Algebra Prelim Solutions, August 2015

1. (a)

   Let U denote the upper unitriangular matrices of G and let D denote the diagonal matrices of G. Then it is easily checked that U⊲G, | U| = 5³, and that | D| = 4³. Therefore G has a normal Sylow 5-subgroup, which means it is unique and so P₅ = U.

   (b)

   Let

   Z = {

   (

   	1	0	z
   	0	1	0
   	0	0	1

   )

   | z∈𝔽₅}

   and let

   N = {

   (

   	1	a	0
   	0	1	b
   	0	0	1

   )

   | a, b∈𝔽₅}.

   Then N, Z⊲G (in fact Z is the center of G) and 1⊲Z⊲N⊲G is a composition series (in fact a chief series) for G, with corresponding quotients isomorphic to ℤ/5ℤ.

   (c)

   Set P₂ = D (note that P₂ is a Sylow 2-subgroup of G, however it is not normal and thus there are other choices for a Sylow 2-subgroup of G). Then P₂∩P₅ = 1, and since | P₂|·| P₅| = | G|, it follows that G is isomorphic to the semidirect product P₂⋉P₅.

2. (a)

   Let 0≠u∈U. Then u_d≠ 0 (dth entry of u) for some d, where 1≤d≤n. Let E_ij denote the matrix unit that has 1 in the (i, j)th position and zeros elsewhere. Then E_idu = u_dv_i, where v_i is the column vector that has 1 in the ith position and zeros elsewhere. It follows easily that U contains 𝔽ⁿ and hence U is simple as required.

   (b)

   Note that S is a direct sum of n copies of V as an S-module. Thus V is a projective S-module, because it is a direct summand of S. Also if I is a left ideal of S, then it has a composition series as an S-module such that each composition factor is isomorphic to V. Since V is projective, it follows that S is a direct sum (of a finite number) of copies of S and the result follows.

3. Let ω = (-1±i√3)/2, a primitive cube root of 1. Then the roots of x¹² - 1 are i^aω^b, where 0≤a≤3 and 0≤b≤2. Also the roots of x² - 2x + 2 are 1±i. It follows easily that a splitting field K for f (x) over ℚ is ℚ(i,√2). Now ℚ(i) and ℚ(√3) are Galois extensions of ℚ of degree 2. Thus [K : ℚ]≤4. Also i∉√3 and it follows that [K : ℚ] = 4. We conclude that Gal(K/ℚ)≌ℤ/2ℤ×ℤ/2ℤ. Let α,β∈Gal(K/ℚ) be defined by αi = -i, α√3 = √3, βi = i, β√3 = -√3. Then α,β have order 2 and Gal(K/ℚ) = ⟨α⟩×⟨β⟩.

4. Let A be a 5×5 matrix of order 3. Then its minimal polynomial divides x³ - 1 and is not x - 1.

   (a)

   We use the rational canonical form to determine the conjugacy classes in GL₅(ℚ) (assume this is what the question means). Here the minimal polynomial must be (x - 1)(x² + x + 1) and there are two possibilities for the invariant factors, namely {x - 1, x - 1, x³ -1} and {x² + x + 1, x³ -1}. It follows that there are two conjugacy classes of matrices of order 3. The corresponding matrices are

   (

   	1	0	0	0	0
   	0	1	0	0	0
   	0	0	0	0	1
   	0	0	1	0	0
   	0	0	0	1	0

   )

   and

   (

   	0	-1	0	0	0
   	1	-1	0	0	0
   	0	0	0	0	1
   	0	0	1	0	0
   	0	0	0	1	0

   )

   (b)

   We use the Jordan canonical form to determine the conjugacy classes in GL₅(ℂ) (again, assume this is what the question means). Since A has finite order, its Jordan canonical form will be a diagonal matrix and hence the conjugacy classes will be determined by the eigenvalues of A (including multiplicities). Let ω = e^2πi/3, a primitive cube root of 1. Now the eigenvalues are the cube roots of 1, there must be 5 eigenvalues, and not all the eigenvalues can be 1 because A is not the identity. It follows that a set of representatives for the conjugacy classes over ℚ are {diag(1,...,ω,...,ω²)}, where there is at most four 1's, and otherwise arbitrary.

   If one wants to find out precisely how many conjugacy classes, note that the number without the restriction that there are at most four ones is the coefficient of x⁵ in

   (1 + x + x² + ...)³ = (1 -x)^-3,

   that is 7!/(2!·5!) = 21. Therefore the number of conjugacy classes is 20.

5. (a)

   Clearly if M = 0, the M_P = 0 for all prime ideals P. Conversely suppose M_P = 0 for all prime ideals P and let 0≠m∈M; we need to show that no such m exists. Define I = {r∈R | rm = 0}. Then I is a proper ideal of R and therefore it is contained in a maximal ideal P. Since maximal ideals are prime, P is a prime ideal. Now M_P = 0 tells us that sm = 0 for some s∈R\P, and we now have a contradiction as required.

