- We have
*f*(*x*) = (*x*+ 1)(*x*^{4}+ 3); since -1**Q**, the splitting field for*x*^{4}+ 3 is also*K*. Let w = (1 +*i*)/, a primitive 8th root of 1. Then w^{4}= - 1 and we see that the four roots of*x*^{4}+ 3 are w^{r}for*r*= 1, 3, 5, 7. Therefore*K*=**Q**(w,w^{3},w^{5},w^{7}). Since*x*^{4}+ 3 is irreducible by Eisenstein for the prime 3, we see that [**Q**(w) :**Q**] = 4. Let g denote complex conjugation. Since*x*^{4}+ 3 is a polynomial with real coefficients, we see that g Gal(*K*/**Q**). Thus*K*, because = w + g(w). Now satisfies*x*^{4}- 12, which is irreducible by Eisenstein for the prime 3. Therefore [**Q**() :**Q**] = 4. Note that**Q**() =/=**Q**(w), because the former is contained in**R**while the latter is not. We deduce that*K*=/=**Q**(w). Also*i*= w^{3}/w, which shows that*i**K*. Since w^{2r + 1}=*i*^{r}w, we conclude that*K*=**Q**(*i*,w). Therefore [*K*:**Q**(w)] = 2 and hence [*K*:**Q**] = 8.Of course a consequence of this is that

*x*^{4}+ 3 remains irreducible over**Q**(*i*). Let q Gal(*K*/**Q**(*i*)) satisfy q(w) = w^{3}. Then q(w^{3}) = w^{5}, q(w^{5}) = w^{7}and q(w^{7}) = w, in particular q has order 4. Furthermore gqg(*i*) =*i*and gqg(w) = w^{7}, which shows that gqg = q^{-1}. We now see that Gal(*K*/**Q**) = {q^{r}g^{s}|*r*= 0, 1, 2, 3,*s*= 0, 1} and is isomorphic to the dihedral group of order 8. - This is false. Consider the group
**Z**_{4}**Z**_{2}, where**Z**_{n}denotes the integers modulo*n*. Then (2, 0) and (0, 1) both have order 2 (when we write (2, 0), the 2 means 2 modulo 4). Suppose q is an automorphism such that q(2, 0) = (0, 1). Then 2(q(1, 0)) = q(2, 0) = (0, 1). On the other hand 2(q(1, 0)) is of the form 2(*a*,*b*) = (2*a*, 0), and so cannot be equal to (0, 1). Thus we have a contradiction and we conclude that there is no such q. - Let
*n*denote the number of Sylow 2-subgroups. Since 2002 = 2*1001, we see that a Sylow 2-subgroup has order 2 and*n*| 1001. Therefore each Sylow 2-subgroup has exactly one element of order 2 and*n*is odd. Also any element of order 2 is in exactly one Sylow 2-subgroup, consequently the number of elements of order 2 is*n*. Since the number of elements in the set {*h**H*|*h*^{2}=*e*} is*n*+ 1 (the ``+1" for the identity), we conclude that this number is even. - Suppose
*G*is nonabelian, so there exist*a*,*b**G*such that*ab*=/=*ba*. Set*g*=*aba*^{-1}*b*^{-1}, so*g*=/= 1. By hypothesis there exists*K*<|*G*such that*G*/*K*is abelian and*g**K*. But*KaKb*(*Ka*)^{-1}(*Kb*)^{-1}=*Kg*=/= 1, which shows that*KaKb*=/=*KbKa*and hence*G*/*K*is nonabelian, which is a contradiction. The result follows. - Certainly if
