August 2015 Qualifying Examination Solutions


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1 August 2015 Qualifying Examination Solutions If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems, in principle, will be accessible during the entire duration of the exam. 1
2 Abstract Algebra In what follows, if n is a positive integer, the ring of integers modulo n is denoted by Z n ; if m is an arbitrary integer, its class modulo n is denoted by m Z n. For a group G, Aut(G) denotes the group of automorphisms of G, under composition of maps. 1. Prove or give a counterexample. (i) (2 points) Let G be a finite group and H, K two normal subgroups of G. Then G/H K if and only if G/K H. (ii) (2 points) The integral domain S = Z[ 5] = {a + b 5 a, b Z} is a Unique Factorization Domain. (iii) (2 points) The polynomial q(x) = x 4 + x 3 + x Z 3 [x] is irreducible over Z (i) (3 points) Let m 2 and n be two positive integers such that m n. Show that Aut(Z m Z n ) is a nonabelian group by exhibiting two noncommuting isomorphisms of Z m Z n. (HINT: Begin by showing that the correspondence f : Z m Z n Z m Z n, f(s, t) = (s + t, t), (s, t) Z m Z n, is a welldefined map and a nonidentity isomorphism of Z m Z n.) (ii) (4 points) Show that for a finite abelian group G, the group of automorphisms Aut(G) is abelian if and only if G is cyclic. 3. Consider the Unique Factorization Domain Z[ 2] = {a + b 2 a, b Z}. (You do not need to prove that Z[ 2] is a UFD.) (i) (2 points) Show that { 1, 1} is the set of units of Z[ 2] and that 2 is an irreducible element of Z[ 2]. (ii) (2 points) Let x Z. Describe the elements of Z[ 2] that divide both x + 2 and x 2. (iii) (4 points) Use part (i) and (ii) to find all integer solutions to the equation: x = y Let Z[x] R = (x 4 + x 3 + x 2 + 1), and denote by 3 the image of 3 in R, i.e. 3 = 3 + (x 4 + x 3 + x 2 + 1) R. (i) (3 points) Show that R is an integral domain but not a field. (ii) (3 points) Show that (3), the ideal of R generated by 3, is a maximal ideal of R. (iii) (3 points) Let K = R/(3). Show that K is a finite field and find the number of elements of K.
3 Abstract Algebra Solutions In what follows, if n is a positive integer, the ring of integers modulo n is denoted by Z n ; if m is an arbitrary integer, its class modulo n is denoted by m Z n. For a group G, Aut(G) denotes the group of automorphisms of G, under composition of maps. 1. Prove or give a counterexample. (i) (2 points) Let G be a finite group and H, K two normal subgroups of G. Then G/H K if and only if G/K H. (ii) (2 points) The integral domain S = Z[ 5] = {a + b 5 a, b Z} is a Unique Factorization Domain. (iii) (2 points) The polynomial q(x) = x 4 + x 3 + x Z 3 [x] is irreducible over Z 3. Solution 1. (i) The answer is NO. Let G = Q 8 = {±1, ±i, ±j, ±k}, H = {±1, ±i}, and K = {±1}. One easily checks that H and K are normal subgroups of G and that G/H is (cyclic) of order 2 and G/K = {1, i, j, k} is the Klein group of order 4. So, G/H K but G/K H. (ii) The answer is NO. Let r = 3 S. We will show that r is irreducible but not prime in S. Using the norm N : S Z, N(a + b 5) = a 2 + 5b 2, a + b 5 S, one immediately checks that r is irreducible. On the other hand, (2 + 5)(2 5) = 9 (r) but it can be easily seen that neither nor 2 5 belongs to (r). This shows that r is irreducible but not prime in S. Consequently, S can not be a UFD. (iii) The answer is YES. Assume for a contradiction that q is reducible over Z 3. Since q(0) = 1, q(1) = 1, q(2) = 2, q does not have any linear factors over Z 3. So, we can write: q(x) = (ax 2 + bx + c)(dx 2 + ex + f) in Z 3 [x]. Equating coefficients, we get: ad = 1, ae + bd = 1, af + be + dc = 1, bf + ce = 0, cf = 1. Since ad = cf = 1, a = d {1, 2} and c = f {1, 2}. Then, the second equation gives that b + e is either 1 or 2, while the 4th equation gives that b + e is zero (contradiction)! So, q must be irreducible over Z (i) (3 points) Let m 2 and n be two positive integers such that m n. Show that Aut(Z m Z n ) is a nonabelian group by exhibiting two noncommuting isomorphisms of Z m Z n. (HINT: Begin by showing that the correspondence f : Z m Z n Z m Z n, f(s, t) = (s + t, t), (s, t) Z m Z n, is a welldefined map and a nonidentity isomorphism of Z m Z n.)
