Math 1530 Solutions for Final Exam Page 1

Transcription

1 Math 1530 Solutions for Final Exam Page 1 Problem 1. (10 points) Let G be a finite group and let H 1 and H 2 be subgroups of G. (a) Suppose that H 1 H 2 = {e}. H 1 H 2 = G. Prove that every element g of G can be written uniquely in the form g = h 1 h 2 with h 1 H 1 and h 2 H 2. (b) Give an example of a group G with subgroups H 1 and H 2 satisfying the conditions specified in (a), but such that G is not isomorphic to H 1 H 2. Solution. (a) Consider the map f : H 1 H 2 G, f(h 1, h 2 ) = h 1 h 2. I claim that f is one-to-one. To see this, suppose that f(h 1, h 2 ) = f(h 1, h 2) with h 1, h 1 H 1 and h 2, h 2 H 2. The definition of f says that h 1 h 2 = h 1h 2, so we find that (h 1) 1 h 1 = h 2h 1 2. But (h 1) 1 h 1 H 1 and h 2h 1 2 H 2, and we are given that H 1 H 2 = {e}. Hence (h 1) 1 h 1 = e and h 2h 1 2 = e. This proves that h 1 = h 1 and h 2 = h 2, which shows that f is one-to-one. On the other hand, we are given that H 1 H 2 = H 1 H 2 = G, so we have a one-to-one map between two finite sets having the same number of elements. It follows that f is also onto. We ve now proven that f is a bijection, which is equivalent to the statement we re asked to prove. But I ll write it out in more detail. First, let g G. Then the fact that f is onto means that there is some (h 1, h 2 ) H 1 H 2 such that f(h 1, h 2 ) = g. In other words, h 1 h 2 = g, so g can be written as a product in the desired form. Next suppose that g can be written as such a product in two ways, say g = h 1 h 2 = h 1h 2. This means that g = f(h 1, h 2 ) = f(h 1, h 2), so the fact that f is one-to-one implies that h 1 = h 1 and h 2 = h 2. Hence g can be written in the desired form in exactly one way. (b) Take G = S 3, the symmetric group on three letters. Let α 1 G be an element of order 3, for example α 1 = (1 2 3), and let α 2 G be an element of order 2, say α 2 = (1 2). Then let H 1 = {e, α 1, α1} 2

2 Math 1530 Solutions for Final Exam Page 2 be the subgroup generate by α 1 and let H 2 = {e, α 2 } be the subgroup generated by α 2. Then G = 6, H 1 = 3, H 2 = 2 and H 1 H 2 = {e}. But G is not isomorphic to H 1 H 2, since G is nonabelian, but H 1 H 2 is abelian. Problem 2. (10 points) Let p and q be distinct odd primes, and let G be a group of order p 2 q. Prove that G has a nontrivial normal subgroup. Solution. Let N p be the number of p Sylow subgroups, and similarly for N q. The Sylow theorems tell us that N p q and N p 1 (mod p) and N q p 2 and N q 1 (mod q). If N p = 1, then the p-sylow subgroup is normal, and similarly if N q = 1 the q-sylow subgroup is normal. So we suppose that N p > 1 and N q > 1 and derive a contradiction. This assumption together with the known divisiblity conditions imply that N p = q and N q = p or p 2. Then the congruence conditions give q 1 (mod p) and p 1 (mod q) or p 2 1 (mod q). The congruence q 1 (mod p) implies in particular that p q 1. We consider first the case that p 1 (mod q). This implies that q p 1, which combined with p q 1 leads to the inequalities q + 1 p q 1. Since p q by assumption, we find that p = q 1 or q 1, which contradicts the assumption that p and q are both odd. We next consider the case that p 2 1 (mod q). This congruence says that q divides p 2 1, so q divides (p 1)(p + 1), so q divides either p 1 or p + 1. We ve already deals with the case that q divides p 1 (this is just another way of saying that p 1 (mod q)), so we consdier the case that q divides p + 1. This means that q p + 1, which, combined with the inequality p q 1 from above, gives q p + 1 q. Hence q = p + 1, again contradicting the assumption that p and q are both odd.

