- 1.
- Since
*f*is an epimorphism from*G*to*H*, the fundamental isomorphism theorem tells us that*G*/ker*f**H*, so in particular |*G*/ker*f*| = |*H*|. Therefore |*G*|/| ker*f*| = |*H*|, hence | ker*f*| = |*G*|/|*H*| and we deduce that | ker*f*| = 3^{3}11^{2}. Using the fundamental theorem for finitely generated abelian groups, we see that there are up to isomorphism six abelian groups of order | ker*f*| = 3^{3}11^{2}, namely/27 X /121 /9 X /3 X /121 (/3) ^{3}X /121/27 X (/11) ^{2}/9 X /3 X (/11) ^{2}(/3) ^{3}X (/11)^{2} - 2.
- (i) Since
*A**B*is a subgroup of*G*whose order divides |*A*| =*p*^{4}and |*B*| =*q*^{5}, we see that |*A**B*| = 1 and hence*A**B*= 1. Next if*a**A*and*b**B*, then*a*^{-1}*b*^{-1}*ab*= (*a*^{-1}*b*^{-1}*a*)*b**B*, because*B**G*. Similarly*a*^{-1}*b*^{-1}*ab**A*and we deduce that*a*^{-1}*b*^{-1}*ab**A**B*= 1. Therefore*a*^{-1}*b*^{-1}*ab*= 1, consequently*ab*=*ba*for all*a**A*and*b**B*. We can now define a map :*A*X*B*- >*G*by (*a*,*b*) =*ab*. Then((

(because*a*_{1},*b*_{1})(*a*_{2},*b*_{2})) = (*a*_{1}*a*_{2},*b*_{1}*b*_{2}) =*a*_{1}*a*_{2}*b*_{1}*b*_{2}= (*a*_{1}*b*_{1})(*a*_{2}*b*_{2})*a*_{2}commutes with*b*_{1}), hence ((*a*_{1},*b*_{1})(*a*_{2},*b*_{2})) = (*a*_{1},*b*_{1})(*a*_{2},*b*_{2}) and we deduce that is a homomorphism. If (*a*,*b*) ker , then*ab*= 1 and so*a*=*b*^{-1}. Thus*a*=*b*^{-1}*A**B*= 1 and we deduce that*a*=*b*= 1. Therefore ker = 1 and so is a monomorphism. Since |*G*| = |*A*| X |*B*| we conclude that is also onto, consequently is an isomorphism and the result follows.(ii) Since

*A*is a*p*-group, it has a normal subgroup*P*of order*p*. Similarly*B*has a normal subgroup*Q*of order*q*. Since*P*X*Q*is a normal subgroup of*A*X*B*of order*pq*, we see that (*P*X*Q*) is a normal subgroup of*G*of order*pq*, and so we may set*N*=*P*X*Q*to satisfy the requirements of the problem. - 3.
- Let
*G*be a simple group of order 2^{2}3 11^{2}. The number of Sylow 11-subgroups is congruent to 1 modulo 11 and divides 12, so the possibilities are 1 and 12. If there is 1 Sylow 11-subgroup, then it would have to be normal, which is not possible because*G*is simple. Therefore there are 12 Sylow 11-subgroups. If*N*is the normalizer of a Sylow 11-subgroup, then [*G*:*N*] is the number of Sylow 11-subgroups, so [*G*:*N*] = 12. By considering the permutation representation of*G*on the left cosets of*N*in*G*and using the fact that*G*is simple, we see that there is a monomorphism of*G*into*A*_{12}, the alternating group of degree 12. This means that*G*is isomorphic to a subgroup of*A*_{12}. This is not possible because 121 divides |*G*|, but 121 does not divide |*A*_{12}|. We now have a contradiction and we deduce that no such*G*exists, as required. - 4.
- Since
( + )
^{3}- 9( + ) = 2 , we see that [ + ] and we deduce that [ + ] = [,]. Now [[] : ] = 2, and [[,] : []] = 1 or 2, because satisfies*x*^{2}- 3, a degree 2 polynomial over . Therefore [[ + ] : ] = 4 or 2, depending on whether or not [].Suppose []. Then we may write =

