Galois automorphisms as polynomials If $K/\mathbb{Q}$ is the splitting field of an irreducible $P(x) \in \mathbb{Q}[x]$, can the elements of $\mathrm{Gal}(K/\mathbb{Q})$ be realized as polynomials in $\mathbb{Q}[x]$ when only considering the roots of $F$? In other words, is there a polynomial $f_\sigma(x) \in \mathbb{Q}[x]$ for each $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ such that $\sigma(\alpha) = f_\sigma(\alpha)$ for all of the roots $\alpha$ of $P$?
Even this is not clear to me: if $\alpha$ is a root of $P$ and $f(\alpha) = \sigma(\alpha)$ for some $f \in \mathbb{Q}[x]$ and $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$, then does $f(f(\alpha)) = \sigma^2(\alpha)$?
 A: This is not quite true. The problem is that the property you describe would mean that $\sigma(\alpha) = f_\sigma(\alpha)$ is in $\mathbb Q(\alpha)$ for all $\sigma$. Since Galois acts transitively on the roots, that means that all of the conjugates of $\alpha$ are in $\mathbb Q(\alpha)$, which is not generally the case.
On the other hand, when you have such $f$, then properties like the one you wrote are true. Suppose $\sigma(\alpha) = f(\alpha)$ where $f$ has rational coefficients. Then:
$$\sigma^2(\alpha) = \sigma(\sigma(\alpha)) = \sigma(f(\alpha)) = f(\sigma(\alpha)) = f(f(\alpha)) = f^2(\alpha)$$
That the coefficients are rational is important, since that is how we turn $\sigma(f(\alpha))$ into $f(\sigma(\alpha))$.
Now, even in the special case where that $\mathbb Q(\alpha)$ contains all of the conjugates of $\alpha$, then you cannot find the $f_\sigma$ you are looking for. However, we can get a little close. If all the roots are in there, then $\sigma(\alpha)$ must be in $\mathbb Q(\alpha)$ and so we can write it in terms of our favorite $\mathbb Q$-basis for this extension: the powers of $\alpha$. This gives you
$$\sigma(\alpha) = a_n\alpha^n + a_{n-1}\alpha^{n-1} + ... + a_1\alpha + a_0$$
and the right hand side is evidently a polynomial $f(\alpha)$ with coefficients $a_0,...,a_n$.
The problem is that this $f$ depends on $\alpha$ and will not in general have the property that $f(\alpha') = \sigma(\alpha')$ for other conjugates of $\alpha$. To help see this, write $\alpha' = \tau(\alpha)$ and then we have
$$f(\alpha') = f(\tau(\alpha)) = \tau(f(\alpha)) = \tau \sigma(\alpha) = \tau \sigma \tau^{-1} (\alpha')$$
We wanted this to equal $\sigma(\alpha')$ but what we got differs from that by conjugation by $\tau$, so in general it won't equal $\sigma(\alpha')$, although in some special cases, like when $\sigma$ and $\tau$ commute it always will.
