Algebra Prelim Solutions, Summer 2002

1. (a)

   H acts on the set of conjugates of Q according to the formula gQg^-1 | - > hgQg^-1h^-1 for h $\in$ H and g $\in$ G. Note that gQg^-1 < G for all g, so O_Q is a set of subgroups. Let S = {h $\in$ H | hQh^-1 = Q}, the stabilizer of Q in H. Then |O_Q|| S| = | H|. Now h $\in$ S if and only if h $\in$ H /\ N_G(Q) = 1, hence | S| = 1 and the result follows.

   (b)

   Let P be a Sylow p-subgroup of G. We apply the above with H = P. Since P /\ N_G(Q) = P /\ Q = 1, we see that O_Q has | P| = p^m subgroups. Furthermore all the subgroups of O_Q have prime order q, so any two of them must intersect in 1. Now each nonidentity element of a subgroup in O_Q has order q, consequently each subgroup of O_Q yields q - 1 elements of order q and we deduce that G has at least (q - 1)p^m elements of order q. Therefore G has at most pq^m - (q - 1)p^m = p^m elements of order a power of p. Since | P| = p^m and every element of a Sylow p-subgroup has order a power of p, we conclude that P is the only Sylow p-subgroup of G. Therefore P is normal in G and we are finished.

2. Let F = Q($\sqrt{2}$,$\sqrt{3}$) and N = Gal(K/F). Then F $\subseteq$ K and is the splitting field over Q for (x² - 2)(x² - 3). Therefore N is a normal subgroup in Gal(K/Q) of index [F : Q]. Now $\sqrt{2}$ satisfies x² - 2 and $\sqrt{2}$$\notin$Q. Therefore [Q($\sqrt{2}$) : Q] = 2.

   Next we show that $\sqrt{3}$$\notin$Q($\sqrt{2}$). Suppose $\sqrt{3}$ $\in$ Q($\sqrt{2}$). Then we could write $\sqrt{3}$ = a + b$\sqrt{2}$ with a, b $\in$ Q, because every element of Q($\sqrt{2}$) can be written in this form. Squaring, we obtain 3 = a² + 2b² + 2ab$\sqrt{2}$. Since $\sqrt{2}$$\notin$Q, we deduce that a or b = 0. But a = 0 yields $\sqrt{3/2}$ $\in$ Q, while b = 0 yields $\sqrt{3}$ $\in$ Q, neither of which is true. We conclude that $\sqrt{3}$$\notin$Q($\sqrt{2}$). Since $\sqrt{3}$ satisfies x² - 3, we deduce that [F : Q($\sqrt{2}$)] = 2. Therefore

   [F : Q] = [Q($$\sqrt{2}$$,$$\sqrt{3}$$) : Q($$\sqrt{2}$$)][Q($$\sqrt{2}$$) : Q] = 2*2 = 4.
   Thus N is a normal subgroup of index 4 in Gal(K/Q).

   Suppose $\sqrt[8]{2}$ $\in$ K. Since $\sqrt[8]{2}$ satisfies x⁸ - 2 and x⁸ - 2 is irreducible over Q by Eisenstein's criterion for the prime 2, we see that 8 divides [K : Q], consequently 8 divides | Gal(K/Q)|. But | Gal(K/Q)| = 4| N|, so this is not possible if | N| is odd.

3. (a)

   If X and Y are right R-modules and q : X - > Y is an R-module homomorphism, then there is a unique group homomorphism q $\otimes$ 1 : X $\otimes_{R}^{}$ C - > Y $\otimes_{R}^{}$ C such that q(x $\otimes$ c) = (qx) $\otimes$ c for all x $\in$ X and c $\in$ C.

   First we apply this to the maps

   $$\oplus$$

   $$\oplus$$

   
   

   We obtain a group homomorphism

   q : (A $$\oplus$$ B) $$\otimes_{R}^{}$$ C - - > A $$\otimes_{R}^{}$$ C $$\oplus$$ B $$\otimes_{R}^{}$$ C

   such that q((a, b) $\otimes$ c) = (a $\otimes$ c, b $\otimes$ c). Next we apply it to the maps

   $$\otimes \otimes_{R}^{} \oplus \otimes_{R}^{}$$

   $$\otimes \otimes_{R}^{} \oplus \otimes_{R}^{}$$

   
   

   We obtain a group homomorphism

   f : (A $$\otimes_{R}^{}$$ C) $$\oplus$$ (B $$\otimes_{R}^{}$$ C) - - > (A $$\oplus$$ B) $$\otimes_{R}^{}$$ C

   such that f(a $\otimes$ c, b $\otimes$ d )= (a, 0) $\otimes$ c + (0, b) $\otimes$ d.

   Finally we show that fq is the identity on (A $\oplus$ B) $\otimes_{R}^{}$ C, and that qf is the identity on (A $\otimes_{R}^{}$ C) $\oplus$ (B $\otimes_{R}^{}$ C). We have

   fq(a, b) $$\otimes$$ c = f(a $$\otimes$$ c, b $$\otimes$$ c) = (a, b) $$\otimes$$ c.

   Since (A $\oplus$ B) $\otimes_{R}^{}$ C is generated as an abelian group by elements of the form (a, b) $\otimes$ c, we see that fq is the identity. Also

   qf(a $$\otimes$$ c, b $$\otimes$$ d )= q(a, 0) $$\otimes$$ c + q(0, b) $$\otimes$$ d = (a $$\otimes$$ c, b $$\otimes$$ d ).

