- Let
*n*= |*G*|. Then*G*has an element*x*of order*n*. However if*H*is any proper subgroup of*G*, then every element of*H*has order strictly less than*n*. Thus*x*cannot be in any proper subgroup of*G*and the result follows. - Suppose
*G*be a simple group of order 6435. Then the number of Sylow 5-subgroups is congruent to 1 mod 5 and divides 9·11·13. Furthermore this number is not 1 because*G*is not simple. Therefore this number must be 11 and it follows that*G*has a subgroup of index 11. Since*G*is simple, we deduce that*G*is isomorphic to a subgroup of*S*_{11}(even*A*_{11}). This is not possible because 13, and hence 6435, does not divide 11! = |*S*_{11}|. We conclude that there is no simple group of order 6435 as required. - Define
θ : M
_{2}(**Q**)×M_{2}(**Q**) -> M_{2}(**Q**) by θ(*A*,*B*) =*AB*. It is easily checked that θ is an M_{2}(**Z**)-balanced map. Therefore θ induces a group homomorphismφ : MIt is easy to see that this map is a (M_{2}(**Q**)⊗_{M2(Z)}M_{2}(**Q**) -> M_{2}(**Q**)._{2}(**Q**), M_{2}(**Q**))-bimodule map. It remains to prove that φ is bijective, and we do this by producing the inverse map. Define ψ : M_{2}(**Q**) -> M_{2}(**Q**)⊗_{M2(Z)}M_{2}(**Q**) by ψ(*A*) =*A*⊗1. It is clear that φψ is the identity, so it remains to prove that φψ is the identity. Since φ and ψ are both group homomorphisms, it will be sufficient to show that ψφ is the identity on simple tensors, that is ψφ(*A*⊗*B*) =*A*⊗*B*. Therefore we need to prove that*AB*⊗1 =*A*⊗*B*.Choose a positive integer

*n*such that*Bn*∈M_{2}(**Z**). Then*AB*⊗1 =*A*/*n*(*Bn*)⊗1 =*A*/*n*⊗*Bn*= (*A*/*n*)*n*⊗*B*=*A*⊗*B*, - By the structure theorem for finitely generated modules over a PID,
*M*is a direct sum of modules of the form*R*and*R*/*p*^{n}where*p*is a prime in*R*and*n*is a positive integer. If*M*is nonzero, then it must contain a summand which is either isomorphic to*R*or*R*/*p*^{n}*R*, where*p*is a prime in*R*. Since*M*is injective and*R*is a domain,*rM*=*M*for all*r*∈*R*\ 0, in particular*pM*=*M*for primes*p*in*R*. Thus*M*cannot contain a summand isomorphic to*R*/*p*^{n}*R*. On the other hand if*M*contains a summand isomorphic to*R*, let*p*be a prime in*R*, which exists because*R*is not a field. Since*pR*≠*R*, we see that*pM*≠*M*, a contradiction and the result follows. - (a)
- Obviously
**Q**(ζ_{p})⊆**Q**(ζ_{2p}), because ζ_{2p}^{2}= ζ_{p}. On the other hand ζ_{2p}= -ζ_{p}, hence**Q**(ζ_{2p})⊆**Q**(ζ_{p}) and the result follows. - (b)
- Set
*f*(*x*) = 1 +*x*^{2}+ ... +*x*^{2p-2}. Since*f*(*x*)(1 -*x*^{2}) = 1 -*x*^{2p}and ζ_{p},ζ_{2p}≠±1, we see that ζ_{p}and ζ_{2p}both satisfy*f*(*x*). Thus the minimal polynomial of both these divides*f*(*x*). Now ζ_{p}satisfies*g*(*x*) : = 1 +*x*+ ... +*x*^{p-1}. By making the substitution*y*=*x*+ 1, we see that*g*(*x*) is irreducible in**Z**[*x*] by Eisenstein for the prime*p*. Since deg(*g*) =*p*- 1≥1, it is also irreducible in**Q**[*x*]. It follows that*g*(*x*) is the minimal polynomial of ζ_{p}over**Q**. Also by considering the automorphism of**Q**[*x*] induced by*x*|--> -*x*, we see that*g*(-*x*) is the minimal polynomial of ζ_{2p}over**Q**, so*g*(-*x*) divides*f*(*x*). It follows that*f*(*x*) =*g*(*x*)*g*(-*x*), the product of two irreducible polynomials.