*a*+*b*where*a*,*b*. Clearly*a*,*b*0. Squaring we obtain 3 =*a*^{2}+ 2*ab*+ 2*b*^{2}and we deduce that is rational, which is not so. Therefore [] and consequently [ + ] = 4. Note we also have that {1,} is a -basis for [], and {1,} is a []-basis for [,]. Recall that if*e*_{i}is an*F*-basis for*E*over*F*and*f*_{j}is an*E*-basis for*K*, then*e*_{i}*f*_{j}is an*F*-basis for*K*. It follows that {1,,,} is a -basis for [,]. - 5.
- Let
*P*be a prime ideal of*D*and suppose*P*was not maximal. Then there would exist*M**D*such that*M**D*and*M*properly containing*P*. Since*D*is a PID, we may write*P*=*pD*and*M*=*mD*for some*m*,*p**D*with*p*0. Then*p*=*mx*for some*x**D*because*M*contains*P*. Since*P*is a prime ideal, we must have*m*or*x**P*. We cannot have*m**P*because*M*properly contains*P*. Therefore we must have*x**P*and then we may write*x*=*py*for some*y**D*. This yields*p*=*mpy*and since*D*is a domain, we see that 1 =*my*. This shows that*mD*=*D*, which contradicts the fact that*M*is a proper ideal of*D*and the first part of the problem is proven.Suppose now that

*f*:*D*- >*K*is a ring epimorphism onto the integral domain*K*with ker*f*0. Then*D*/ker*f**K*, so ker*f*is a prime ideal because*D*/ker*f*is an integral domain. Using the first part of the problem, we see that ker*f*is a maximal ideal of*D*. Therefore*D*/ker*f*is a field and it follows that*K*is a field as required.For the last part a counterexample is

*R*= /6 . Let*P*be the prime ideal 2/6 ,*Q*the prime ideal 3/6 , and*S*=*R**P*. Note that the ideals of*R*are precisely 0,*R*,*P*and*Q*, and the prime ideals of*R*are precisely*P*and*Q*. Then*S*^{-1}*P*= 0 because*s*= 6 + 3*S*and*sp*= 0 for all*p**P*, and*S*^{-1}*Q*=*R*_{P}because*Q**S*. Since all ideals of*R*_{P}are of the form*S*^{-1}*I*for some*I**R*, we see that 0 and*R*_{P}are the only ideals of*R*_{P}and it follows that*R*_{P}is a field. Similarly*R*_{Q}is a field. Since*R*is not an integral domain, we have now established that*R*is a counterexample. - 6.
- Suppose
is a prime ideal of
*R*and*R*_{}has a nonzero nilpotent element. Then we may assume that*R*_{}has a nonzero element such that = 0. If*S*=*R*, then we may write the nilpotent element as*r*/*s*where*r**R*and*s**S*. Since (*r*/*s*)^{2}= 0, we see that*r*^{2}*t*= 0 for some*t**S*, and*rt*0 because*r*/*s*0. Also (*rt*)^{2}=*r*^{2}*tt*= 0, so*rt*is a nonzero nilpotent element of*R*.Suppose

*r*is a nonzero nilpotent element of*R*. It remains to prove that*R*_{}has a nonzero nilpotent element for some prime ideal of*R*. We may assume that*r*^{2}= 0. Let*I*= {*s**R*|*rs*= 0} , an ideal of*R*which does not contain 1. By Zorn's lemma, there is a maximal ideal of*R*containing*I*; of course will also be a prime ideal. Then the image*r*/1 in*R*_{}is nonzero because*rt*0 for all*t**R*. Since (*r*/1)^{2}=*r*^{2}/1^{2}= 0/1^{2}= 0, we see that*r*/1 is a nonzero nilpotent element of*R*_{}. This completes the proof. - 7.
- The submodule of
*M*generated by*A*and*B*is*A*+*B*; this is the set {*a*+*b*|*a**A*and*b**B*} . Thus we need to prove that*A**B**A*+*B*. We define a map :*A**B*- >*M*by (*a*,*b*) =*a*+*b*. Clearly this is an*R*-module homomorphism of*A**B*onto*A*+*B*. If (*a*,*b*) ker , then*a*+*b*= 0, consequently*a*= -*b*. This shows that*a*and -*b*are both in*A**B*= 0. Therefore*a*=*b*= 0 and hence ker = 0. It follows that is an isomorphism and so*A**B**A*+*B*as required. - 8.
- (i) Since there is a one-one correspondence between the subgroups of
Gal(
*E*/*F*) (the Galois group of*E*over*F*) and the proper subfields of*E*containing*F*, we need to show that*S*_{6}has at least 35 proper subgroups. One way to do this is to note that*S*_{6}has 144 5-cycles which gives 36 subgroups of order 5.(ii) The subfield

*L*required is Fix(*A*_{6}), the subset of*E*which is fixed pointwise by all elements of the alternating subgroup*A*_{6}of*S*_{6}. Since*A*_{6}is a normal subgroup of*S*_{6}, we see that*L*is a Galois extension of*F*, and that Gal (*E*/*L*)*A*_{6}. Since*A*_{6}is a simple group, there is no subfield between*E*and*L*which is Galois over*L*.(iii) The dimension of

*L*over*F*is |*S*_{6}/*A*_{6}| = 2.