4 (ii) (4 points) Show that for a finite abelian group G, the group of automorphisms Aut(G) is abelian if and only if G is cyclic. Solution 2. (i) Let s s and t t. Then s s = am and t t = bn for some m, n Z. Since m n, we get that s + t (s + t) is divisible by m, i.e. s + t = s + t Z m. This shows that the correspondence f is a welldefined map. It is also immediate to check that f is an injective group endomorphism of Z m Z n. Therefore, it has to be an automorhism. It is clear that f is not the identity automorphism since f(0, 1) = (1, 1) (0, 1). If n 2 then g : Z m Z n Z m Z n, g(s, t) = (s, t) is a nonidentity automorphism, g f, and (g f)(0, 1) = (1, 1) ( 1, 1) = (f g)(0, 1). If n = 2 then h : Z m Z n Z m Z n, h(s, t) = (t, s) is a nonidentity automorphism, h f, and (h f)(0, 1) = (1, 1) (1, 0) = (f h)(0, 1). The discussion above shows that Aut(Z m Z n ) is not abelian. (ii) ( ) From the Fundamental Theorem of Finite Abelian Groups, we know that G Z m1... Z mn, where m 1 m 2... m n and m 1 2. If n 2 than Aut(Z m1 Z m2 ) is clearly (isomorphic to) a subgroup of Aut(G) and Aut(Z m1 Z m2 ) is not abelian by (i). Therefore, Aut(G) can not be abelian if n 2. So, n = 1, i.e. G Z m1 is cyclic. ( ) Assume that G = Z m for some integer m 2. If f Aut(G) then f(1) is a generator of Z m. So, there exists a unique integer 1 n f n 1 with (n f, n) = 1 such that f(1) = n f. Moreover, f(1) completely determines f. If f, g Aut(G) then (f g)(1) = n f n g = (g f)(1) which implies that f g = g f. So, Aut(G) is an abelian group. 3. Consider the Unique Factorization Domain Z[ 2] = {a + b 2 a, b Z}. (You do not need to prove that Z[ 2] is a UFD.) (i) (2 points) Show that {±1} is the set of units of Z[ 2] and that 2 is an irreducible element of Z[ 2]. (ii) (2 points) Let x Z. Describe the elements of Z[ 2] that divide both x + 2 and x 2. (iii) (4 points) Use part (i) and (ii) to find all integer solutions to the equation: x = y 3. Solution 3. (i) Let N : Z[ 2] Z, N(a + b 2) = a 2 + 2b 2, a + b 2 Z[ 2], be the standard norm on Z[ 2]. If a + b 2 is a unit of Z[ 2] then N(a + b 2) = a 2 + 2b 2 = 1 and hence a = ±1 and b = 0. So, 1 and 1 are the only units of Z[ 2]. Now, let us check that 2 is irreducible. If c + d 2 divides 2 then, after applying the norm, we get that c 2 + 2d 2 divides N( 2) = 2. So, either c = ±1
5 and d = 0 or else c = 0 and d = ±1. This shows that if 2 = x y in Z[ 2] then either x = ±1 or y = ±1 since otherwise we would get that 2 = ± 2 2 = ±2 which is impossible. Hence 2 is irreducible in Z[ 2]. (ii) Let d = a + b 2 Z[ 2] be a common divisor of x + 2 and x 2. Then d divides 2 2 and hence N(d) = a 2 + 2b 2 divides N(2 2) = 8. If a 2 + 2b 2 = 1 then a = ±1 and b = 0, i.e. d = ±1. If a 2 + 2b 2 = 2 then a = 0 and b = ±1, i.e. d = ± 2. This can happen precisely when x is even. If a 2 + 2b 2 = 4 then a = ±2 and b = 0, i.e. d = ±2. However, 2 can never divide an element of the form x ± 2. If a 2 + 2b 2 = 8 then a = 0 and b = ±2, i.e. d = ±2 2. Again, 2 2 can never divide an element of the form x ± 2. In conclusion, d is either a unit (when x is odd) or ± 2 (when x is even). (iii) Let x and y be integers such that x = y 3. In Z[ 2], we have: (x + 2)(x 2) = y 3. It follows form part (ii) that the greatest common divisor of x + 2 and x 2 is either ±1 or ± 2. If p ± 2 is an irreducible factor of x + 2 of multiplicity n p 1 then p is an irreducible factor of y 3 of multiplicity n p. So, n p is divisible by 3. Note that for this argument to work we need to know that Z[ 2] is a UFD. Now, if 2 is an irreducible factor of x + 2 of multiplicity n then 2 is an irreducible factor of x 2 of the same multiplicity n. So, the multiplicity of 2 as an irreducible factor of y 3 is 2n. We conclude that 2n, and hence n, must be divisible by 3. The discussion above shows that x + 2 is a perfect cube in Z[ 2]. So, we can write: x + 2 = (a + b 2) 3 with a, b Z. From this, we get that a 3 6ab 2 = x and (3a 2 2b 2 )b = 1, which implies that a = b = ±1. Hence, the only integral solutions to x = y 3 are x = ±5 and y = Let R = Z[x] (x 4 + x 3 + x 2 + 1), and denote by 3 the image of 3 in R, i.e. 3 = 3 + (x 4 + x 3 + x 2 + 1) R. (i) (3 points) Show that R is an integral domain but not a field. (ii) (3 points) Show that (3), the ideal of R generated by 3, is a maximal ideal of R. (iii) (3 points) Let K = R/(3). Show that K is a finite field and find the number of elements of K.
6 Solution 4. (i) Set p(x) = x 4 +x 3 +x 2 +1 Z[x] and q(x) = x 4 +x 3 +x 2 +1 Z 3 [x]. From Problem 1(iv), we know that q(x), the reduction of p(x) mod 3, is irreducible over Z 3. Therefore, p(x) remains irreducible over Z. So, the ideal (p(x)) is a prime ideal of Z[x]. Therefore, R is an integral domain. To show that R is not a field, we show that I := (p(x)) is not a maximal ideal of Z[X]. Recall that for any maximal ideal M of Z[x] has the property that M Z is a nonzero ideal. So, assuming that I is maximal, I Z would be of the form (p) with p a prime. But I contains no nonzero integers. Therefore, I can not be maximal. Alternatively, one can invoke part (ii) to justify that I is not maximal since it has a proper ideal, namely (3). (ii) Using the 3rd Isomorphism Theorem, we have: R/(3) = Z[x] (p(x)) 3Z[x]+(p(x)) (p(x)) Z[x] 3Z[x] + (p(x)) Z 3[x]/(q(x)). (iii) We have seen that K Z 3 [x]/(q(x)) with q(x) irreducible over Z 3. So, K/Z 3 is a field extension of degree 4 with {1, α, α 2, α 3 } a Z 3 basis of K. (Here α := x + (q(x)) Z 3 [x]/(q(x)).) We conclude that K = 3 4 = 81.
