August 1997 Algebra Prelim Solutions

  1. (a) Obviously 0 $ \in$ IJ. Now suppose x, y $ \in$ IJ and r $ \in$ R. We want to prove that x + y, rx $ \in$ IJ. Write x = Si = 1naibi and y = Si = 1mcidi, where ai, ci $ \in$ I and bi, di $ \in$ J. Then

    x + y = Si = 1naibi + Si = 1mcidi

    which shows that x + y $ \in$ IJ. Also rx = Si = 1n(rai)bi, and since I is an ideal of R, we see that rai $ \in$ I for all i. Therefore rx $ \in$ IJ and we have proven that IJ $ \lhd$ R.

    (b) Since I $ \lhd$ R, we have IJ $ \subseteq$ I. Similarly IJ $ \subseteq$ J and we deduce that IJ $ \subseteq$ I $ \cap$ J.

    (c) Since I + J = R, we may write i + j = 1 where i $ \in$ I and j $ \in$ J. If x $ \in$ I $ \cap$ J, then x = xi + xj $ \in$ JI + IJ = IJ (because R is commutative). Therefore x $ \in$ IJ and we have proven that I $ \cap$ J $ \subseteq$ IJ. The result now follows from (b).

    (d) Let a $ \in$ R\ 0. We need to prove that a has a multiplicative inverse. Using IJ = I $ \cap$ J with I = J = aR, we see that aRaR = aR $ \cap$ aR = aR, hence aras = a for some r, s $ \in$ R. Since a $ \neq$ 0 and R is an integral domain, we may cancel a to obtain ars = 1. We have now shown that all nonzero elements of R have a multiplicative inverse, hence R is a field.

  2. (a) Since F is a finite Galois extension of K with Galois group S5, there is a one-one correspondence between the fields strictly between F and K, and the proper nontrivial subgroups of S5. Therefore we need to show that S5 has more than 40 subgroups other than 1 and S5. Now S5 has 24 elements of order 5 which gives 6 subgroups of order 5; 20 elements of order 3 which gives 10 subgroups of order 3; 10 2-cycles which gives 10 subgroups of order 2; 15 permutations which are a product of two disjoint 2-cycles which gives 15 more subgroups of order 2; and now we have 6 + 10 + 10 + 15 subgroups which is already more than 40, as required.

    (b) The subfields E of F containing K which are Galois extensions of K correspond to the normal subgroups of Gal(F/K). Specifically if H is a normal subgroup of Gal(F/K), then the corresponding subfield is Fix(H), the elements of F which are fixed by all automorphisms of H. Furthermore we have Gal(Fix(H)/K) $ \cong$ Gal(F/K)/H and [Fix(H) : K] = [Gal(F/K) : H]. Since S5 has a unique nontrivial normal subgroup, namely the alternating group A5, it follows that the subfield E required is Fix(A5). Then [E : K] = 2 and Gal(E/K) $ \cong$ S5/A5 $ \cong$ $ \mathbb {Z}$/2$ \mathbb {Z}$.

  3. (a) 455 = 5·7·13. We determine the number of Sylow 13-subgroups. This is congruent to 1 modulo 13 and divides 35. The only possibility is 1, which means that G has a normal subgroup A of order 13 and so G is not simple.

    (b) Since G/A is a group of order 35, we can apply Sylow's theorems to see that G/A has exactly one subgroup of order 7, which by the subgroup correspondence theorem we may call H/A. Then H/A $ \lhd$ G/A, so H $ \lhd$ G. Now H is a group of order | A|| H/A| = 13·7, and we may apply Sylow's theorems for the prime 7 to deduce that H has exactly one subgroup of order 7; we shall call this subgroup B. Then B $ \lhd$ H; in fact we can assert more, namely that B $ \lhd$ G. To see this, let g $ \in$ G. Then gBg-1 is a subgroup of gHg-1 = H because H $ \lhd$ G, and since | gBg-1| = | B| = 7, we see that gBg-1 = B which establishes that B $ \lhd$ G. Similarly G has a normal subgroup of order 5, which we shall call C.

