we have b c b cec by the first part, and since the second part says b b ce we take the contraction of both sides to conclude b c b cec. The second equality is obtained in the same manner. iii.) Let a C. Then a is a contracted ideal so a = b c for some ideal b of B. Then a ec = b cec = b c = a. The fact that such a with this property are in C follows since a = a ec guarantees that a is a contraction. The proof for E is nearly identical. To prove that the map a a e is bijective we note that it is surjective since if a e E then a e = a ece which is the extension of the contraction of an ideal. To see that it is injective, suppose that 12

13 b ce 1 = b ce 2. Take the contraction of both sides to obtain b cec 1 = b cec 2 which by ii). implies b c 1 = b c 2 as expected. AM feels the need here to include a ridiculous number of properties of e and c so we list them and maybe we ll prove the hard ones sometime. Proposition 3.2. If a 1, a 2 are ideals of A and 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 operations. On second thought, I just worked all of these out and they are completely trivial and are not difficult at all. 4 Exercises 1 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. Suppose that x n = 0. Then wlog, let n be even (take n + 1 if you like) and then (1 + x)(1 x + x 2 x n 1 ) = 1 x n = 1 so (1 + x) is a unit. Now let c be a unit. Then (c + x) = c(1 + x/c) which is a product of two units since x/c is nilpotent if x is. 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. 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. 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. i.) : Suppose that f is a unit with inverse g = m i=0 b ix i. Then since the constant term in fg must be 1, we see that b 0 a 0 = 1 so a 0 is a unit. Further, by examining the highest order terms we get the relations: a n b m = 0 a n 1 b m + a n b m 1 = 0 a n 2 b m + a n 1 b m 1 + a n b m 2 = 0 13

14 Multiplying the second equation by a n we obtain a n 1 b m a n + a 2 nb m 1 = 0 The first term vanishes from the first equation and we have a 2 nb m 1 = 0. Inductively we can prove that a r+1 n b m r = 0. Thus a r+m n b 0 = 0 so a n is nilpotent. Now note that a n x n is nilpotent as well. It follows then that a 0 a n x n is a unit and we can argue as before to show that a n 1 is nilpotent and we are done. : Since a 1,..., a n are nilpotent so is n i=1 a ix i. (Just take a huge power so that each turn vanishes). Thus f = a 0 +nilpotent which is a unit by problem 1. ii.) : Clear : Suppose that f m = 0. Then a m n = 0 by expansion. Then f a n x n as a difference of two nilpotents is also nilpotent and we can repeat the argument. iii.) : extremely clear : Suppose that fg = 0 with g = m i=0 b ix i of minimal degree. Then a n b m = 0. Also by commutativity and associativity, f(a n g) = 0 and a n g has degree less than g. Thus we must have that a n g = 0. But then this means that a n b k = 0 for every k. Thus fg = 0 really means: 0 = (a a n x n )(b b m x m ) = (a a n 1 x n 1 )(b b m x m ) Thus we argue as before to see that a n 1 g = 0 and in general can do this for a n k. So then just pick any nonzero b k and b k f = i b ka i = 0. iv.) : Suppose that fg is primitive. Now suppose that f were not primitive. Then (a 0, a 1,..., a n ) (1) so there is a common factor to all of the terms. But then fg would have a common factor as well. Thus f is primitive, and so must g be. : The classical Gauss Lemma, which is kind of fancy. INCOMPLETE 4. Show that in the ring A[x] the Jacobson Radical = Nilradical. It is clear that since all maximal ideals are prime, that the Jacobson Radical contains the Nilradical. Now suppose that f is in the Jacobson Radical. Then 1 fg is a unit for every g A[x]. Let g = x. Then 1 fx is a unit, so every non constant term s coefficient is nilpotent by exercise 2. But these are exactly the original coefficients of f so f is nilpotent and thus in the nilradical. 6. A ring A is such that every ideal not contained in the nilradical contains a non-zero idempotent (an element e with e 2 = e 0). Prove that the nilradical and Jacobson radical of A are equal. As in exercise 4, we only need to show that the Jacobson radical is contained in the nilradical. Suppose that m is a maximal ideal that is not contained in 14

