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?


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.

  • $\begingroup$ 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

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