7 Linear Algebra In what follows, F stands for a field of characteristic 0; V stands for an n dimensional vector space over the field F with 0 n ; T : V V stands for a linear operator on V with f T (X) and p T (X) its characteristic and minimal polynomials, and Im(T ) and Ker(T ) its image and kernel. In problems 1 and 2, determine which of the statements (i), (ii), (iii) is TRUE and which is FALSE. If true, provide a proof to the statement; if false, provide a counterexample. 1. (6 points) Let U and W be subspaces of V such that V = U W, and dim F U 1 and dim F W 1. Let T : V V be surjective. (i) Then T (V ) = T (U) T (W ). (ii) If T (U) W, and T (W ) U, then there is an ordered basis B for V, and two l l matrices A and( B over F such ) that the representing matrix [T ] B for T 0l l B with respect to B is A =. A 0 l l (iii) If T (U) W, then T (W ) U. 2. (6 points) Let T : V V be a nonzero linear operator, and F be algebraically closed. (i) If T has 0 F as its only eigenvalue, then T is not diagonalizable. (ii) Let [α 1, α 2, α 3,.., α l ] be a basis for Ker(T ), and [α 1, α 2, α 3,.., α l, α l+1,.., α n ] be a basis for V. Then [α l+1, α l+2, α l+3,.., α n ] is a basis for Im(T ). (iii) Let g 1 (X), g 2 (X) F [X] be such that g 1 (T ), g 2 (T ) are nilpotent. Then g 1 (T ) + g 2 (T ) is nilpotent also. 3. (3 points) Let f T (X) = (X + 1)X 2. Write down all the possible Jordan canonical forms for T. Justify your answer. 4. (4 points) Let m and n be positive integers and F be a field. Let V be the vector space of all m n matrices over F. Let A be a fixed m m matrix over F. Let C, D be two vectors in V. Define f A : V V F by f A (C, D) = tr(c t AD) where tr(b) stands for the trace of an n n matrix B, and C t stands for the transpose of C. Check whether f A is a bilinear form on V. 5. (6 points) Let f T (X) be an irreducible polynomial; let h(x) F [X]; let W be the subspace h(t )(V ) of V. Determine what W is if (i) h(x) and f T (X) are relatively prime, (ii) f T (X) divides h(x). 6. (5 points) Let f T (X) be an irreducible polynomial. Let W be a nonzero invariant subspace of V. Prove that there is a polynomial h(x) F [X] such that W = h(t )(V ).
8 Linear Algebra Solutions In what follows, F stands for a field of characteristic 0; V stands for an n dimensional vector space over the field F with 0 n ; T : V V stands for a linear operator on V with f T (X) and p T (X) its characteristic and minimal polynomials, and Im(T ) and Ker(T ) its image and kernel. In problems 1 and 2, determine which of the statements (i), (ii), (iii) is TRUE and which is FALSE. If true, provide a proof to the statement; if false, provide a counterexample. 1. (6 points) Let U and W be subspaces of V such that V = U W, and dim F U 1 and dim F W 1. Let T : V V be surjective. (i) Then T (V ) = T (U) T (W ). (ii) If T (U) W, and T (W ) U, then there is an ordered basis B for V, and two l l matrices A and( B over F such ) that the representing matrix [T ] B for T 0l l B with respect to B is A =. A 0 l l (iii) If T (U) W, then T (W ) U. Solution 1. Since T is onto, and dim F V = n, then T must be an isomorphism. It follows that T (V ) = V, dim F T (U) = dim F U, and dim F T (W ) = dim F W. (i) TRUE: Since U W = V = T (V ) = T (U + W ) = T (U) + T (W ), then dim F U+dim F W = dim F V = dim F T (V ) = dim F T (U)+dim F T (W ) dim F [T (U) T (W )] = dim F U +dim F W dim F [T (U) T (W )]. Thus, dim F [T (U) T (W )] = 0, and, hence, T (U) T (W ) = {0 V }. Hence T (V ) = T (U) T (W ). (ii) TRUE: By the inclusions in the hypothesis, dim F U = dim F T (U) dim F W. Similarly, dim F W = dim F T (W ) dim F U. So dim F T (U) = dim F U = dim F W = dim F T (W ). Thus T (U) W, and T (W ) U imply that T (U) = W, and T (W ) = U. Then, there are B 1 = [α 1, α 2, α 3,.., α l ] and B 2 = [α l+1, α l+2, α l+3,.., α l2 ], bases for U and W, respectively. Then B = [α 1, α 2, α 3,.., α l, α l+1, α l+2, α l+3,.., α l2 ] is an ordered basis for V. Since T (U) W, and T (W ) U, the result ( ) follows. 1 (iii) FALSE: Let V = F 2 with U the subspace generated by, and W 0 ( ) ( ) ( ) ( ) the subspace generated by. Let T map into, and into ( ) 1. Then T (U) = W, but T (W ) U (6 points) Let T : V V be a nonzero linear operator, and F be algebraically closed. (i) If T has 0 F as its only eigenvalue, then T is not diagonalizable. (ii) Let [α 1, α 2, α 3,.., α l ] be a basis for Ker(T ), and [α 1, α 2, α 3,.., α l, α l+1,.., α n ] be a basis for V. Then [α l+1, α l+2, α l+3,.., α n ] is a basis for Im(T ).