15 the nilradical. Then m has an idempotent, e 0. Then since e is in a maximal ideal, 1 e is a unit. Finally, 1 e = 1 e 2 = (1 e)(1 + e). Cancelling, we see that 1 + e = 1 and e = 0 which is a contradiction. 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 in A. Now let y be an element not in p. Then y n y = 0 p for some n. Thus y(y n 1 1) p and since p is prime, we know that y n 1 1 p. Thus it follows that y n 1 1 = x for some x p and p + (y) = (1) so that p is maximal. 8.Let A be a ring 0. Show that the set of prime ideals of A has minimal elements with respect to inclusion. We first recall what a minimal element is. A minimal element is an element m such that m n implies m = n. We will fancifully use Zorn s Lemma. Suppose we have a chain of prime ideals p 1 p 2. Then consider their intersection P = i=1 p i. This is clearly an ideal, and we now show that it is prime. Suppose that xy P but that x, y / P. Then there exists some i, j such that x / p i and y / p j. Since xy P then it must happen that x p j and y p i. WLOG, assume i > j. Then x p j p i which is a contradiction. Thus this chain has a lower bound and by Zorn s lemma, has a minimal element. 9. Let a be an ideal (1) in a ring A. Show that a = r(a) a is an intersection of prime ideals. : Recall that r(a) is the preimage of the nilradical of A/a, which is an intersection of the prime ideals of A/a. This preimage is therefore the intersection of all prime ideals that contain a : a is always contained in r(a). Suppose a is an intersection of prime ideals. Let x r(a). Then for some n, x n a = p i. The primeness of p i guarantees that x a. 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; iii) A/R is a field. i) ii): Let x be a nonunit. Then it is contained in some maximal ideal which must necessarily be R so x is nilpotent. ii) iii): Let x + R be an element of A/R. Then if x is nilpotent, it is the 0 element, and if not, then it is a unit so A/R is a field. iii) i): If A/R is a field, then R is maximal. Let p 1, p 2 be two prime ideals of A. Then p 1 R which contradicts that R is maximal. 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. 15

16 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. i): (x + 1) 2 = x + 1 = x 2 + 2x + 1 = x + 1 = x + 2x + 1 = x + 1 = 2x = 0. ii): The first part was shown in exercise 7. Let a + p be an element of A/p and a / p. This element is invertible since A/p is a field. Then since a 2 = a we have (a + p) 2 = a + p. This implies (a + p) = 1 + p and thus the field has only two elements. iii): We claim that (x, y) = (xy + x + y). This is clear since x(xy + x + y) = xy+x+xy = 2xy+x = x and similarly for y. The result follows from induction. 12. A local ring contains no idempotent 0, 1. If x A is not contained in the maximal ideal, then it is a unit and therefore if it is idempotent, it is 1. Suppose that x 2 = x for some x in the maximal ideal m of A. Then 1 x is not in m so it is a unit. But then (1 x) 2 = 1 2x + x 2 = 1 x which implies that 1 x = 1 or that x = 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). Suppose that some combination of f(x f ) is equal to 1. Then for some polynomials g f in A we have g f f(x f ) = 1 f Σ It is easy to convince yourself that this cannot be the case by examining what would have to cancel out on the left side. (Of course this isn t very satisfactory. More to come later). 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 this process with K 1 instead of K to get a new field K 2, and so on. Let L = n=1 K n. Then L is a field in which each f Σ splits completely into linear factors. Let K be the set of all elements in L which are algebraic over K. Then K is the algebraic closure of K. 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 and let T be a chain. Then it is easy to see that the union of all elements in the chain is in Σ and that it is an upper bound. Now just apply Zorn s Lemma. Let m be a maximal element. Then we show it is prime. Suppose that xy m, but x, y / m. Since m is maximal, we need to have m + (x) and m + (x) 16

17 to have nonzero divisors. Suppose that m 1 +ax m+(x) and m 2 +by m+(y) are not zero divisors. Then their product is not a zero divisor, and cannot be in m. But look: (m 1 + ax)(m 2 + by) = m 1 (m 2 + by) + m 2 (ax) + abxy which is an element of m. This is a contradiction. 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)). 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). i): V (E) = V (a) is obvious. Since r(a) a we have V (r(a)) V (a). Finally, suppose p is a prime ideal containing a. Then let x r(a). We will show x p so that p contains r(a). x n a p for some n, and since p is prime, x p. ii): Clear iii): Let p contain the union. Then in particular, p contains E i for each i. Thus p V (E i ). Arguing in reverse easily proves the other direction. iv): The first equality: Since ab a b, V (ab) V (a b). Conversely, if p contains ab let x a b. Then x 2 ab p and since p is prime, x p. The second equality: : Let p contain both a and b. Then it must contain the smaller ideal ab. : Let p contain ab. If p contains a we are done. So suppose not. Then let y b, x a (x / p). Now we have xy ab p and since x / p and p is prime we must have y p so that p contains b. 16. Draw pictures of Spec(Z), Spec(R), Spec(C[x]), Spec(R[x]), Spec(Z[x]). 5 Modules 5.1 Modules and Module Homomorphisms Let A be a commutative ring. Then an A-module is an abelian group M written additively on which A acts linearly. What this means is that it is a pair (M, µ) 17