   (b)

   It is obvious that if f : M→N is surjective then f_P : M_P→N_P is surjective, so we need to prove the converse. Now localization is an exact functor, in particular M_P→N_P→(M/N)_P→ 0 is exact. Therefore if f_P is surjective for all prime ideals P, we see that (M/N)_P = 0 for all prime ideals P, and then we deduce from (a) that M/N = 0. This completes the proof.

6. Note that a Sylow p-subgroup has order p, in particular a Sylow p-subgroup is a nontrivial proper subgroup of G. We'll consider the cases a = 1, 2, 3 separately. First suppose a = 1. Then the number of Sylow p-groups is congruent to 1 mod p and divides 2 and we see that there is exactly one Sylow p-subgroup. Thus the Sylow p-subgroup is normal and we see that G is not simple.

   Next suppose that a = 2. Then the number of Sylow p-groups is congruent to 1 mod p and divides 4 and we see that there is exactly one Sylow p-subgroup unless p = 3 and we conclude that G is not simple. On the other hand if p = 3 and G is simple, then G is isomorphic to a subgroup of A₃ because G has a subgroup of index 3, namely a Sylow 2-subgroup. This is clearly not possible because | G| = 12 and | A₃| = 3. We deduce that in all cases, G is not simple.

   Finally suppose that a = 3. Then the number of Sylow p-groups is congruent to 1 mod p and divides 8 and we see that there is exactly one Sylow p-subgroup unless p = 3 or 7. If p = 3, then the Sylow 2-subgroup has index 3 in G, so if G is simple, we see that G is isomorphic to a subgroup of A₃. This is not possible because | G| = 24 and | A₃| = 3. Now suppose that p = 7. Then the number of Sylow 7-subgroups is congruent to 1 mod 7 and divides 8. If there is 1, then the Sylow 7-subgroup is normal, so if G is simple, then there are 8 Sylow 7-subgroups. Since two distinct subgroups of order 7 intersect in the identity, we see that there are 48 elements of order 7 in G. Also if the Sylow 2-subgroup is not normal, there are at least 9 elements of order a power of 2, so G has at least 48 + 9 = 57 elements, which is not possible. We conclude that in all cases, G is not simple.

7. It is obvious that each statement implies the next, because at each stage given a solution, we use the images of that solution for the next stage.

   (c) implies (b). Write n = p₁^e₁p₂^e₂...p_d^e_d, where the p_i are distinct primes, and d, e_i∈ℕ. Suppose we have solutions (a_{1, i},...a_{m, i}) in ℤ/p_i^e_iℤ for all i. By the Chinese remainder theorem, we may choose a₁,..., a_m∈ℤ/nℤ such that a_i≡a_ijmodℤ/p_i^e_iℤ for all i. Then (a₁,..., a_m) is a solution in ℤ/nℤ.

   (c) doesn't imply (a). Consider the polynomial f (x) = (x² + x + 1)(x³ -7)(x⁵ - 2). Clearly f has no root in ℤ. We need to show that f has a root in ℤ/pⁿℤ, for all primes p and n∈ℕ. Recall that the multiplicative group U(pⁿ) of nonzero elements of ℤ/pⁿℤ has order p^n-1(p - 1). If 3 | p - 1, then U(pⁿ) has an element α of order 3. If α≡1 mod p, then α^pn = 1 which is not the case. It follows that α - 1 is a unit ℤ/pⁿℤ and since (α -1)(α² + α + 1) = 0, we deduce that α is a root of x² + x + 1 and hence also a root of f. On the other hand if 3 | p - 2, then (3,| U(pⁿ)|) = 1 and 7∈U(pⁿ), and therefore there exists β∈U(pⁿ) such that β³ = 7. It again follows that f has root in ℤ/pⁿℤ. If p = 3, then 2∈U(3ⁿ) and (| U(3ⁿ), 5) = 1 and therefore there exists γ∈U(3ⁿ) such that γ⁵ = 2. We have now shown that f (x) has a root in ℤ/pⁿℤ for all primes p and all n∈ℕ.

   (d) doesn't imply (c). Consider the polynomial f (x) = (x² + x + 1)(x³ - 2). We first show that f has a root in ℤ/pℤ for all primes p. If 3 | p - 1, then U(p) has an element of order 3 and we see that x² + x + 1 and hence also f (x) has a root. On the other hand if 3 | p - 2 and p≠2, then (| U(p)|, 3) = 1 and since 2∈U(p), we find that x³ - 2 and hence also f (x) has a root. Finally f (0) = 0 in ℤ/2ℤ and f (1) = 0 in ℤ/3ℤ, and we have now shown that f has a root in ℤ/pℤ for all primes p. However f (x)≠ 0 for all x∈ℤ/4ℤ (just plug in x = 0, 1, 2, 3).