- We use the structure theorem for finitely generated modules over
a PID. This tells us that
*M*is a direct sum of modules of the form*R*/*Rt*where*t*is either zero or a power of an irreducible element of*R*. If this direct sum has more than one factor, then we may write*M*=*A**B*for some nonzero submodules*A*,*B*of*M*, and then*A**B*= 0. Therefore there is exactly one factor in the direct sum, and the result follows. - Since the action is transitive, all the stabilizers are
conjugate and we see that the stabilizer of any element of
*X*has no element of finite order other than 1. Thus if*f**G*has finite order larger than 1, then*f*cannot be in the stabilizer of any element of*X*, and it follows that*f*cannot fix any element of*X*. - Suppose now that
*f*has prime order*q*. Then the orbits of <*f*> have order dividing*q*. Since*q*is prime, these orbits have order 1 or*q*. But if one of the orbits has order 1, then*f*fixes the element in that orbit which contradicts the above. Therefore all orbits of <*f*> have order*q*, and it follows that*q*divides |*X*| as required.

- Since the action is transitive, all the stabilizers are
conjugate and we see that the stabilizer of any element of
- Since
*S*/(*P*_{1}...*P*_{t}) is finite, we see that*S*/*P*_{i}is finite for all*i*. Now*S*/*P*_{i}is an integral domain because*P*_{i}is prime, and finite integral domains are fields. Therefore*S*/*P*_{i}is a field and we deduce that*P*_{i}is a maximal ideal for all*i*, as required. - Suppose
*K*is a nontrivial extension field of*F*with degree which is not a power of*p*. Note that if*L*is any finite extension field of*K*, then the degree of*L*over*F*is also not a power of*p*, because [*L*:*F*] = [*L*:*K*][*K*:*F*]. Since we are in characteristic zero, everything is separable so by taking a splitting field containing*K*, we may assume that*K*is a Galois extension of*F*. Let*G*= Gal(*K*/*F*) and let*P*be a Sylow*p*-subgroup of*G*. Then [*K*^{P}:*F*] = [*G*:*P*] and [*K*:*F*] = |*G*|. Since [*G*:*P*] has order prime to*p*, it follows that [*K*^{P}:*F*] has order prime to*p*and we conclude that*K*^{P}is a nontrivial extension field of*F*with degree prime to*p*. - This question depends on what we are allowed to assume; some
people take the given property as the definition of projective module.
Also the question does not require that
*R*be commutative. Let us use the definition that an*R*module*P*is projective if and only if it is a direct summand of a free*R*-module. Suppose first that we have the given property. Choose an epimorphism*f*:*F*- >*M*where*F*is a free*R*-module. Since the map*f*_{*}: Hom(*M*,*F*) - > Hom(*M*,*M*) is surjective, there exists an*R*-module map*g*:*M*- >*F*such that*fg*is the identity map on*M*. Then*g*is a monomorphism and so*M**gM*. Also*F*= ker*f**gM*, which shows that*gM*and hence also*M*are projective.Conversely suppose

*M*is a direct summand of a free module*F*. First we show that*F*satisfies the given condition. Let*f*:*N*- >*N'*be a surjection of*R*-modules and let*h*:*F*- >*N'*be any*R*-map. Let {*e*_{i}|*i**I*} be an*R*-basis for*F*, where*I*is some indexing set. Since*h*is surjective, we may choose*n*_{i}*N*such that*f*(*n*_{i}) =*h*(*e*_{i}) for all*i*. Now we can define*g*Hom_{R}(*F*,*N*) by*g*(*e*_{i}) =*n*_{i}for all*i*, and then*fg*(*e*_{i}) =*h*(*e*_{i}) for all*i*. Thus*fg*=*h*, and we have proved the result in the case*M*=*F*.For the general case, write

*F*=*M**P*as*R*-modules, and let y:*M*- >*F*be the natural monomorphism Then we have a commutative diagramHom _{R}(*F*,*N*)*f*_{*}- >Hom _{R}(*F*,*N'*)y ^{*}y ^{*}Hom _{R}(*M*,*N*)*f*_{*}- >Hom _{R}(*M*,*N'*)^{*}*g*)(*m*) =*g*(y*m*) for all*m**M*and*g*Hom_{R}(*F*,*N*) or Hom_{R}(*F*,*N'*). Note that the right hand (and also the left hand) y^{*}is surjective: if*h*Hom_{R}(*M*,*N*), we may extend*h*to an*R*-map*F*- >*N*by defining it to be 0 on*P*and then y^{*}*h*=*g*. By the previous paragraph the top*f*_{*}is surjective and we deduce that the bottom*f*_{*}is surjective as required. - The dihedral group has a cyclic subgroup
*C*of index 2. Let c denote the character of the regular representation of*C*. Since*C*is abelian, we may write c=a_{1}+ ... +a_{n}for some integer*n*, where the a_{i}are degree one characters of*C*. Let y denote the character of*V*. Since c^{D}is the character of the regular representation of*D*, we see that <y,c^{D}> 0. Therefore <y,a_{i}^{D}> 0 for some*i*. Since y is irreducible, we deduce that a_{i}^{D}=y+f for some character f of*D*. Therefore y has degree at most that of a_{i}^{D}. But a_{i}has degree 1, consequently a_{i}^{D}has degree 2 and the result follows. - Let
*G*denote the Galois group of*X*^{10}- 1 over , and let w be a primitive 10th root of 1 (so w =*e*^{pi/5}). Then the roots of*X*^{10}- 1 are w^{r}, where*r*= 0, 1,..., 9, which shows that the splitting field for*X*^{10}- 1 is [w]. Since*X*^{10}- 1 = (*X*^{5}- 1)(*X*+ 1)(*X*^{5}+ 1)/(*X*+ 1) and w does not satisfy (*X*^{5}- 1)(*X*+ 1), we see that w is a root of (*X*^{5}+ 1)/(*X*+ 1). By making the substitution*Y*=*X*+ 1 and using Eisenstein's criterion for the prime 5, we see that (*X*^{5}+ 1)/(*X*+ 1) is irreducible over . This shows that [[w ] : ] = 4 and we deduce that |*G*| = 4. Finally we can define q*G*by q(w) =w^{3}, because w^{3}is also a primitive 5th root of 1, and since q^{2}(w) =w^{9}w, we deduce that q^{2}1. We conclude that*G*has an element of order 4 and hence*G*is cyclic of order 4.