18 where M is an abelian group and µ : A M M such that if we write ax for µ(a, x) (a A, x M), the following axioms are satisfied for all a, b A and all x, y M. a(x + y) = ax + ay (a + b)x = ax + bx (ab)x = a(bx) 1x = x All we are really concerned with the module structure however, is how multiplying by an element of the ring changes moves elements of M. In effect for each a A there is a corresponding endomorphism of M, µ a given by µ a (x) = ax. Thus to completely understand the module structure, we need to know the function A End(M) sending a to µ a. It is easy to see that the above properties make this map into a ring homomorphism and even more succinctly: An A-module is an abelian group M together with a ring homomorphism A End(M). Example ) Any ideal of A is an A-module. In particular, A itself is an A-module. 2) If A is a field k then an A-module is precisely a k-vector space. 3) If A = Z then a Z-module is the same as an abelian group. (Just define the action nx = x + + x). Question for prof. Polini. Every Z-module is an abelian group, but must it have the trivial ring action? 4) If A = k[x] where k is a field, then an A-module is a k-vector space v with a linear transformation T. Define µ f : V V by µ f (v) = f(t )(v). Let M and N be A-modules. We say that a function f : M N is an A-module homomorphism if f(x + y) = f(x) + f(y) and f(ax) = af(x). If A is a field, then these are just the properties of a linear transformation. Note that the composition of A-module homomorphisms is again an A-module homomorphism. Thus by defining addition and multiplication in an obvious way, we can turn the set of all A-module homomorphisms from M to N into an A-module. This A-module is denoted Hom A (M, N). Sometimes we might just write Hom(M, N) and you ll have to guess what the ring is. Now is where the fun starts. Suppose all the capital letters are A-modules, and we have homomorphisms u : M M and v : N N. Then these induce maps: u : Hom(M, N) Hom(M, N) and v : Hom(M, N) Hom(M, N ) defined in exactly how you d think they were: u(f) = f u v(f) = v f. To illustrate this, consider the following commutative diagram: (I admit it commutes quite trivially, but it still commutes) Notice that there is a natural isomorphism Hom(A, M) = M. Indeed, any module homomorphism f : A M is determined exactly by f(1) which can be any element of M. 18

19 5.2 Submodules and Quotient Modules A submodule M of M is a subgroup of M which is closed under multiplication by elements of A. The abelian group M/M then inherits an A-module structure by a(x + M ) = ax + M. This is well defined, and as for ideals, there is a one-to-one order preserving correspondence between submodules of M/M and submodules of M which contain M. The kernel of a module homomorphism is the set of all x M such that f(x) = 0. It is a submodule of M. The image of f is the set of all f(x) N with x M. The cokernel of f is Coker(f) = N/Im(f). If M is a submodule of M and M ker(f) then we have an induced map. f : M/M N f(x + M ) = f(x). This is well defined (you should check this) and in particular if M = kerf then by the 1st isomorphism theorem, (or explicit calculation, we have that M/kerf = Im(f) We define the sum and intersection of modules in the same way we did for rings. Proposition 5.1. If L M N then (L/N)/(M/N) = L/M. Proof. Consider the map f : L/N L/M, f(x + N) = x + M. This is well defined (check if you don t believe this!) and has kernel M/N so by the 1st isomorphism theorem, we have the result. Proposition 5.2. If M 1, M 2 are submodules of M then Proof. Consider the composite map (M 1 + M 2 )/M 1 = M2 /(M 1 M 2 ). M 2 M 1 + M 2 (M 1 + M 2 )/M 1. This map is surjective (think about it). The kernel of the composite is M 1 M 2 and the result follows. If a A is an ideal then am is exactly what you think it is, and is a submodule of M. We define (N : P ) = {a A : ap N}. This is an ideal of A. In particular, (0 : M) = {a : am = 0} is called the annihilator of M and we write Ann(M). This gives rise to a curious result: If a Ann(M) then M is an A/a module. Indeed, if b + a A/a then define (b + a)x = bx. This is well-defined: Suppose that b + a = c + a then b c a implies (b c)x = 0 for every x M and that bx = cx. We say that an A-module is faithful if Ann(M) = 0. 19