- (a) Obviously 0
*IJ*. Now suppose*x*,*y**IJ*and*r**R*. We want to prove that*x*+*y*,*rx**IJ*. Write*x*= S_{i = 1}^{n}*a*_{i}*b*_{i}and*y*= S_{i = 1}^{m}*c*_{i}*d*_{i}, where*a*_{i},*c*_{i}*I*and*b*_{i},*d*_{i}*J*. Then*x*+*y*= S_{i = 1}^{n}*a*_{i}*b*_{i}+ S_{i = 1}^{m}*c*_{i}*d*_{i}*x*+*y**IJ*. Also*rx*= S_{i = 1}^{n}(*ra*_{i})*b*_{i}, and since*I*is an ideal of*R*, we see that*ra*_{i}*I*for all*i*. Therefore*rx**IJ*and we have proven that*IJ**R*.(b) Since

*I**R*, we have*IJ**I*. Similarly*IJ**J*and we deduce that*IJ**I**J*.(c) Since

*I*+*J*=*R*, we may write*i*+*j*= 1 where*i**I*and*j**J*. If*x**I**J*, then*x*=*xi*+*xj**JI*+*IJ*=*IJ*(because*R*is commutative). Therefore*x**IJ*and we have proven that*I**J**IJ*. The result now follows from (b).(d) Let

*a**R*\ 0. We need to prove that*a*has a multiplicative inverse. Using*IJ*=*I**J*with*I*=*J*=*aR*, we see that*aRaR*=*aR**aR*=*aR*, hence*aras*=*a*for some*r*,*s**R*. Since*a*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. - (a) Since
*F*is a finite Galois extension of*K*with Galois group*S*_{5}, there is a one-one correspondence between the fields strictly between*F*and*K*, and the proper nontrivial subgroups of*S*_{5}. Therefore we need to show that*S*_{5}has more than 40 subgroups other than 1 and*S*_{5}. Now*S*_{5}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*) Gal(*F*/*K*)/*H*and [Fix(*H*) :*K*] = [Gal(*F*/*K*) :*H*]. Since*S*_{5}has a unique nontrivial normal subgroup, namely the alternating group*A*_{5}, it follows that the subfield*E*required is Fix(*A*_{5}). Then [*E*:*K*] = 2 and Gal(*E*/*K*)*S*_{5}/*A*_{5}/2. - (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**G*/*A*, so*H**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**H*; in fact we can assert more, namely that*B**G*. To see this, let*g**G*. Then*gBg*^{-1}is a subgroup of*gHg*^{-1}=*H*because*H**G*, and since |*gBg*^{-1}| = |*B*| = 7, we see that*gBg*^{-1}=*B*which establishes that*B**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**A*X*B*X*C*. Let*a**A*be an element of order 13, let*b**B*be an element of order 7, and let*c**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*a*^{35}*b*^{35}*c*^{35}= 1. But*b*^{35}=*c*^{35}= 1, hence*a*^{35}= 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. - We shall use the fundamental theorem for finitely generated
modules over a PID. Thus we may write
*A*= *R*^{a}(*R*/*q*_{i}*R*)^{ai}*B*= *R*^{b}(*R*/*q*_{i}*R*)^{bi}

where*a*,*b*,*a*_{i},*b*_{i},*m*are nonnegative integers and the*q*_{i}are distinct prime powers. Since*A*^{n}*B*^{n}, we have*R*^{na}(*R*/*q*_{i}*R*)^{nai}*R*^{nb}(*R*/*q*_{i}*R*)^{nbi}.

The fundamental theorem now gives that*na*=*nb*and*na*_{i}=*nb*_{i}for all*i*, hence*a*=*b*and*a*_{i}=*b*_{i}for all*i*and the result follows. - Since
*P*is a projective module, it is a submodule of a free module*F*. The mapping q :*P*- >*F*defined by q*p*= 2*p*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*= 2f*p*for all*p**P*and hence*P*2*P*. Therefore*P*2^{n}*P*and we deduce that*P*2^{n}*F*for all positive integers*n*. But 2^{n}*F*= 0 because*F*is a free module and the result follows. (Note: the hypothesis*P*is finitely generated has not been used.) - Let
a :
*mB*- >*B*denote the natural inclusion, and let b :*B*- >*B*/*mB*denote the natural surjection. Then the exact sequence*mB*^{a}- >*B*^{b}- >*B*/*mB*- > 0 yields an exact sequence*A**mB*^{1 a}- >*A**B*^{1 b}- >*A**B*/*mB*- - > 0*A*(*B*/*mB*) (*A**B*)/im(1 a), where im denotes the image of a map. Now im(1 a) is the -submodule of*A**B*generated by {*a**mb*|*a**A*and*b**B*} and since*a**mb*=*m*(*a**b*), this is the same as the -submodule generated by {*m*(*a**b*) |*a**A*and*b**B*}. This submodule is precisely*m*(*A**B*), hence im(1 a) =*m*(*A**B*) and the proof is complete. - 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**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*B*_{1}and*B*_{2}are two distinct Sylow 3-subgroups, then*B*_{1}*B*_{2}is a proper subgroup of*B*_{1}whose order divides 3 by Lagrange's theorem, hence*B*_{1}*B*_{2}= 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. - Let
*L*= (,,). Then clearly*K**L*. Also ( + )^{3}- 9( + ) = 2, hence*K*and we deduce that*L*=*K*. Thus*K*= (,,). It follows that*K*is the splitting field for the polynomial (*x*^{2}- 2)(*x*^{2}- 3)(*x*^{2}- 5) and we deduce that*K*is a Galois extension of . In particular the number of fields between*K*and equals the number of subgroups of Gal(*K*/).Now any element of Gal(

*K*/) must send to ±, to ±, and to ±. It follows that every nonidentity element of Gal(*K*/) has order 2, and that Gal(*K*/) 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 () = (), then*a*=*b*. To see this, write =*r*+*s*where*r*,*s*. Then*a*=*r*^{2}+ 2*rs*+*s*^{2}*b*. Clearly*s*0 and*rs*= 0, consequently*r*= 0 and*a*=*s*^{2}*b*which establishes the result.It follows immediately that there are at least 8 subfields between and

*K*, namely () where*c*,*d*,*e*are 0 or 1. Now if | Gal(*K*/)|4, then there would be at most 5 subgroups of Gal(*K*/), consequently there would be at most 5 fields between*K*and . This is a contradiction, so we must have | Gal(*K*/)| = 8 and therefore Gal(*K*/) (/2)^{3}.