1
$\begingroup$

I think that the problem statement on this practice question is incorrect, but I don't want to assume that.

The statement is: Let F be a field whose characteristic does not divide n. Prove that f(x) = $x^n − 1$ is separable over F and that the Galois group of f is isomorphic to a subgroup of $(Z/nZ)^×$.

Proving that f is separable: f'(x) = $nx^n-1$ is relatively prime to f(x) since the factors of f' are just n-1 copies of x. If a function is relatively prime to its derivative in a field, then it is separable over that field.

While proving that the Galois group is isomorphic to a subgroup of $(Z/nz)^x$ I ran into some trouble. I think it is just isomorphic to $(Z/nz)^x$ not a subgroup.

My proof is: Since f is separable in F, we must have that L the splitting field for f is a subfield of F. Since all splitting fields are unique up to isomorphism we know L $\cong $ $\mathbb Q$[w, $w^2$, ..., $w^n$,] where w is the primitive nth root of unity.

These roots form a cyclic group of order n, and Aut(Z/nZ) $\cong $ $(Z/nZ)^×$ So the Galois Group (which is the group of permutations of the roots) is isomorphic to $(Z/nZ)^×$.

My issue is that it is isomorphic to the multiplicative group itself, not a subgroup and in a followup problem I am asked to construct a field for $x^15-1$ where the Galois group is a proper subgroup of (Z/15Z)$^x$.

Any insight on this would be awesome! Thanks.

$\endgroup$
  • $\begingroup$ A group is always a subgroup of itself. $\endgroup$ – Lord Shark the Unknown Mar 26 '18 at 18:22
1
$\begingroup$

It is true that the Galois group $G_n$ of the splitting field of $x^n-1$ is isomorphic to $(\Bbb Z/n\Bbb Z)^*$. That's not so easy (see for instance Washington's book on cyclotomic fields), but is is much more elementary to prove that $G$ embeds in $(\Bbb Z/n\Bbb Z)^*$. Given an $n$-th root of unity $\zeta$, $\sigma\in G_n$ maps $\zeta$ to $\zeta^a$ for some $a$, and $\sigma\mapsto a$ is the embedding you seek. It's more difficult to show it's surjective.

To get a smaller Galois group than say $(\Bbb Z/15\Bbb Z)^*$ take the fixed field of a subgroup of the Galois group of $\Bbb Q(\zeta_{15})$ but I suppose that would give an extension whose Galois group was a quotient rather than a subgroup of $(\Bbb Z/15\Bbb Z)^*$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.