- First we write 380 as a product of prime powers, namely 2
^{2}*5*19. Suppose by way of contradiction*G*is a simple group of order 380. The number of Sylow 19-subgroups is congruent to 1 mod 19 and divides 20, hence is 1 or 20. But 1 is ruled out because then*G*would have a normal subgroup of order 19, which would contradict the hypothesis that*G*is simple. Therefore*G*has 20 Sylow 19-subgroups. Next we consider the Sylow 5-subgroups. The number is congruent to 1 mod 5 and divides 4*19. Thus there are 1 or 76 Sylow 5-subgroups.Now we count elements. If

*P*and*Q*are distinct Sylow 19-subgroups, then*P*∩*Q*≠*P*and*P*∩*Q*__<__*P*. Since |*P*∩*Q*| divides |*P*| = 19 by Lagrange's theorem, we deduce that*P*∩*Q*= 1. It follows that*G*has at least 20*18 = 360 elements of order 19. Similarly two distinct Sylow 5-subgroups intersect trivially and we deduce that*G*has at least 76*4 = 304 elements of order 5. We conclude that*G*has at least 360 + 304 = 664 > 380 elements, which is a contradiction. Therefore there is no simple group of order 380. - Let
ω = ∈
**F**_{7}, so ω≠1 = ω^{3}. We have (*f*-*g*)(*f*- ω*g*)(*f*- ω^{2}) =*h*^{3}. Since*f*,*g*are coprime, we see that*f*-*g*,*f*- ω*g*,*f*- ω^{2}*g*are pairwise coprime. Now use the fact that*k*[*x*_{1},...,*x*_{n}] is a UFD; remember that the units of*k*[*x*_{1},...,*x*_{n}] are precisely the nonzero elements of*k*. Write*h*=*up*_{1}^{r1}...*p*_{m}^{rm}where 0≠*u*∈*k*,*m*is a nonnegative integer,*p*_{i}is prime for all*i*, and*r*_{i}is a positive integer for all*i*. Since*f*-*g*,*f*- ω*g*,*f*- ω^{2}*g*are pairwise coprime, we see that if*p*_{i}divides one of these three polynomials, then*p*_{i}doesn't divide the other two polynomials, and it follows that*p*_{i}^{3ri}is the precise power of*p*_{i}which divides this polynomial. We deduce that each of*f*-*g*,*f*- ω*g*,*f*- ω^{2}*g*is of the form*uq*^{3}for some unit*u*and some polynomial*q*, and the result follows. - We use the structure theorem for finitely generated modules over a
PID, elementary divisor form. We may write
*M*= ⊕_{i∈I}(*R*/*p*^{i}*R*)^{ei}⊕⊕_{i}*C*_{i}, where*e*_{i}∈**N**,*I*is a finite subset of**N**, and the*C*_{i}are modules of the form*R*or*R*/*q*^{h}*R*, where*h*∈**N**and*q*is a prime which is not associate to*p*. This expresses*M*in a unique way as a direct sum of indecomposable*R*-modules. The hypothesis that*pm*= 0≠*m*implies*Rm*is not a direct summand of*M*tells us that 1∉*I*. Similarly we may write*N*= ⊕_{i∈J}(*R*/*p*^{i}*R*)^{fi}⊕⊕_{i}*D*_{i}, where*f*_{i}∈**N**,*J*is a finite subset of**N**not containing 1, and the*D*_{i}are modules of the form*R*or*R*/*q*^{f}*R*, where*f*∈**N**and*q*is a prime which is not associate to*p*. Note that*pC*_{i}≌*C*_{i}and*pD*_{i}≌*D*_{i}for all*i*. Also*p*(*R*/*p*^{i}*R*)≌*R*/*p*^{i-1}*R*≠ 0 for*i*≥2. Thus*pM*≌⊕_{i∈I}*R*/*p*^{i-1}*R*⊕⊕_{i}*C*_{i}and*pN*≌⊕_{i∈J}*R*/*p*^{i-1}⊕*R*⊕_{i}*D*_{i}, and these expressions are direct sums of indecomposable modules. Since*pM*≌*pN*, the uniqueness statement in the structure theorem for modules over a PID yields*I*=*J*and after renumbering if necessary,*C*_{i}≌*D*_{i}for all*i*. The result follows. - Let
*G*denote the Galois group of*K*over**Q**. Then |*G*| = 27 and there is a one-to-one correspondence between subfields of*K*and subgroups of*G*which is reverse including. Also [**Q**(α) :**Q**] = 9 because*f*is irreducible with degree 9. Therefore**Q**(α) corresponds to a subgroup*H*of order 3. To find a subfield of**Q**(α) which has degree 3 over**Q**, we need to find a subgroup of order 9 which contains*H*. Since*G*is a nontrivial finite 3-group, it contains a central subgroup*Z*of order 3. If*Z*is not contained in*H*, then*H*∩*Z*= 1, hence|and we see that*HZ*|/3 = |*HZ*|/|*Z*| = |*HZ*/*Z*| = |*H*/*H*∩*Z*| = |*H*| = 3*HZ*is a subgroup of order 9 containing*H*. On the other hand if*Z*⊆*H*, then*H*=*Z*and hence*H*<|*G*. Thus*G*/*H*is a group of order 9, and hence has a subgroup of order 3, which by the subgroup correspondence theorem we may write as*K*/*H*, where*K*is a subgroup of*G*containing*H*. The order of*K*is 3|*H*| = 9, which finishes the proof. - We note that given
*a*,*b*∈*A*, there exists*n*∈**N**and*c*∈*A*such that*p*^{n}*a*= 0 and*p*^{n}*c*=*b*. This shows that for*a*,*b*∈*A*,*a*⊗*b*=*a*⊗*p*^{n}*c*=*p*^{n}*a*⊗*c*= 0⊗*c*= 0.*A*⊗*A*is generated as an abelian group by ``simple tensors"*a*⊗*b*, we deduce that every element of*A*⊗_{Z}*A*is zero, in other words*A*⊗_{Z}*A*= 0. - The minimal polynomial divides the characteristic polynomial, so is
*x*,*x*^{2}or*x*^{3}. Also since the minimal polynomial factors into linear factors over*k*, the Jordan canonical form for*A*is defined over*k*.If the minimal polynomial is

