Parity of the order of the Galois group of a polynomial basing on its discriminant

Question

Let $K$ be a field with $char(K)=0$ and $f\in K[t]$ an irreducible polynomial which Galois group $G_{K}(f)$ is cyclic. Show the discriminant $\Delta(f)$ of $f$ is a square of an element of $K$ if and only if $o(G_{K}(f))$ is odd.

I have been able to prove $o(G_{K}(f))$ odd $\Rightarrow$ $\Delta(f)$ is a square of an element of $K$. I do not know how to prove the other implication.

Answer 1

If $\Delta(f)$ is a square of an element of $K$, then the Galois group, regarded as a group of permutations on the roots, is a subgroup of the alternating group. Hence a generator $g$ of the group must be an even permutation. Now $g$ is a single cycle (as $f$ is irreducible, so that the Galois group is transitive on the roots), so $g$ has to be a cycle of odd length.

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The following are actually equivalent in characteristic different from $2$:

• $\Delta(f)$ is a square of an element of $K$, and
• the Galois group is a subgroup of the alternating group,

so you can use this for both implications.