3 Math 1530 Solutions for Final Exam Page 3 Problem 3. (10 points) Let S n be the symmetric group on {1, 2,..., n}, and let A n be the alternating subgroup consisting of even permutations. We may view S n as being the subgroup of S n+1 consisting of elements that fix n + 1. With this viewpoint, prove that A n = A n+1 S n. Solution. By definition, A n = {σ S n : sgn(σ) = 1}. First let σ A n. This means that σ = τ 1 τ 2 τ k with τ i S n transpositions and k even. The τ i are also transpositions in S n+1 (that happen to fix n + 1), so when we view σ as an element of S n+1, it is still a product of an even number of transpositions. Hence σ A n+1. This proves that A n A n+1 S n. For the other inclusion, let σ A n+1 S n. Since σ S n, we know that we can write it as a product of transpositions in S n, say σ = τ 1 τ 2 τ k with τ i S n transpositions. And the number of transposition k satisfies sgn(σ) = ( 1) k. But we are also given that σ A n+1, so whenever we write σ as a product of transpositions in S n+1, we know that there are an even number of transpositions in the product. In particular, since we have σ = τ 1 τ k, we must have k even. Hence sgn(σ) = ( 1) k = 1, so σ A n. Problem 4. (12 points) Let G be a group, let R be a commutative ring with identity, and let F be a field. Complete the following definitions (write neatly): (a) A subgroup H of G is a normal subgroup if... (b) A subset I of R is an ideal if... (c) A polynomial f(x) F [x] is irreducible if... (d) An ideal I of R is principal if... (e) The group G is simple if... (f) Let K be an extension field of F. An element α K is algebraic over F if... Solution. (a) A subgroup H of G is a normal subgroup if... ghg 1 = H for every g G. (b) A subset I of R is an ideal if... for every a, b I and every r R we have a + b I and ra I.

4 Math 1530 Solutions for Final Exam Page 4 (c) A polynomial f(x) F [x] is irreducible if... f(x) 0 and if the only way to factor f(x) as f(x) = g(x)h(x) with g, h F [x] is to have deg(g) = 0 or deg(h) = 0. (d) An ideal I of R is principal if... it has the form (a) = {ra : r R} for some a R. (e) The group G is simple if... its only normal subgroups are {e} and G. (f) Let K be an extension field of F. An element α K is algebraic over F if... there is a nonzero polynomial f(x) F [x] such that f(α) = 0. Problem 5. (9 points) In this problem, all rings are commutative and have a multiplicative identity (as usual). (a) Let R be a ring, let F be a field, and let ϕ : F R be a non-trivial homomorphism. Prove that ϕ is one-to-one. (b) Give an example of a ring R, a field F, and a non-trivial homomorphism ϕ : R F such that ϕ is onto, but is not one-to-one. (c) Give an example of fields E and F and a non-trivial homomorphism ϕ : E F such that ϕ is not onto. Solution. (a) Let I be the kernel of ϕ, that is, I = {α F : ϕ(α) = 0}. Since I is the kernel of a homomorphism, it is an ideal. We are told that ϕ is not the trivial homomorphism, so ϕ doesn t send everything to 0. This tells us that I F. We next note that in general, if an ideal I in any ring R contains a unit, then it is the whole ring. (Proof: Let u I R. Then for any r R we have r = (ru 1 )u I.) So the fact that our I is not all of F means that I contains no units of F. But every nonzero element of F is a unit, hence I = (0). Thus the kernel of ϕ is the 0-ideal, which we know is equivalent to ϕ being one-to-one. (b) This is easy. Indeed, take F to be any field, let R = F [x], let α F be any element, and let ϕ : R F be the evaluation-at-α homomorphism, ϕ : F [x] F, ϕ ( f(x) ) = f(α). Applying ϕ to constant polynomials shows that ϕ is onto, but it s certainly not one-to-one, since ϕ(x α) = 0. (c) This part is even easier that (b), any nontrivial field extension will do. So for example, ϕ : R C, the usual inclusion of R in C. Or we could use the inclusion of Q into R, or into C.