    We now have that G has normal subgroups A, B, C of orders 13, 7 and 5 respectively. Since 13, 7 and 5 are coprime and their product is 455, we deduce that G $ \cong$ A X B X C. Let a $ \in$ A be an element of order 13, let b $ \in$ B be an element of order 7, and let c $ \in$ C be an element of order 5. We want to show that abc is an element of order 455. Since the order of an element divides the order of the group, we certainly have the order of abc divides 455. Suppose the order of abc was less than 455. Then the order of abc would have to divide 455/13, or 455/7, or 455/5. Suppose the order of abc divided 455/13 = 35. Then (abc)35 = 1 and since a, b, c commute, we see that a35b35c35 = 1. But b35 = c35 = 1, hence a35 = 1. This is not possible because a has order 13. Similarly the order of abc cannot divide 455/7 and 455/5. We deduce that abc has order 455, hence < abc > = G and the result is proven.

  4. We shall use the fundamental theorem for finitely generated modules over a PID. Thus we may write

    A = Ra $\displaystyle \oplus$ $\displaystyle \bigoplus_{i=1}^{m}$(R/qiR)ai    
    B = Rb $\displaystyle \oplus$ $\displaystyle \bigoplus_{i=1}^{m}$(R/qiR)bi    

    where a, b, ai, bi, m are nonnegative integers and the qi are distinct prime powers. Since An $ \cong$ Bn, we have

    Rna $\displaystyle \oplus$ $\displaystyle \bigoplus_{i=1}^{m}$(R/qiR)nai $\displaystyle \cong$ Rnb $\displaystyle \oplus$ $\displaystyle \bigoplus_{i=1}^{m}$(R/qiR)nbi.    

    The fundamental theorem now gives that na = nb and nai = nbi for all i, hence a = b and ai = bi for all i and the result follows.

  5. Since P is a projective module, it is a submodule of a free module F. The mapping q : P - > F defined by qp = 2p is a monomorphism, so by using the hypothesis that P is injective, we see that it has a left inverse f : F - > P. Since fq is the identity mapping on P, we see that p = 2fp for all p $ \in$ P and hence P $ \subseteq$ 2P. Therefore P $ \subseteq$ 2nP and we deduce that P $ \subseteq$ 2nF for all positive integers n. But $ \bigcap$2nF = 0 because F is a free module and the result follows. (Note: the hypothesis P is finitely generated has not been used.)

  6. Let a : mB - > B denote the natural inclusion, and let b : B - > B/mB denote the natural surjection. Then the exact sequence mBa- > Bb- > B/mB - > 0 yields an exact sequence

    A $\displaystyle \otimes$ mB1 $\scriptstyle \otimes$ a- > A $\displaystyle \otimes$ B1 $\scriptstyle \otimes$ b- > A $\displaystyle \otimes$ B/mB - - > 0

    where 1 indicates the identity map. Therefore A $ \otimes$ (B/mB) $ \cong$ (A $ \otimes$ B)/im(1 $ \otimes$ a), where im denotes the image of a map. Now im(1 $ \otimes$ a) is the $ \mathbb {Z}$-submodule of A $ \otimes$ B generated by {a $ \otimes$ mb | a $ \in$ A and b $ \in$ B} and since a $ \otimes$ mb = m(a $ \otimes$ b), this is the same as the $ \mathbb {Z}$-submodule generated by {m(a $ \otimes$ b) | a $ \in$ A and b $ \in$ B}. This submodule is precisely m(A $ \otimes$ B), hence im(1 $ \otimes$ a) = m(A $ \otimes$ B) and the proof is complete.

  7. Let G be the group of order 588 and write 588 as a product of prime powers: 588 = 4·3·49. The number of Sylow 7-subgroups is congruent to 1 modulo 7 and divides 12, hence there is a unique Sylow 7-subgroup A which must be normal in G. Since A has order 49, it is abelian and so certainly solvable. Thus we need only prove that G/A is solvable, because G/A and A solvable implies G solvable. Since G/A has order 12, this means we need to prove that all groups of order 12 are solvable.

