Solutions Ark3 From the book: Number 1, 2, 3, 5 and 6 on page 43 and 44. Number 1: Let S be a multiplicatively closed subset of a ring A, and let M be a finitely generated A-module. Then S 1 M = 0 if and only if there exists an element s S such that sm =0. Solution: If sm = 0 for an element s S it follow immediatly from the equivalence relation used to define S 1 M that S 1 M =0. Assume S 1 M and let m 1,...,m r be a generating set of elements in M. Eachone of them maps to zero in S 1 M meaning that for each i, with 1 i r, there exists an s i S such that s i m i =0.Takes = s 1 s 2 s r. Then s S since S is closed under multiplication, and sm i = 0 for all i. But the m i s forming a generating set for M, this implies that sm = 0. Number 2: Let a be an ideal of a ring A, and let S =1+a. Show that S 1 a is contained in the Jacobson radical of S 1 A. Use this to give a proof of Nakayama s lemma, version (2.5). Solution: The prime ideals in S 1 A are all of the form S 1 p for p a prime ideal in A disjoint from S, and maximal ideals correspond to maximal ideals. The pime ideals p satisfying p (1 + a) = are those for which we may find a y p with y =1+a for an a A, i.e., those which are comaximal with a. Hence the maximal ideals in S 1 A are all of the form S 1 m where m is a maximal ideal in A not comaximal with a. But a not being comaximal with a maximal ideal m means that a m. Hence S 1 a is contained in all maximal ideals in S 1 A, i.e.,, it is contained in the Jacobson radical. We shall show that Nakayama version (2.6) implies Nakayama (2.5): Assume am = M for a finitely generated A-module M. LetS =1+a. Clearly S 1 as 1 M = S 1 M (if m = a i m i with a i a, them s 1 m = s 1 a i m i where s 1 a i S 1 a.) We just checked that S 1 a lies in the Jacobson radical of S 1 A,soNakayama version (2.6) gives us that S 1 M = 0. By exercise (1) it follows that there is an element x in S =1+a i.e., that satisfies x 1 mod a with xm =0. Number 3: Let A be a ring and let S and T be two multiplicatively closed subsets of A, and let U be the image of T in S 1 A. Show that the ring (ST) 1 A and U 1 (S 1 A) are isomorphic. Solution: This is an exercise in the yoga of universal properties and in the jui-juisti of diagram chasing 1, using the following diagram, where the arrow at the top going to 1 It is also an exercise in the aikido toho iai of writing diagrams in L A TEX
Solutions Ark3 MAT4200 autumn 2011 the right and the two arrows pointing down are canonical localisation maps: A S 1 A (ST) 1 A U 1 (S 1 A) The four other maps are all induced by the universal property of an appropriate localisation. The two horisontal maps in the lower row one each way are the ones that interest us the most. They give the required isomorphism (do the details yourself!). Alternatively if your inclination is less universal you may do this directly on elements; i.e., check that sending ( u 1 ) 1 (s 1 a)to(us) 1 a and vice versa are two well defined maps being mutually inverses. Number 5: Let A be a ring. Suppose that for each prime ideal p, the local ring A p has no nilpotent element = 0. The A has no nilpotent element = 0.IfeachA p is an integral domain, is it true that A is an integral domain? Solution: Assume that all the localisations A p are reduced (newspeak for not having non-zero nilpotents), and assume that there is an x A with x = 0 and x n =0.We may as well assume that x n 1 =0. Let m be a maximal ideal containing the ideal Ann(x n 1 ) (which is proper because x n 1 = 0.) Then x n 1 = 0 in A m ; indeed if it were zero, there would be an s m with sx n 1 = 0 in A which is impossible since Ann(x n 1 ) m. But clearly x n = 0 also holds in A m,soa m has non-zero nilpotents! Even if all localisations A p are integral domains, it is not true that A is. The simplest example is the product of two fields: A = k 1 k 2 which is not an integral domain since (x, 0)(0,y)=0foranyx k 1 and y k 2. This ring has exactly two ideals which both are maximal, namely m 1 = { (x, 0) x k 1 } and m 2 = { (0,y) y k 2 } indeed every element (x, y) with both x and y different from zero, is invertible. Let us argue that A m2 k 1. Then by symmetry A m2 k 1, and we have our example since they both are without zero-divisors. All elements (0,y) map to 0 in A m2 since s =(1, 0) m 2 and s(0, 1) = 0. This shows that the inclusion map k 1 A (which is a ring homomorphism and sends x to (x, 0)) composed with the localisation map A A m2 is surjective indeed, (x, y) = (x, 0) + (0,y) and (x, 0) have a common image in A m2. Hence it is an isomorphism. Number 6: Let A be a ring and let Σ be the set of all multiplicatively closed subsets s of A such that 0 S. Then S has maximal elements, and S is maximal in S if and 2