5 Math 1530 Solutions for Final Exam Page 5 Problem 6. (10 points) Let R be an integral domain, and let R[x, y] be the ring of polynomials in two variables with coefficients in R, i.e., R is the ring of polynomials n m f(x, y) = a ij x i y j with coefficients a ij R. i=0 j=0 Prove that the ring R[x, y] is an integral domain. Solution. We proved in class that if D is an integral domain, then the ring D[x] of polynomials in one variable with coefficients in D is an integral domain. Applying this to the integral domain D = R and the variable x, we conclude that R[x] is an integral domain. Now applying the same theorem using the integral domain D = R[x] and the variable y, we conclude that the ring R[x][y] is an integral domain. But this is the same as the ring R[x, y], since any polynomial in two variables can be written as a polynomial in the second variable with coefficients being polynomials in the first variable. Problem 7. (10 points) Let R be a ring (commutative with identity, as usual), and let R[x, y] be the ring of polynomials in two variables with coefficients in R, as in Problem 6. Prove that the ring R[x, y] is not a principal ideal domain. (Hint. Consider the ideal generated by x and y.) Solution. Consider the ideal I generated by x and y, that is, I = { f(x, y)x + g(x, y)y : f, g R[x, y] }. Suppose that I is principal, say I = ( h(x, y) ). Since x I and y I, this means we can write x = r(x, y)h(x, y) and y = s(x, y)h(x, y) for some r, s R[x, y]. The first equation x = rh implies that neither r nor h can have any terms in which y appears. (More formally, pick out the largest power of y appearing in r, say y k, and the largest power of y appearing in h, say y l ; then when we multiply rh, there will be a term with y k+l.) Similarly, since y = sh, neither s nor h can have any terms in which x appears. We ve shown that h(x, y) involves neither x nor y, so h is in R. On the other hand, h is in the ideal I, which is generated by x and y, so we can find f, g R[x, y] so that h = f(x, y)x + g(x, y)y. The right-hand side has not constant term, i.e., every term looks like x i y j with either i 1 or j 1. But we know that h R does t have

6 Math 1530 Solutions for Final Exam Page 6 any powers of x and y. The only way for this to happen is to have h = f = g = 0. But then I = (h) = (0) is the zero ideal, which is absurd, since x and y are in I. The contradiction proves that I is not principal, so R[x, y] is not a principal ideal domain. Problem 8. (9 points) Prove that each of the following polynomials is irreducible in the indicated ring. Be sure to clearly state your reasoning. (a) f(x) = x x 13 in Q[x]. (b) f(x) = x 3 + x 2 + x 1 in Z 3Z [x]. (c) f(x) = 3x x 5 126x 2 84 in Q[x]. Solution. (a) From a theorem in class, we know that a cubic is reducible if and only if it has a root. There are two ways to do this problem. One is to show that f(x) has no roots mod p for some prime p, since then it is irreducible in (Z/pZ)[x], and hence is irreducible in Q[x] by a theorem from class. However, one quickly checks that f(x) has a root modulo 2, and modulo 3, and modulo 5. (If one persists, then you ll find that there are no roots modulo 7, but that s a lot of work.) The other way to do the problem is to use the Rational Roots Theorem, which says that any root r Q of f(x) satisfies r 13 and s 1. Since we may also assume that s > 0, this gives ±13 as the only possible roots. So we just evaluate at these two values. Yes, I know you don t have a calculator, but it s really not so hard to cube 13. Or, even easier, note that f(13) = = 13( ) = 13( ) > 0 and f( 13) = ( 13) ( 13) 13 = 13( ) < 0., so f(13) 0 and f( 13) 0. Hence f(x) has no roots in Q, so it is irreducible in Q[x]. (b) As in (a), we use the fact that a cubic is reducible in F [x] if and only if it has a root in F. In this case, the field F = Z/3Z only has three elements, so we can check them: f(0) = 1 = 2, f(1) = 1 = 2, f(2) = 13 = 1. Thus f(x) has no roots in Z/3Z, so it is irreducible. (c) We use the Eisenstein criterion. The coefficients of f(x) = 3x x 5 126x 2 84