9 (iii) Let g 1 (X), g 2 (X) F [X] be such that g 1 (T ), g 2 (T ) are nilpotent. Then g 1 (T ) + g 2 (T ) is nilpotent also. Solution 2. (i) TRUE: Since F is algebraically closed, and since 0 F is the sole eigenvalue for T, then f T (X) = X n. By CaleyHamilton, p T (X) = X s for some 1 s n. If s = 1, then T 0, contradicting the hypothesis that T is nonzero. Thus 1 s, and, therefore, T is not diagonalizable. ( ) ( ) 1 0 (ii) FALSE: Let V = F 2, and let T map α 1 := into, and 0 0 ( ) ( ) 0 1 α 2 := into. Then [α ] is an ordered basis for Ker(T ), and [α 1, α 2 ] is an ordered basis for V. Now α 2 / Im(T ), and hence [α 2 ] is not an ordered basis for Im(T ). (iii) TRUE: Note that g 1 (T g 2 (T ) g 2 (T ) g 1 (T ). Thus we have that (g 1 (T ) + g 2 (T )) r ( ) r r i=0 g i 1 r i (T ) g2(t i ). Let g r 1 1 (T ) 0 g r 2 2 (T ). If r := r 1 + r 2, then either r i r 1, and therefore g 1 (T ) r i 0; or i r 2, and therefore g2(t i ) 0. Thus g1 r i (T ) g2(t i ) 0, and hence (g 1 (T ) + g 2 (T )) r 0, showing g 1 (T ) + g 2 (T ) to be nilpotent. 3. (3 points) Let f T (X) = (X + 1)X 2. Write down all the possible Jordan canonical forms for T. Justify your answer. Solution 3. If T is diagonalizable, then p T (X) = (X +1)X. So, the primary decomposition of T is {p 1 (X) := (X + 1)X = p T (X), p 2 (X) := X}, and hence the Jordan and rational canonical forms are J = , and R = , respectively. If T is not diagonalizable, then p T (X) = (X +1)X 2 = f T (X). So, the primary decomposition of T is {p 1 (X) := (X + 1)X 2 = p T (X)}, and hence the Jordan and rational canonical forms are J = , and R = , respectively. 4. (4 points) Let m and n be positive integers and F be a field. Let V be the vector space of all m n matrices over F. Let A be a fixed m m matrix over F. Let C, D be two vectors in V. Define f A : V V F by f A (C, D) = tr(c t AD) where tr(b) stands for the trace of an n n matrix B, and C t stands for the transpose of C. Check whether f A is a bilinear form on V.