*x*, then*A*= 0, so we could take*B*= 0, since then*B*^{2}= 0 =*A*.If the minimal polynomial is

*x*^{2}, then the invariant factors of*A*are*x*,*x*^{2}. Consider the matrix*C*: =( 0 0 1 0 0 0 0 0 0 ) Then

*C*≠ 0 and*C*^{2}= 0, so the invariant factors of*C*are*x*,*x*^{2}and therefore*C*is similar to*A*. Thus it will be sufficient to find a matrix*B*such that*B*^{2}=*C*; here we could take*B*to be( 0 1 0 0 0 1 0 0 0 ) Then

*B*^{2}=*C*as required.Finally suppose

*A*has one invariant factor, which will necessarily be*x*^{3}. Then the Jordan canonical form of*A*is( 0 1 0 0 0 1 0 0 0 ) Suppose there is a matrix

*B*such that*B*^{2}=*A*. Then*B*^{6}=*A*^{3}= 0. Therefore the minimal polynomial of*B*divides*x*^{6}and since*B*is a 3 by 3 matrix, we deduce that the minimal polynomial of*B*divides*x*^{3}. Therefore*B*^{3}= 0 and we conclude that*A*^{2}=*B*^{4}= 0. But*A*^{2}=( 0 0 1 0 0 0 0 0 0 ) which is nonzero, and the result follows.

- Let
*K*denote the field of fractions of*R*and let*I*denote the ideal of*K*[*x*_{1},...,*x*_{n}] generated by*S*. Then*Z*(*S*) = {(*r*_{1},...,*r*_{n})∈*R*^{n}|*f*(*r*_{1},...,*r*_{n}) = 0} for all*f*∈*I*. By Hilbert's basis theorem, there is a finite subset*T*of*S*which generates the ideal*I*. Then*Z*(*S*) =*Z*(*T*).

Peter Linnell 2011-08-19