*A*and*B*are commutative rings, then*A*X*B*is also a commutative ring. We need to show that if*A*and*B*are in addition Noetherian, then so is*A*X*B*. Suppose*I*_{1},*I*_{2},... is an ascending chain of ideals in*A*X*B*. Then (0 X*B*) /\*I*_{1},(0 X*B*) /\*I*_{2},... is an ascending chain of ideals in 0 X*B*. But 0 X*B*@*B*and*B*is Noetherian, hence there exists a positive integer*M*such that 0 X*B*/\*I*_{n}= 0 X*B*/\*I*_{M}for all*n*__>__*M*. Also (*A*X*B*)/(0 X*B*) @*A*as rings, so (*A*X*B*)/(0 X*B*) is Noetherian. Therefore the ascending chain of ideals (0 X*B*) +*I*_{1},(0 X*B*) +*I*_{2},... of (*A*X*B*)/(0 X*B*) becomes stationary, that is there is a positive integer*N*such that (0 X*B*) +*I*_{n}= (0 X*B*) +*I*_{N}for all*n*__>__*N*. Let*P*be the maximum of*M*and*N*. We claim that*I*_{n}=*I*_{P}for all*n*__>__*P*. Obviously*I*_{n}*I*_{P}for all*n*__>__*P*, so we need to show the reverse inclusion. Let*x**I*_{n}. Since (0 X*B*) +*I*_{n}= (0 X*B*) +*I*_{P}, we may write*x*=*b*+*i*where*b*0 X*B*and*i**I*_{P}. Since*x*,*i**I*_{n}, we see that*b*(0 X*B*) /\*I*_{n}and hence*b*(0 X*B*) /\*I*_{P}, because (0 X*B*) /\*I*_{n}= (0 X*B*) /\*I*_{P}. This shows that*x**I*_{P}and hence*I*_{n}=*I*_{P}for*n*__>__*P*. - Obviously
*P*/*IP*is an*R*/*I*-module; we need to prove that it is projective. Suppose we are given an*R*/*I*-epimorphism m :*M*- - > >*N*of*R*/*I*-modules and an*R*/*I*-map q :*P*/*IP*- >*N*. We need an*R*/*I*-map b :*P*/*IP*- >*M*such that q = mb. Let p :*P*- >*P*/*IP*denote the natural epimorphism. We can also view*M*and*N*as*R*-modules, and then m is also an*R*-map. Since*P*is a projective*R*-module, certainly there exists an*R*-map a :*P*- >*M*such that ma = qp. If*i**I*and*p**P*, then a(*ip*) =*i*a*p**IM*= 0. Therefore*IP*kera and we deduce that a induces an*R*/*I*-map b :*P*/*IP*- >*M*satisfying bp = a. Then mbp = ma = qp and since p is onto, we conclude that mb = q.Sketch of alternate proof. Since

*P*is projective, we may write*P**Q*@*F*for some*R*-modules*Q*,*F*with*F*free. Then*P*/*IP**Q*/*IQ*@*F*/*IF*and since*F*/*IF*is a free*R*/*IR*-module, we see that*P*/*IP*is a projective*R*/*IR*-module. - (a)
- Certainly
*k*+*I*is a subgroup of*k*[*x*] under addition; we need to show that it is closed under multiplication. However if*a*,*b**k*and*i*,*j**I*, then (*a*+*i*)(*b*+*j*) =*ab*+ (*aj*+*ib*+*ij*)*k*+*I*, because*aj*,*ib*,*ij**I*by using*I*<|*k*[*x*]. - (b)
- Note that if
*i**I*, then*i*a = 1*i*a = 0 for all a*k*[*x*]/*I*. Since*I*is a nonzero ideal of*k*[*x*], we know that*k*[*x*]/*I*is finite dimensional over*k*. Thus there is a positive integer*n*such that {*I*+ 1,*I*+*x*,...,*I*+*x*^{n}} spans*k*[*x*]/*I*as a*k*-module. It follows that any simple tensor b a (where b*k*[*x*] and a*k*[*x*]/*I*) is a*k*-linear sum of elements of the form*x*^{r}(*I*+*x*^{s}), where*r*,*s*__<__*n*. Since every element of*k*[*x*]*k*[*x*]/*I*is a sum of simple tensors, we deduce that*k*[*x*]*k*[*x*]/*I*is generated as a*k*-module by elements of the form*x*^{r}(*I*+*x*^{s}) where*r*,*s*__<__*n*and the result follows.