10 Solution 4. Yes, f A is a bilinear form by the following: Let c F and C 1, D 1 V. Then (i) f A (cc 1 +C, D) = tr[(cc 1 +C) t AD] = tr[(cc t 1+C t )AD] = tr(cc t 1AD+C t AD) = tr(cc t 1AD) + tr(c t AD) = ctr(c t 1AD) + tr(c t AD) = cf A (C 1, D) + f A (C, D), and (ii) f A (C, cd 1 + D) = tr[c t A(cD 1 + D)] = tr(cc t AD 1 + C t AD) = tr(cc t AD 1 ) + tr(c t AD) = ctr(c t AD 1 ) + tr(c t AD) = cf A (C, D 1 ) + f A (C, D). 5. (6 points) Let f T (X) be an irreducible polynomial; let h(x) F [X]; let W be the subspace h(t )(V ) of V. Determine what W is if (i) h(x) and f T (X) are relatively prime, (ii) f T (X) divides h(x). Solution 5. If f T (X) is irreducible, then by CayleyHamilton f T (X) = p T (X), and, so V is a T cyclic space, that is, V = {g(t )(α) g(x) F [X]} for some α V. Now, W is T invariant since T (W ) = T (h(t )(V )) = h(t )(T (V )) h(t )(V ) = W. (i) Since h(x) and f T (X) are relatively prime, there exist k(x), l(x) F [X] such that h(x)k(x)+l(x)f T (X) = 1. Since f(t )(V ) = {0 V }, then h(t )(k(t )(β)) = h(t )(k(t )(β)) + l(t )(f T (T )(β)) = β for all β V. But h(t )(k(t )(β)) W. So β W. So V W, and thus V = W. Alternate proof: Since h(x) and f T (X) are relatively prime, there exist k(x), l(x) F [X] such that h(x)k(x)+l(x)f T (X) = 1. Hence, h(t )(k(t )(α)) = h(t )(k(t )(α))+ l(t )(f T (T )(α)) = α. But h(t )(k(t )(α)) W. So α W. But, W is Tinvariant, and, hence, T i (α) W for all i N. Since V = {g(t )(α) g(x) F [X]}, then V W, and so W = V. Alternate proof: Since W is T invariant, then f T W (X) divides f T (X). But f T (X) is irreducible, and so f T W (X) = 1 or f T W (X) = f T (X). The former indicates that h(t )(V ) = W = {0 V }, in which case f T (X) divides h(x). That is a contradiction. The latter indicates that dim(w ) = deg(f T W (X)) = deg(f T (X)) = dimv. Since W V, then W = V. (ii) If f T (X) divides h(x), then h(t ) 0. Hence W = h(t )(V ) = {0 V }. 6. (5 points) Let f T (X) be an irreducible polynomial. Let W be a nonzero invariant subspace of V. Prove that there is a polynomial h(x) F [X] such that W = h(t )(V ). Solution 6. By 5., V = {g(t )(α) g(x) F [X]}, and therefore there are polynomials l 1 (X), l 2 (X),..., l d (X) F [X] such that W =< l 1 (T )(α), l 2 (T )(α),..., l d (T )(α) >. Let h(x) be the greatest common divisor of {l 1 (X), l 2 (X),..., l d (X)}. Then there are relatively prime polynomials k i (X) F [X] such that l i (X) = k i (X)h(X) for i {1, 2,..., d}. Let β := h(t )(α). Then W =< k 1 (T )(β), k 2 (T )(β),..., k d (T )(β) >. Since W is T invariant, then g(t )(k i (T )(β)) W for all g(x) F [X], for all i = 1, 2,..., d. Let g i (X) F [X] be such that 1 = d i=1 g i(x)k i (X). Then β = d i=1 g i(t )(k i (T )(β)) W. Since W is T invariant, then < T i (β) i N > W =< k 1 (β), k 2 (β),..., k d (β) > < T i (β) i N >. Hence W =< T i (β) i N >.
11 So W =< h(t )(T i (α)) i N >= h(t )(V ). Alternate solution: As in the second Alternate solution of 5(ii), W = {0 V }, or W = V. The former contradicts the fact that W is a nonzero subspace of V. The latter implies that W = h(t )(V ) where h(x) = 1.
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