- Set
*f*(*x*) =*x*^{5}- 5*x*- 1. Then*f'*(*x*) = 5*x*^{4}-5 = 5(*x*^{2}+ 1)(*x*- 1)(*x*+ 1). Thus*f*(*x*) has a maximum at -1, a minimum at 1. Since*f*(1) > 0 and*f*(- 1) < 0, we find that*f*has exactly 3 real roots and 2 complex roots. We want to prove that*f*is irreducible (as a polynomial in**Q**[*x*]). By Gauss's lemma, if*f*is not irreducible, then we may write*f*=*gh*where*g*,*h*∈**Z**[*x*], deg*g*, deg*h*≥1, and*g*,*h*are monic. Neither of*g*,*h*has degree one, because ±1 is not a root of*f*. Therefore we may without loss of generality assume that deg*g*= 3 and deg*h*= 2, say*g*=*x*^{3}+*ax*^{2}+*bx*+*c*and*h*=*dx*+*e*, where*a*,*b*,*c*,*d*,*e*∈**Z**. By equating coefficients, we find that*a*+*d*= 0,*ad*+*bc*+*e*= 0,*ae*+*bd*+*c*= 0,*be*+*cd*= 1,*ce*= 1. Thus*c*,*e*= 1 or*c*,*e*= -1, and we find that*a*^{2}±*a*+ 1 = 0. This last equation has no root in**Z**and we conclude that*f*is irreducible.Let

*G*denote the Galois group of*f*over**Q**. We consider*G*as a subgroup of*S*_{5}(by permuting the 5 roots of*f*). Since*f*is irreducible and 5 is prime, we see that*G*contains a 5-cycle. Also*G*contains a transposition, namely complex conjugation. Since*S*_{5}is generated by a 5-cycle and a transposition, we deduce that*G*≌*S*_{5}. - (a)
- We may write the general element of
**Q**[*x*,*y*] as*f*_{0}+*f*_{1}*y*+*f*_{2}*y*^{2}+ ... +*f*_{n}*y*^{n}, where*n*is a positive integer and*f*_{i}∈**Q**[*x*] for all*i*. Then modulo the ideal (*x*^{3}-*y*^{2}), we may replace*y*^{2}with*x*^{3}everywhere and we see that every element of**Q**[*x*,*y*] can be written in the form*f*+*gy*+ (*x*^{3}-*y*^{2})*h*, where*f*,*g*∈**Q**[*x*] and*h*∈**Q**[*x*,*y*]. The result follows. - (b)
- Define a ring homomorphism
θ :
**Q**[*x*,*y*] ->**Q**[*t*] by θ(*x*) =*t*^{2}, θ(*y*) =*t*^{3}and θ(*q*) =*q*for*q*∈**Q**. Then imθ =**Q**[*t*^{2},*t*^{3}]. Also if*h*∈kerθ, write*h*=*k*+*f*+*yg*, where*k*∈(*x*^{3}-*y*^{2}), and*f*,*g*∈**Q**[*x*]. Then θ(*h*) =*f*(*t*^{2}) +*t*^{3}*g*(*t*^{2}). Since*f*(*t*^{2}) is a polynomial involving only even powers of*t*and*t*^{3}*g*(*t*^{2}) is a polynomial involving only odd powers of*t*, we see that θ(*h*) can only be zero if*f*,*g*= 0. It follows that kerθ = (*x*^{3}-*y*^{2}) and the result now follows from the fundamental homomorphism theorem. Note that we have also proven that if*h*(*x*^{2},*x*^{3}) = 0, then*h*∈(*x*^{3}-*y*^{2}). - (c)
- Note that
*t*^{2}and*t*^{3}are irreducible in**Q**[*t*^{2},*t*^{3}] (use unique factorization in**Q**[*t*]). Since*t*^{6}= (*t*^{2})^{3}= (*t*^{3})^{2}, two different ways of factoring*t*^{6}, we see that**Q**[*t*^{2},*t*^{3}] is not a UFD. - (d)
- Note that
Z(
*x*^{3}-*y*^{2}) = {(*t*^{2},*t*^{3}) |*t*∈**Q**}. Indeed (*t*^{2},*t*^{3})∈Z(*x*^{3},*y*^{2}), because (*t*^{2})^{3}- (*t*^{3})^{2}= 0. On the other hand if (*p*,*q*)∈Z(*x*^{3}-*y*^{2}), write*t*=*q*/*p*(*t*= 0 if*p*= 0). Since*p*^{3}=*q*^{2}, we see that*p*=*t*^{2}and*q*=*t*^{3}. Now suppose*f*is a polynomial vanishing on*V*. Then*f*(*t*^{2},*t*^{3}) = 0 for all*t*∈**Q**. Since**Q**is infinite, we see that*f*(*x*^{2},*x*^{3}) = 0 and it follows from (b) that*f*∈(*x*^{3}-*y*^{2}). It follows that the coordinate ring**Q**[*V*] of*V*is**Q**[*x*,*y*]/(*x*^{3}-*y*^{2})≌**Q**[*t*^{2},*t*^{3}] by (b). - (e)
- Since
**Q**is an infinite field,**A**^{1}has coordinate ring**Q**[*x*], a UFD. But**Q**[*V*] is not a UFD by (c), in particular**Q**[**A**^{1}] is not isomorphic to**Q**[*V*]. Since isomorphic affine algebraic sets have isomorphic coordinate rings, we deduce that*V*is not isomorphic to**A**^{1}.

Peter Linnell 2012-01-14