- (a)
- By using the elementary divisor decomposition,
up to isomorphism, there are three abelian groups of order
*p*^{3}*q*, namely**Z**_{p3}×**Z**_{q},**Z**_{p2}×**Z**_{p}×**Z**_{q}, and**Z**_{p}×**Z**_{p}×**Z**_{p}×**Z**_{q}. - (b)
- The first group above is generated by one element, while the third
requires 3 elements. Therefore
*G*≌**Z**_{p2}×**Z**_{p}×**Z**_{q}≌**Z**_{p2}×**Z**_{pq}, because**Z**_{p}×**Z**_{q}≌**Z**_{pq}, as required.

- We have
[
*k*(α) :*k*] = deg*f*and [*K*:*k*(α)][*k*(α);*k*] = [*K*:*k*]. Thus deg*f*| [*K*:*k*] and the result follows. - By Gauss's lemma,
*f*is irreducible in*Q*[*x*]. Since*Q*[*x*] is a PID, this tells us that*fQ*[*x*] is a maximal ideal of*Q*[*x*]. The result follows. *A*_{4}is a normal subgroups of*S*_{4}and*V*: = {(1),(12)(34),(13)(24),(14)(23)} is a normal subgroup of*A*_{4}(even normal in*S*_{4}). Since |*S*_{4}/*A*_{4}| = 2, |*A*_{4}|/*V*= 3, |*V*| = 4, the groups*S*_{4}/*A*_{4},*A*_{4}/*V*and*V*are all abelian, because groups of order 2,3 or 4 are abelian. This proves that*S*_{4}is solvable.- We have a short exact sequence
0 --> ker
*f*-->*P*--^{f}-->*Q*--> 0. Since*Q*is projective, the sequence splits, so*P*≌*Q*⊕ker*f*. This proves the result, because direct summands of projective modules are projective. - First observe that
*Q**Q*≌*Q*as*Q*-modules. To do this, define*f*:*Q*×*Q*-->*Q*by*f*(*p*,*q*) =*pq*. Clearly this is*R*-bilinear, so induces an*R*-map*g*:*Q**Q*-->*Q*satisfying*g*(*p**q*) =*pq*. Also we can define a*Q*-map*h*:*Q*-->*Q**Q*by*h*(*q*) =*q*1. Since*gh*(*q*) =*g*(*q*1) =*q*, we see that*gh*is the identity on*Q*. Now consider*hg*(*p**q*) =*pq*1. Write*q*=*a*/*b*where*a*,*b*∈*R*with*b*≠ 0. Then*pq*1 =*pa*/*b*1 =*p*/*b**ab*/*b*=*pb*/*b**a*/*b*=*p**q*and it follows that*hg*is the identity on*P**Q*, because we only need to check that*hg*is the identity on the ``simple tensors". Thus*h*is one-to-one and onto, and our observation is established.Now observe that

*Q**V*≌*V*. Indeed we can define a*Q*-bilinear map θ :*Q*×*V*-->*V*by θ(*q*,*v*) =*qv*, and this induces a*Q*-map φ :*Q**V*-->*V*satisfying φ(*q**v*) =*qv*. Also we can define a*Q*-map ψ :*V*-->*Q**V*by ψ(*v*) = 1*v*. Then φψ(*v*) = φ(1*v*) =*v*, so φψ is the identity on*V*. Since ψφ(*q**v*) = ψ(*qv*) = 1*qv*=*q**v*and ψφ is the identity on*Q**V*provided it is the identity on the simple tensors, we see that ψφ is the identity on*Q**V*, and the result follows.Note that this proof does not use the hypothesis that

*V*is finite dimensional. - Let
*G*denote the Galois group of*F*over*K*. Since*G*is a*p*-group for the prime*p*= 11, it has a sequence of normal subgroups 1 =*G*_{4}<|*G*_{3}<|*G*_{2}<|*G*_{1}<|*G*_{0}=*G*, such that*G*_{i}<|*G*and |*G*_{i+1}/*G*_{i}| = 11 for all*i*. Now let*K*_{i}be the fixed subfield of*G*_{i}in*K*, for*i*= 0, 1,..., 4. Then*K*_{i}is a Galois extension of*F*for all*i*, because*G*_{i}<|*G*. Since [*K*_{i}:*K*_{i-1}] = |*G*_{i-1}/*G*_{i}| = 11, the result is proven. - Suppose
*R*/*I*is a projective*R*-module. Then we may write*R*=*I*⊕*J*for some*R*-submodule*J*of*R*. Of course*R*-submodules of*R*are the same as ideals, so*J*is an ideal of*R*. Since*M*is the unique maximal ideal of*R*and*I*⊆*M*, we must have*J*=*R*. But then*J*⊇*I*and thus*I*+*J*is not a direct sum. We now have a contradiction and the result follows.

Peter Linnell 2007-12-09