Solutions A ring A is called a Boolean ring if x 2 = x for all x A.


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1 1. A ring A is called a Boolean ring if x 2 = x for all x A. (a) Let E be a set and 2 E its power set. Show that a Boolean ring structure is defined on 2 E by setting AB = A B, and A + B = (A B c ) (B A c ) (superscript c denotes complementation). Solution: Easy. (b) Prove that every Boolean ring is commutative and such that x + x = 0 for all x A. (Hint. First consider (x + x) 2 and then (x + y) 2.) Solution: (x+x) = (x+x) 2 = x 2 +xx+xx+x 2 = x+x+x+x = (x+x)+(x+x). Hence, x + x = 0 for all x A. Now, (x + y) = (x + y) 2 = x 2 + xy + yx + y 2 = x + y + xy + yx. This implies that xy + yx = 0. But, yx = yx from what we just proved. Therefore, xy yx = 0 and hence xy = yx for all x, y A. (c) Prove that every Boolean ring is commutative and such that x + x = 0 for all x A. (Hint. First consider (x + x) 2 and then (x + y) 2.) Solution: (x+x) = (x+x) 2 = x 2 +xx+xx+x 2 = x+x+x+x = (x+x)+(x+x). Hence, x + x = 0 for all x A. Now, (x + y) = (x + y) 2 = x 2 + xy + yx + y 2 = x + y + xy + yx. This implies that xy + yx = 0. But, yx = yx from what we just proved. Therefore, xy yx = 0 and hence xy = yx for all x, y A. (d) Prove that if a Boolean ring contains no divisors of 0 it is either {0} or is isomorphic to Z/(2) (show that xy(x + y) = 0 for all x, y A). Deduce that in a Boolean ring every prime ideal is maximal. Solution: Since, A is Boolean it is commutative and x + x = 0 for all x A. Let x, y A be nonzero. Now, xy(x + y) = x 2 y + xy 2 = xy + xy = 0. Hence, either A has a divisor of zero or x + y = 0 for every nonzero x, y A. In the latter case, x = y = y and A can have only one nonzero element. Hence, A = Z/(2). If P A is a prime ideal. Then, A/P is also a Boolean ring with no zero divisors (because P is prime). But then, A/P = Z/(2) which is a field and hence P must be maximal. (e) Prove that in a Boolean ring every ideal I A is the intersection of the prime ideals containing I. Solution: Since x 2 = x for every x A, we have that x n = x for all n > 0. Hence, every ideal I of A is equal to its radical. Now apply the result of the previous problem. 2. Let A be a ring such that x 3 = x for all x A. The goal is to prove that A is commutative. (a) Show that 6A = {0} and that 2A and 3A are twosided ideals such that 2A+3A = A and 2A 3A = {0}. Deduce that it can be assumed that either 2A = {0} or 1
2 3A = {0} (for the purpose of this problem). Solution: We have that (x + x) = (x + x) 3 = x 3 + 3x 2 x + 3xx 2 + x 3 = 8x. Hence, 6x = 0 for all x A. It is very easy to show that 2A and 3A are twosided ideals. To show that 2A + 3A = A, consider any element x A. Now, 6x = 0. Hence, x = 5x = 5( x) = 2( x) + 3( x). To show that 2A 3A = {0}, consider any element x 2A 3A. Then, there exists a, b A such that x = 2a = 3b. But, then 3x = 6a = 2x = 6b = 0. Hence, x = 3x 2x = 0. In order to show that A is commutative we have to show that xy = yx for all x, y A. Suppose that x 2A, y 3A. Since, 2A, 3A are twosided ideals xy, yx 2A 3A and hence both equal to 0. Therefore, we need only to consider products of the form xy with x, y 2A or x, y 3A. But, 2A is a subring with x 3 = x for all x 2A and such that 3(2A) = 0. Similarly, 3A is a subring with x 3 = x for all x 3A and such that 2(3A) = 0. Thus, if we prove commutativity of all rings with x 3 = x and 2A = 0 and the same with 3A = 0 then we are done. (b) If 2A = 0, calculate (1 + x) 3 to deduce that x 2 = x for all x A and conclude by means of the previous problem. Solution: (1 x) = (1 x) 3 = 1 3x + 3x 2 x 3 = 1 + (x + 2x) (x 2 + 2x 2 ) x. From which we get that, x x 2 = 0 or x = x 2. Now apply the result of the previous problem. (c) If 3A = {0}, calculate (x + y) 3 and (x y) 3 to show that x 2 y + xyx + yx 