   Since (A $\otimes$ C) $\oplus$ (B $\otimes$ C) is generated as an abelian group by elements of the form (a $\otimes$ c, b $\otimes$ d ), we deduce that qf is the identity. It now follows that (A $\oplus$ B) $\otimes_{R}^{}$ C @ (A $\otimes_{R}^{}$ C) $\oplus$ (B $\otimes_{R}^{}$ C).

   (b)

   Since M is a finitely generated Z-module, we may express it as a finite direct sum of cyclic Z-modules, say M = $\bigoplus_{i}^{}$Z/a_iZ, where we may assume that a_i =/= ±1 for all i. Then by the first part, we see that

   M $$\otimes_{\mathbf {Z}}^{}$$ M @ $$\bigoplus_{i,j}^{}$$Z/a_iZ $$\otimes_{\mathbf {Z}}^{}$$ Z/a_jZ.

   Therefore it will be sufficient to prove Z/a_iZ $\otimes_{\mathbf {Z}}^{}$ Z/a_iZ =/=  0 for all i (of course even for just one i will be sufficient). However we can define a bilinear map

   q : Z/a_iZ X Z/a_iZ - > Z/a_iZ

   by q(x, y) = xy. This induces a Z-module homomorphism Z/a_iZ $\otimes_{\mathbf {Z}}^{}$ Z/a_iZ - > Z/a_iZ, which is obviously onto. We conclude that Z/a_iZ $\otimes_{\mathbf {Z}}^{}$ Z/a_iZ =/=  0, as required.

4. Note that Ann(m) is an ideal of R. Since R is Noetherian, we may choose w $\in$ M such that Ann(w) is maximal (that is Ann(w) is as large as possible, but not R). Suppose Ann(w) is not prime. Then there exists a, b $\in$ R\Ann(w) such that ab $\in$ Ann(w), i.e. abw = 0. But then a $\in$ Ann(bw) and Ann(w) $\subseteq$ Ann(bw). Furthermore Ann(bw) =/= R because bw =/=  0, so the maximality of Ann(w) has been contradicted and the result follows.

5. Suppose R is a field. Then an R-module is the same thing as an R-vector space, and since every vector space has a basis this means that every R-module is free; in particular every R-module is projective.

   Conversely suppose every R-module is projective. Since R is an integral domain, to prove R is a field we only need show that every nonzero element of R is invertible. Suppose to the contrary that x is a nonzero element of R which is not invertible. Then R/Rx is a nonzero R-module, so it has a nonzero element u. Note that xu = 0. Consider the exact sequence

   0 - - > Rx - - > R - - > R/xR - - > 0.
   Since R/xR is projective, the sequence splits, in particular R/xR is isomorphic to a submodule of R. Now R is an integral domain, so xv =/=  0 for all nonzero v $\in$ R and we deduce that xu =/=  0. We now have a contradiction and the result follows.

6. Since Z_N = Z_{N + 1}, we see that Z(G/Z_N) = 1. Therefore K $\subseteq$ $\mathcal {Z}$_N.

   Now suppose L is a normal subgroup of G such that Z(G/L) = 1 and L does not contain Z_N. Then there is a nonnegative integer n such that

   Z_n $$\subseteq$$ L,    Z_{n + 1} $$\nsubseteq$$ L.
   Choose x $\in$ Z_{n + 1}\L. Then xL =/= 1 in G/L. Also xgx^-1g^-1 $\in$ Z_{n + 1} for all g $\in$ G, because xZ_{n + 1} $\in$ Z(G/Z_{n + 1}). Therefore xgx^-1g^-1 $\in$ L and we deduce that xL $\in$ Z(G/L). This is a contradiction, and so the result is proven.

7. Let I be the set of matrices in M₂(Q[x]/(x² - 1)) of the form

   (

   	a(x + 1) + (x2 - 1)	0
   	b(x + 1) + (x2 - 1)	0

   )

   with a, b $\in$ Q. Note that if f $\in$ Q[x], then (x - 1) divides f (x) - f (1), consequently f (x)(x + 1) + (x² - 1) = f (1)(x + 1) + (x² - 1) in Q[x]/(x² - 1).

   Now we verify that I is a left ideal of Q[x]/(x² - 1). Clearly I is an abelian group under addition. Since

   (

   	f (x) + (x2 - 1)	g(x) + (x2 - 1)
   	h(x) + (x2 - 1)	k(x) + (x2 - 1)

   )

   (

   	a(x + 1) + (x2 - 1)	0
   	b(x + 1) + (x2 - 1)	0

   )

   =

   (

   	af (1)(x + 1) + (x2 - 1)	0
   	bh(1)(x + 1) + (x2 - 1)	0

   )

   we see that I is closed under left multiplication by elements of Q[x]/(x² - 1), and it now follows that I is a left ideal.

   Finally we need to show that I is a minimal ideal. Obviously I =/=  0 (note x + 1$\notin$(x² - 1)). Suppose J is a nonzero left ideal contained in I. We need to show that J = I. By multiplying on the left by the matrix

   (

   	0	1 + (x2 - 1)
   	1 + (x2 - 1)	0

   )

   if necessary, we may assume that I contains a matrix of the form

   (

   	a(x + 1) + (x2 - 1)	0
   	b(x + 1) + (x2 - 1)	0

   )

   with a =/=  0. Then by multiplying on the left by

   (

   	c/a + (x2 - 1)	0
   	d /a + (x2 - 1)	0

   )

   we see that J must be the whole of I and the result follows.