1
$\begingroup$

Suppose $K/\mathbb Q$ is a Galois extension of degree $3$. I know that this implies: $K/\mathbb Q$ is a splitting field of a separable polynomial $f(x)$ with coefficients in $\mathbb Q$. I can see why $\deg f(x) \neq 1$ and $\neq 2$. I think it can be that $\deg f(x) \ge 4$, but I also think that this can only happen if $f(x)$ already has roots in $\mathbb Q$; I am not sure though.

Can we extract from $f(x)$ a degree $3$ polynomial whose splitting field is $K$? Anyhow, does the question in the title have a positive answer?

$\endgroup$
5
$\begingroup$

Yes. It has to be the splitting field of an irreducible cubic.

By the primitive element theorem $K=\Bbb{Q}(\alpha)$ for some $\alpha\in K$. The minimal polynomial $m(x)$ of $\alpha$ is then necessarily a cubic. But, as $K/\Bbb{Q}$ is normal, all the zeros of $m(x)$ are in $K$.


Extras for the question about an arbitrary polynomial $f(x)$ with splitting field $K$. Consider a non-linear irreducible factor of $f(x)$. It cannot be a quadratic for then $[K:\Bbb{Q}]$ would be even. It cannot be $\ge4$ for then $[K:\Bbb{Q}]$ would also be $\ge4$. Ergo, that irreducible factor must be cubic.

$\endgroup$
  • $\begingroup$ Although I am perfectly okay with this, the primitive element theorem is a bit advanced for me (a few pages away). Is there perhaps a more "elementary" way to see this? $\endgroup$ – Cauchy Aug 16 '17 at 21:32
  • 3
    $\begingroup$ @Cauchy. We can also take advantage of the fact that $3$ is a prime. The argument goes as follows. Let $\alpha\in K\setminus \Bbb{Q}$ be arbitrary. Let $L=\Bbb{Q}(\alpha)$. Then $[L:\Bbb{Q}]\mid [K:\Bbb{Q}]=3$ by the field tower law. Clearly $[L:\Bbb{Q}]>1$, so the only alternative is $[L:\Bbb{Q}]=3$. Therefore $K=L=\Bbb{Q}(\alpha)$. The rest goes as in my answer. $\endgroup$ – Jyrki Lahtonen Aug 16 '17 at 21:36

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.