2 = 0. Now left multiply by x to deduce that xy yx = 0. Solution: (x + y) = (x + y) 3 = (x + y)(x 2 + xy + yx + y 2 ) = x 3 + x 2 y + xyx + xy 2 + yx 2 + yxy + y 2 x + y 3. Thus, x 2 y + xyx + xy 2 + yx 2 + yxy + y 2 x = 0. Similarly, (x y) = (x y) 3 = (x y)(x 2 xy yx + y 2 ) = x 3 x 2 y xyx + xy 2 yx 2 + yxy + y 2 x y 3, from which we get that, x 2 y xyx + xy 2 yx 2 + yxy + y 2 x = 0. Subtracting second from the first equation, we get 2(x 2 y + xyx + yx 2 ) = 0. Hence, x 2 y + xyx + yx 2 = 3(x 2 y + xyx + yx 2 ) 2(x 2 y + xyx + yx 2 ) = 0. Left multiplying by x we have that, x 3 y + x 2 yx + xyx 2 = xy + (x 2 y + xyx)x = 0. Now, x 2 y + xyx = yx 2. Substituting back we get that, xy yx 2 x == xy yx = Let A be a commutative ring. Recall that A is a local ring if it has a unique maximal ideal. Prove that the following is an alternative definition. A is local iff for every element a A either a or 1 + a is invertible. Solution: Let A be a local ring and let M be its unique maximal ideal. We first show 2
3 that x A is invertible iff x M. Let x M. If ax = 1 for some a A, it would imply that 1 M and thus M = A. This is impossible and hence x is not invertible. If x A \ M. We claim that in this case (x) = A and hence x is invertible. Suppose not. Consider the set S of ideals containing x but not equal to A partially ordered by inclusion. Then, S is nonempty and every chain has a maximal element. Then S has a maximal element by Zorn s lemma. This maximal element is a maximal ideal and by uniqueness has to be M. but then x M, a contradiction. Finally note that for any element a A, either a M and hence is invertible. Else, a M but in this case 1 + a M because otherwise 1 M. So in this case 1 + a is invertible. Now, let A be a commutative ring with the property that for every element a A either a or 1 + a is invertible. Let M be the subset of A consisting of all noninvertible elements. We claim that M is the unique maximal ideal and hence A is local. We first check that M is an ideal. If a, b M, then we claim a + b is also noninvertible. Assume otherwise, Then, there must exist c A such that (a + b)c = 1 and hence ac = 1 bc. But, b and hence bc is noninvertible. Therefore 1 bc must be invertible which would make a invertible  a contradiction. It is clear that M is the unique maximal ideal since for any other ideal N and x N, x M would imply that x is invertible and hence N = (1) = A. 4. Let k be a field and k[[x]] denote the ring of formal power series in x with coefficients in k. Show that k[[x]] is a local ring. What is its unique maximal ideal? (Hint. Use the result proved in the previous problem). Solution: The invertible elements of k[[x]] are exactly those with constant terms not equal to 0. Hence, its clear that for every element a A either a or 1 + a is invertible, making k[[x]] a local ring. The unique maximal ideal of k[[x]] consists of all the noninvertible elements, namely the elements with constant term Consider the following commutative diagram of Amodules and homomorphisms such that each row is exact. M 1 M 2 M 3 M 4 M 5 a b c d e N 1 N 2 N 3 N 4 N 5 Prove (by diagram chasing) that if a, b, d, e are isomorphisms then so is c. Solution: Let φ i (resp. ψ i ) for 1 i 4 be the horizontal homomorphisms M i M i+1 (resp. N i N i+1 ). We first show that c is injective. Let c(x 3 ) = 0 for some x 3 M 3. Then dφ(x 3 ) = 0 φ 3 (x 3 ) =, because d is an isomorphism. Hence, x 3 ker(φ 3 ) = Im(φ 2 ). Let x 2 M 2 3
