Algebra Prelim Solutions, January 2019

  1. Let G be a group of order 992. The number of Sylow 31-subgroups is congruent to 1 mod 31 and divides 32 and is therefore 1 or 32. First suppose G has 1 Sylow 31-subgroup N. Then N⊲G and G/N is a group of order 32. Since a nontrivial p-group has nontrivial center, we see that G/N has a central element of order 2 and therefore it has a normal subgroup M/N of order 2, where M⊲G, by the subgroup correspondence theorem. Then M is a normal subgroup of order 62.

    Now suppose that the number of Sylow 31-subgroups is 32. Then the number of elements of order 31 is 32·30 = 960. It follows that G has at most 32 elements that are a power of 2. Let H be a Sylow 2-subgroup of G. Then H has 32 elements that are a power of 2. If K is another Sylow 2-subgroup, then there exists k∈K∖H, and since k has order a power of 2, we have that G has a least 33 elements that have order a power of 2. This means that G has at least 960 + 33 = 993 > 992 elements, a contradiction. Therefore G has only one Sylow 2-subgroup and it follows that H⊲G. This completes the proof.

  2. Let A denote the set of prime ideals of R. Since R≠ 0, it has maximal ideals. Furthermore every maximal ideal is a prime ideal, consequently A≠Ø. Partially order the prime ideals of A by reverse inclusion; that is P≤Q means Q⊆P. Suppose {Pj | j∈J} is a chain in A (where J is an indexing set). Let Q = ∩jPj. Then Q is certainly an ideal of R (the intersection of ideals is always an ideal), so we need to check that it is prime. Suppose a, b∈R∖Q. Then a∉Pj and b∉Pk for some j, k∈J. Since {Pj} is a chain, without loss of generality we may assume that Pj⊆Pk. Then a, b∉Pj and since Pj is a prime ideal, we deduce that ab∉Pj and hence ab∉Q. Therefore Q∈A and is an upper bound for the chain. We conclude by Zorn's lemma that A has maximal elements. This means that R has minimal prime ideals with respect to inclusion.

  3. Suppose R is not a field. Then R has a nonzero maximal ideal M. Since R/M is irreducible, it is free a free R-module by hypothesis. Choose m∈M∖ 0 and x∈R∖M. Since R/M is free, we see that m(x + M)≠ 0 in R/M. On the other hand m(x + M) = mx + M = 0 because M is an ideal, a contradiction, and the result follows.

  4. Suppose first that dimkM < ∞. If N is a proper submodule of N, then dimkN < dimkM and we cannot have N≌M. This proves the ``only if" part of the statement.

    Now suppose dimkM = ∞. We use the structure theorem for finitely generated modules over the PID k[x] to write M≌k[x]n⊕i=1dk[x]/(fi), where the fi are monic polynomials with positive degree, and n and d are nonnegative integers. Since dimkk[x]/(fi) < ∞, this implies that n > 0 and therefore we may write M≌k[x]⊕L for some k[x]-module L. Since xk[x] is a proper k[x]-submodule of k[x] and xk[x]≌k[x], we see that xk[x]⊕L is a proper submodule of k[x]⊕L and xk[x]⊕L≌k[x]⊕L. The result follows.

  5. Let G denote the automorphism group of ℚ(α) over ℚ. Since α and β are roots of the same irreducible polynomial f, there is an isomorphism θ : ℚ(α)→ℚ(β). Thus θ∈G and therefore G≠1. Since [ℚ(α) : ℚ] = deg f, because f is irreducible, we see that [ℚ(α) : ℚ] = p, a prime, and it follows that the fixed field of G is ℚ. We conclude that ℚ(α) is a Galois extension of ℚ.

  6. (a)
    Let a, b∈K and c∈k. Then θ(a + b) = θa + θb by Freshman's dream, and θ(ca) = θcθa = cθa because θc = c. This proves that θ is a k-linear map.

    (b)
    Let ι : K→K denote the identity map. Note that θn(a) = apn. Since apn = a for all a∈K, we see that θn = ι and we deduce that the minimal polynomial of θ divides Xn - 1.

    (c)
    Since n | p - 1, we see that the roots of Xn - 1 are a subset of the roots of Xp-1 - 1 (including multiplicities). However the roots of Xp-1 - 1 are precisely the p - 1 nonzero elements of k. Therefore minimal polynomial has distinct roots, all lying in k. It follows that θ is diagonalizable over k.

  7. S3 has 3 conjugacy classes with representatives (1), (1 2) and (1 2 3). It has two one-dimensional characters, namely the trivial character, which we shall denote by χ1, and the sign of a permutation, which we shall denote by χ2. Since there are 3 conjugacy classes, there are three irreducible characters; we'll call the third irreducible character χ3. This character can be derived from the orthogonality relations. Therefore character table for S3 is

    Class Size 1 3 2
    Class Rep 1 (1 2) (1 2 3)
    χ1 1 1 1
    χ2 1 -1 1
    χ3 2 0 -1

    The character table for ℤ/2ℤ is

    Class Size 1 1
    Class Rep 0 1
    ψ1 1 1
    ψ2 1 -1

    The conjugacy classes for S3×ℤ/2ℤ are of the form 𝓢×𝓣, where 𝓢 is a conjugacy class for S3 and 𝓣 is a conjugacy class for ℤ/2ℤ. Thus in particular S3×ℤ/2ℤ has 3*2 = 6 conjugacy classes. We get the six irreducible representations from taking the tensor product of irreducible representations of S3 and ℤ/4ℤ, namely the representations χi⊗ψj, which have characters χiψj.

    Class Size 1 1 3 3 2 2
    Class Rep ((1), 0) ((1), 1) ((1 2), 0) ((1 2), 1) ((1 2 3), 0) ((1 2 3), 1)
    χ1⊗ψ1 1 1 1 1 1 1
    χ1⊗ψ2 1 -1 1 -1 1 -1
    χ2⊗ψ1 1 1 -1 -1 1 1
    χ2⊗ψ2 1 -1 -1 1 1 -1
    χ3⊗ψ1 2 2 0 0 -1 -1
    χ3⊗ψ2 2 -2 0 0 -1 1