If $K/F$ is Galois, and the coefficients of the monic $f \in K[x]$ generates $K$, then $\prod_{\sigma \in Gal(K/F)}\sigma f \in F[x]$

Question

Suppose $K/F$ is a Galois extension, and suppose that $f \in K[x]$ is a monic polynomial whose coefficients generate $K$.

Let $\mathcal{F} = \{ \sigma f : \sigma \in Gal(K/F)\}$

Then let $g := \prod_{f' \in \mathcal{F}} f'$

How does one see that $g \in F[x]$? Furthermore, if $g$ is irreducible in $F[x]$ then $f$ is irreducible in $K[x]$.

I'm a bit confused on this one.

Each $f'$'s coefficients should also be able to generate $K$ of course. But then multiplying these up together, not sure how one ends up in $F[x]$. Would appreciate some help.

Answer 1

If you want go gain some intuition, consider $F=\mathbb Q$ and $K=\mathbb Q[\sqrt 2]$, $f=\sqrt 2 x + 1$, then $g=(\sqrt 2x + 1)(-\sqrt 2x + 1) = -2x^2+1\in\mathbb Q[x]$.

The proof is straightforward. Any $\sigma\in\text{Gal}(K/F)$ permutates $\mathcal F$ and hence leaves $g$ unchanged. And by $K/F$ is Galois, the only elements in $K$ that are invariant under $\text{Gal}(K/F)$ are elements of $F$, thus $g\in F[x]$.

Pick a root $\alpha$ of $f$, then $\alpha$ is also a root of $g$. If $g$ is irreducible over $F$, then $$[F(\alpha):F] = \deg(g) = \deg (f) \cdot |\text{Gal}(K/F)|=\deg(f)\cdot [K:F]$$

And $$[F(\alpha):F]\le [K(\alpha):F] = [K(\alpha):K][K:F]$$ $$[K(\alpha):K]\ge [F(\alpha):F]/[K:F]=\deg(f)$$

Since $\alpha$ is a root of $f$, we also have $[K(\alpha):K]\le \deg (f)$, hence $[K(\alpha):K]=\deg(f)$ and $f$ must be the minimal polynomial of $\alpha$ over $K$ hence irreducible.

Note that we don't even need the coefficients of $f$ generate $K$.

Answer 2

I think that one can use the following result in order to show your claim.

Let $K/F$ be a galois extension then $$ K^{\operatorname{Gal}(K/F)}= \{ x\in F : \sigma x =x ,\forall \sigma \in \operatorname{Gal}(K/F)\} = F \tag{$*$}$$ Thus, pick some $\sigma\in \operatorname{Gal}(K/F)$. We have $$ \sigma f' = \sigma \prod_{\tau \in \operatorname{Gal}(K/F)} \tau f = \prod_{\tau \in \operatorname{Gal}(K/F)} \tau f = f'$$ because the map $\operatorname{Gal}(K/F) \to \operatorname{Gal}(K/F), \tau \mapsto \sigma\tau$ induces a permutation of $\operatorname{Gal}(K/F)$ leaving the product unchanged. Hence applying any $\sigma \in \operatorname{Gal}(K/F)$ has left the coefficients of $f'$ fixed, thus by $(*)$ they must lie within $F$.