7 Math 1530 Solutions for Final Exam Page 7 have the following factorizations: 3 = 3, 42 = 2 3 7, 126 = , 84 = We can t use p = 2, since 2 2 divides the constant term. And we can t use p = 3, because 3 divides the coefficient of x 7. But we can use p = 7, since: 7 does not divide the coefficient of x 7 ; 7 divides every other coefficient; 7 2 does not divide the constant term. Hence Eisenstein s criterion with p = 7 implies that f(x) is irreducible in Q[x]. An alternative, but much more painful, way to show that f(x) is irreducible is to find a prime p for which it is irreducible. It turns out that the first such prime is p = 13. Note that it is not enough to show that f(x) has no roots modulo 13, one would have to check for other possible factorizations, so it would be a lot of work. Problem 9. (10 points) Let α = Prove that α is algebraic over Q. (Hint. Don t try to write down a polynomial that has α as a root.) Solution. The individual numbers 2, 3 3, 5 5, and 7 7 are algebraic over Q, since they are roots of the polynomials x 2 2, x 3 3, x 5 5, and x 7 7, respectively. We have a tower of fields Q Q( 2) Q( 2, 3 3) Q( 2, 3 3, 5 5) Q( 2, 3 3, 5 5, 7 7). To ease notation, let s give these fields names, say K 1 = Q, K 2 = Q( 2), K 3 = Q( 2, 3 3), K 5 = Q( 2, 3 3, 5 5), K 7 = Q( 2, 3 3, 5 5, 7 7). Then the multiplication property of degrees tells us that [K 7 : K 1 ] = [K 7 : K 5 ] [K 5 : K 3 ] [K 3 : K 2 ] [K 2 : K 1 ]. But K 2 = K 1 ( 2), and 2 is the root of a quadratic polynomial, so [K 2 : K 1 ] 2. Similarly, K 3 = K 2 ( 3 3), and 3 3 is the root of a cubic

8 Math 1530 Solutions for Final Exam Page 8 polynomial, so [K 3 : K 2 ] 3. And by the same reasoning, we find that [K 5 : K 3 ] 5 and [K 7 : K 5 ] 7. Hence [K 7 : K 1 ] = 210. The exact number doesn t matter, what s important is that this proves that [K 7 : K 1 ] is finite. Hence, from a theorem that we proved in class, the field K 7 is algebraic over the field K 1. By definition, that means that every element of K 7 is algebraic over K 1. In particular, the element α K 7 is algebraic over K 1, where K 1 = Q. Problem 10. (9 points) Let R be a ring (commutative with a multiplicative identity, as usual). An element a R is said to be nilpotent if there is some integer n 1 such that a n = 0. (a) Suppose that a R and b R are both nilpotent. Prove that their product ab is nilpotent. (b) Suppose that a R and b R are both nilpotent. Prove that their sum a + b is nilpotent. (Parts (a) and (b) show that the nilpotent elements form an ideal.) (c) Let p be a prime and let d 1. Prove that every element of the ring Z/p d Z is either a unit or is nilpotent. Solution. (a) The assumption that a and b are nilpotent means that a n = 0 and b m = 0 for some integers n 1 and m 1. Let k = min{n, m}. Then { (ab) k = a k b k 0 b k = 0 if k = n, = a k 0 = 0 if k = m. (b) As in (a), we assume that a n = 0 and b m = 0. This one is a little trickier. The idea is to raise a + b to a high enough power so that when we multiply it out using the binomial formula, every term a i b j has either i n or j m. Since the terms in (a + b) k have i + j = k, we can take k = m + n. (In fact, k = m + n 1 will work.) So m+n ( ) m + n (a + b) m+n = a i b m+n i. i i=0 In each term a i b m+n i, either i n, in which case a i = a i n a n = a i n 0 = 0, or else i < n, in which case m + n i m, so we similarly get b m+n i = 0. Hence every term in the sum is 0, so (a + b) m+n = 0. Therefore a + b is nilpotent.

9 Math 1530 Solutions for Final Exam Page 9 (c) The units in Z/p d Z are the congruence classes [a] with p a. (We proved that in class, but here s the proof again. If p a, then gcd(p d, a) = 1, so we can solve p d u + av = 1 with integers u and v. Hence av 1 (mod p d ), so a is a unit.) So next suppose that [a] is not a unit. This means that p a, say a = pb. But then a d = p d b d 0 (mod p d ), so [a] d = 0 in Z/p d Z. Hence [a] is nilpotent.