    Let H be a group of order 12. The number of Sylow 3-subgroups is 1 or 4. Suppose there is exactly one Sylow 3-subgroup B. Then B $ \lhd$ H and | H/B| = 4. Since groups of order 3 and 4 are abelian, we see that B and H/B are abelian and hence H is solvable. Suppose on the other hand that H has 4 Sylow 3-subgroups. If B1 and B2 are two distinct Sylow 3-subgroups, then B1 $ \cap$ B2 is a proper subgroup of B1 whose order divides 3 by Lagrange's theorem, hence B1 $ \cap$ B2 = 1 and we conclude that H has (at least) 8 elements of order 3. Now the Sylow 2-subgroups of H have order 4, and every element of a Sylow 2-subgroup has order a power of 2. If H had more than one Sylow 2-subgroup, then H would have at least 5 elements of order a power of 2, consequently H would have at least 5 + 8 = 13 elements, which is not possible because | H| = 12. Therefore H has a unique subgroup C of order 4, which must be normal in H. Since | H/C| = 3 and | C| = 4, we see that H/C and C are abelian, and we conclude that H is solvable. This completes the proof.

  8. Let L = $ \mathbb {Q}$($ \sqrt{2}$,$ \sqrt{3}$,$ \sqrt{5}$). Then clearly K $ \subseteq$ L. Also ($ \sqrt{2}$ + $ \sqrt{3}$)3 - 9($ \sqrt{2}$ + $ \sqrt{3}$) = 2$ \sqrt{2}$, hence $ \sqrt{2}$ $ \in$ K and we deduce that L = K. Thus K = $ \mathbb {Q}$($ \sqrt{2}$,$ \sqrt{3}$,$ \sqrt{5}$). It follows that K is the splitting field for the polynomial (x2 - 2)(x2 - 3)(x2 - 5) and we deduce that K is a Galois extension of $ \mathbb {Q}$. In particular the number of fields between K and $ \mathbb {Q}$ equals the number of subgroups of Gal(K/$ \mathbb {Q}$).

    Now any element of Gal(K/$ \mathbb {Q}$) must send $ \sqrt{2}$ to ±$ \sqrt{2}$, $ \sqrt{3}$ to ±$ \sqrt{3}$, and $ \sqrt{5}$ to ±$ \sqrt{5}$. It follows that every nonidentity element of Gal(K/$ \mathbb {Q}$) has order 2, and that Gal(K/$ \mathbb {Q}$) is elementary abelian of order 1,2,4 or 8.

    We shall use the following result: if a and b are products of distinct prime numbers and $ \mathbb {Q}$($ \sqrt{a}$) = $ \mathbb {Q}$($ \sqrt{b}$), then a = b. To see this, write $ \sqrt{a}$ = r + s$ \sqrt{b}$ where r, s $ \in$ $ \mathbb {Q}$. Then a = r2 + 2rs$ \sqrt{b}$ + s2b. Clearly s $ \neq$ 0 and rs = 0, consequently r = 0 and a = s2b which establishes the result.

    It follows immediately that there are at least 8 subfields between $ \mathbb {Q}$ and K, namely $ \mathbb {Q}$($ \sqrt{2^c 3^d 5^e}$) where c, d, e are 0 or 1. Now if | Gal(K/$ \mathbb {Q}$)|$ \le$4, then there would be at most 5 subgroups of Gal(K/$ \mathbb {Q}$), consequently there would be at most 5 fields between K and $ \mathbb {Q}$. This is a contradiction, so we must have | Gal(K/$ \mathbb {Q}$)| = 8 and therefore Gal(K/$ \mathbb {Q}$) $ \cong$ ($ \mathbb {Z}$/2$ \mathbb {Z}$)3.

Peter Linnell