Showing that this polynomial is in F[x] (Galois theory, given K is some extension of F)

Question

Suppose we have some extension $K$ of some field $F$ and the Galois group $G(K/F)$ which consists of automorphisms $\sigma$ of $K$. How could we show the following:

For any $a \in K$, show that $p(x) = \prod_{\sigma \in G(K/F)} (x - \sigma(a)) \in F[x]$

What I did was take an arbitrary automorphism $\sigma_j$ and acted on the polynomial with it. So:

$\sigma_j p(x) = \sigma_j \prod_{\sigma \in G(K/F)} (x - \sigma(a)) = \prod_{\sigma \in G(K/F)} \sigma_j (x - \sigma(a)) = \prod_{\sigma \in G(K/F)} (\sigma_j(x) - \sigma_j( \sigma(a))) = \prod_{\sigma \in G(K/F)} (x - \sigma(a))$

Because if I apply $\sigma_j$ to $\sigma(a)$ it will just give me another $\sigma$ in $G(K/F)$ acting on $a$, so I showed that $\sigma$ fixes the polynomial, thus it must be part of the underling field $F$ adjoined $x$. Does that make sense?

Answer 1

Your argument is fine.

Another way is to consider that the coefficients of $p$ are symmetric polynomials in $\sigma_1(a), \dots, \sigma_n(a)$ and so are fixed by any $\sigma$, since $\sigma \sigma_i$ runs through all $\sigma_j$, as you have noticed.

Answer 2

An important detail that should be noted is that there is a one-to-one correspondence between the distinct elements of the orbit of $ a $ and the cosets of its stabilizer in the Galois group. Since the cosets partition the group into sets of equal cardinality, this means that each conjugate of $ a $ appears an equal number of times in the product, so the polynomial is indeed fixed by the entire Galois group $ \textrm{Gal}(K/F) $. By the fundamental theorem of Galois theory, the subfield fixed by $ \textrm{Gal}(K/F) $ is $ F $, so we conclude that the polynomial is in $ F[X] $. In fact, it can be shown that the minimal polynomial of $ a $ over $ F $ is given by

$$ \prod_{\sigma \in \textrm{Gal}(K/F)} (x - \sigma(a))^{\frac{1}{[K:F(a)]}} $$

as we have that $ [K:F(a)] = |\textrm{Gal}(K/F(a))| = |\textrm{Stab}_{\textrm{Gal}(K/F)}(a)| $.