Cyclic Galois group of even order and the discriminant

I am stuck on the following problem:

Let K be a field of characteristic $\neq 2$ and $f\in K[X]$ a separable irreducible polynomial with roots $\alpha_1,\ldots \alpha_n$ in a splitting field $L$ of $f$ over $K$. The Galois group of $f$ is cyclic of even order. Show that the discriminant $\Delta=\prod_{i<j}(\alpha_i-\alpha_j)^2$ doesn't have a square root in $K$.

To show that the discriminant doesn't have a square root in $K$ is equivalent to showing that the Galois group (viewed as a subgroup of $S_n$) contains an odd permutation (in which case it doesn't fix $\prod_{i<j} (\alpha_i-\alpha_j)$). Also this can only happen if the generator of the Galois group is odd (since $\operatorname{sgn}$ is a group homomorphism). The only other thing I know is that the Galois group acts transitively on the roots since $f$ is irreducible. Hints are much appreciated.

You are nearly there.

Let $\sigma\in S_n$ be a generator of the Galois group $G$ (viewed as acting on the set of roots). Write $\sigma$ as a product of disjoint cycles.

1. Show that any cycle of $\sigma$ forms an orbit of $G$.
2. Show that $\sigma$ must be an $n$-cycle.
3. Apply what you know.
• Thank you for your answer, I think I was able to figure it out. 1. follows from considering the natural action of $\langle \sigma \rangle$ on $\{1,\ldots,n\}$, 2. from transitivity of the action and then finally for 3. we can use the formula for the sign of a cycle and the fact that $n$ is even.
– user228680
Jun 30 '16 at 11:45
• Correct. Well done, @ManfredHohler! Jun 30 '16 at 11:46
• Actually now I'm not sure why $n$ is even, do we in fact have $n=|Gal(f)|$ here?
– user228680
Jun 30 '16 at 12:08
• Yes. Because A) $\sigma$ is an $n$-cycle, hence has order $n$, and B) generates the Galois group. Jun 30 '16 at 12:12
• Ah yes, I was a little confused. Thank you!
– user228680
Jun 30 '16 at 12:18