4 be such that x 3 = φ 2 (x 2 ). But then, ψ 2 b(x 2 ) = 0 b(x 2 ) ker(ψ 2 ) = Im(ψ 1 ). Let y 1 N 1 be such that ψ 1 (y 1 ) = b(x 2 ). Since a is an isomorphism there exists x 1 M 1 such that y 1 = a(x 1 ) and ψ 1 a(x 1 ) = b(x 2 ) = bφ 1 (x 1 ). Since b is an isomorphism this implies that x 2 = φ 1 (x 1 ), and thus x 3 = φ 2 φ 1 (x 1 ) = 0. Next we show that c is surjective. Let y 3 N 3. Since d is surjective there exists x 4 M 4 such that ψ 3 (y 3 ) = d(x 4 ). Now, ψ 4 ψ 3 (y 3 ) = 0 = eφ 4 (x 4 ). Since e is injective this implies that x 4 ker(φ 4 ) = Im(φ 3 ). Let x 3 M 3 be such that x 4 = φ 3 (x 3 ). Then, dφ 3 (x 3 ) = ψ 3 c(x 3 ) = ψ 3 (y 3 ). Hence, c(x 3 ) y 3 ker(ψ 3 ) = Im(ψ 2 ). Let y 2 N 2 be such that ψ 2 (y 2 ) = c(x 3 ) y 3. There exists x 2 M 2 such that ψ 2 b(x 2 ) = c(x 3 ) y 3 = cφ 2 (x 2 ). But then, c(x 3 φ 2 (x 2 )) = y 3 showing that c is surjective. 6. Let k be a field. The ring k[x, y] can be viewed as a kmodule, as a k[x]module, as a k[y]module, or as a k[x, y]module. Illustrate the differences between these structures by providing nontrivial examples of maps from k[x, y] to itself which are: (a) kmodule homomorphism but not a k[x], k[y], k[x, y]module homomorphisms; Solution: Example: the homomorphism that sends f(x, y) to the polynomial f(y, x) is an example of such a homomorphism. (b) a k[x]module homomorphism but not a k[y], k[x, y]module homomorphisms; Solution: Example: the homomorphism that send f(x, y) to the polynomial f(x, y 2 ). (c) a ring homomorphism but not a k[x, y]module homomorphism. Solution: The example in part (a). 7. Let A be a commutative ring and let X = SpecA. Recall that for each subset E of A, V (E) = {p X E p}. For each f A, let X f be the complement of V (f) in X. The sets X f are open in the Zariski topology. Show that, (a) the sets X f form a basis of open sets for the Zariski topology (that is every open set is a union of some of the sets X f ); Solution: An open set U is the complement of a closed set V (E). Now, V (E) = f E V f. Hence, U = f E X f. (b) X f X g = X fg ; Solution: By definition, X f = {p SpecA f p} and X g = {p SpecA g p}. Thus, X f X g = {p SpecA f p, g p}. Also, X fg = {p SpecA fg p}. Now, for prime ideals p, fg = p if and only if f p, g p. Hence, X f X g = X fg. (c) X f = iff f is nilpotent; Solution: X f = implies that f p for every p SpecA. This, implies that f p SpecA = nilradical(a). Hence, f is nilpotent. Converesely, if f is nilpotent then it is contained in every prime ideal of A and hence X f is empty. (d) X f = X iff f is a unit; Solution: X f = X implies that f is not contained in any prime ideal of f. But, 4
5 then (f) = (1), otherwise the maximal ideal containing (f) would be proper, prime and would contain f. Conversely, if (f) = (1) then f cannot be contained in any prime ideal as otherwise that prime ideal would have to contain 1. (e) X f = X g iff (f) = (g); Solution: Suppose X f = X g. Now, (f) is the intersection of all prime ideals containing (f) and equals p Vf p. Similarly, (g) is the intersection of all prime ideals containing (g) and equals p Vg p. But X f = X g implies that V f = V g and hence, (f) = (g). Conversely, suppose that (f) = (g). If p X f then f p. Suppose that g p. Then, p contains (g) = (f), and hence, f p, a contradiction. Hence, g p and hence, p X g. Thus, X f X g and hence X f = X g. (f) X is quasicompact (that is every open covering of X has a finite subcovering). Solution: By part (a) it suffices to consider open covers using sets of the form X f. Suppose that X = α A X fα. Consider the ideal I generated by (f α ) α A. Now, if I (1) then there would exist a proper maximal ideal p containing I, and p would be prime. But, then f α p for every α A, and hence, p X fα for every α A, Thus, p α A X fα, which is a contradiction. Thus, I = (1). Therefore, 1 = i I f ig i, for some finite subset I A. We claim that X = i I X fi. Let p be a prime ideal. Suppose f i p for every i I. Then, 1 p a contradiction. Hence, p f i for some i I, and p X fi. This shows that indeed X = i I X fi. 5
(a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9
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