only if S = A \ p for a minimal prime ideal p of A. Solution: We intend to use Zorn s lemma, so let { S i } i I be a chain in Σ. Clearly i I S i is multiplicatively close, since if s S i and s S i they both belong to S j for any j I with j>iand j>i. Hence Zorn gives us maximal elements in Σ. Now let us check that p = A \ S is an ideal. Let a S and b A. We assume that ab S and take a look at the set S = { a n c n N and c S }. It is clearly a multiplicatively closed subset since S is, and as a S but a S, there is a strict inclusion S S. Now we claim that 0 S ; indeed if a n c =0,weget(ab) n c = 0, but ab S, hence (ab) n c S which gives a contradiction as 0 S by hypothesis. As S is maximal, it follows that ab S, and A \ S is an ideal. It is easy to se that it must be a prime ideal, and by the maximality of S, it must be a minimal prime ideal. Oppgave 1. Let n Z be an integer and let S be the multiplicativvely closed set S = { m Z (m, n) =1}. We denote the ring S 1 Z by Z (n).ifp is prime, then Z (p) is the ring Z localised at the prime ideal (p). This is a local ring with maximal ideal (p). a) Let n = 6. And show that Z (6) has two maximal ideals, namely m 1 =(3)Z 6 and m 2 =(2)Z (6). b) Show that Z (6) /m 1 F 3 and Z (6) /m 2 F 2. c) What is the Jacobson radical to Z 6?IfJ denotes the Jacobson radical, describe Z 6 /J. d) In general, for any n Z, show that Z (n) is a semilocal ring, i.e., is a ring with only finitely many maximal ideals. Describe those ideals and their residue fields. (The residue field to a maximal ideal m in a ring A is A/m.) Solution: a) Let a Z 6 be an ideal. The elements in a are of the form nm 1 where neither 2 nor 3 is a factor of m. If the same applies to n, i.e., (n, 6) = 1, then nm 1 is invertibel in Z 6 and a = Z 6. This shows that the only ideals in Z 6 are the principal ideals (2 a 3 b )Z 6 a and b non-negative integers and among them (2)Z 6 and (3)Z 6 are the only maximal ones. b) We have F 3 = Z/(3)Z Z 6 /(3)Z 6 The inclusion Z Z 6 induces a map Z/(3)Z Z 6 /(3)Z 6 wich is injective since (3)Z 6 Z =(3)Z. If m is an integer with (m, 3) = 1 we may write mx+3a = 1 where x and a are in Z. Hence x m 1 mod (3)Z 6, and hence any element satisfies nm 1 nx mod (3)Z 6 where nx Z, and the map above is also surjective. 3
c) The Jacobson radical R (we all \mathfrak) ofz 6 is (6)Z 6, and by the Chinese remainder theorem, R = Z 6 /(6)Z 6 F 3 F 2. d) This is a straight forward generalisation of what we have done so far. We know that there is a one-to-one correspondence between maximal ideals in Z n = S 1 Z and ideals p in Z maximal among those with p S =, i.e., prime ideals (p) where p is a prime not dividing n. So there is one maximal ideal S 1 (p)z =(p)z n in Z n for each prime divisor p of n. There is a natural map F p = Z/(p)Z Z n /(p)z n (sending the residue class of x mod (p)z to the one mod (p)z n ) which is injective because (p)z n Z =(p)z. It is surjective: If (m, p) = 1 we may find integers x and a with mx + ap = 1. Hence km 1 kx mod (p)z n and consequently km 1 is in the image. Oppgave 2. Let A be a subring of the ring B and let p A be a prime ideal. Let S be the multiplicatively closed set S = { x A x p } = A \ p. a) Show that the primes in S 1 B correspond to the primes q in B such that q A p more presicely, they are of the form S 1 q with q A p. Let now A = Z and B = Z[i 5]. b) If p =(3)Z, show that S 1 Z[i 5] has exactly the two maximal ideals (3, 1+i 5) and (3, 1 i 5). What are the residue fields? c) If p = (2) show that S 1 Z[i 5] is a local ring. What is the residue field? Solution: a) This is the general principle (3.11) on page 41 describing how prime ideals behave when localised, but adapted to the setting of this exercise. b) We first check that the two ideals in question are maximal, and by symmetry, it suffises to do that for one of them. We do that by computing Z[i 5]/(3, 1+i 5) which as a bonus gives us that the residue field is F 3. By letting X correspond to i 5we have an isomorphism Z[i 5]/(3, 1+i 5) Z[X]/(X 2 +5, 3, 1+X). Now (X 2 +5, 3, 1+X) =(3, 1+X) since X 2 +5=(X 1)(X + 1) + 3, and thus Z[X]/(X 2 +5, 3, 1+X) Z[X]/(3, 1+X) F 3. Let p Z[i 5] be a prime ideal surviving in S 1 Z[i 5], i.e., such that S 1 p stays proper. Then a Z =(3),so3 p. Pick another element w = a + ib 5 p. As 3 p, we may replace a and b by any other integer in the same residue class mod 3 and still have an element in p. In other we may assume that a and b both are in {±1, 0 }. Hence if we there is an element with ab 0 mod 3, the prime ideal p contains either 1 + i 5or1 i 5 and we are through! 4
We have to eliminate the case that ab 0 mod 3 for all w p. If for one w we have a 0 mod 3, then i 5 p, hence also (i 5) 2 = 5 and therefore 2 3 5 = 1. This can not be the case since p is a proper ideal. Hence a 0 mod 3 for any element w. Then b 0 mod 3 for all w. It follows that a 0 mod 3 since if not, p would not be proper, and hence p =(3)Z[i 5]. But this is not a prime ideal: 6 = (1+i 5)(1 i 5) (3)Z[i 5]. c) One checks that m =(2, 1+i 5) is a maximal ideal in Z[i 5] with residue field F 2 as above, using the following equalities between ideals Now (X 2 +1, 2,X +1)=((X +1)(X 1) + 6, 2,X +1)=(2,X +1). m 2 =(2 2, 2(1 + i 5), (1 + i 5) 2 )=(4, 2(1 + i 5), 4+2i 5) = (2), so if p is any prime ideal with 2 p, we have m 2 p. Hence m p because p is prime, and knowing that m is maximal we conclude that p = m. This shows that S 1 Z[i 5] is a local ring with residue field F 2. 5