Every automorphism of L:K fixes K? I know that if an monomorphism $L:K$ is an algebraic extension and $\tau: L \rightarrow L$ has the property that fixes $K$, then $\tau$ is 1-1.
Ok then, in the proof of this result we can see that $\tau$ even permutes the roots of $m_\alpha$ ($\alpha$ is an algebraic element).
The question that arises for me is that: If $L:K$ is an Galois is an algebraic extension, what happens with the set $Aut(L)$ ? Every automorphism of $L$ fixes $K$?
Edit
I was conjecturing this because I need to show this:
Let $f(x) \in \mathbb{Q}[X]$, $L = Gal(f,\mathbb{Q})$ - the Galois extension - if $\alpha$ is a root of $f(x)$ and $\tau \in Aut(L)$ then $\tau(x)$ is a root of $f(x)$.
I tried to show this in the following way:
We can write $$f(x) = \sum_{i = 0}^{n} a_i X^i$$ since $\alpha$ is a root, then applying $\tau$ we have
$$\tau( f(\alpha) ) = \tau(0) = 0 = \sum_{i = 0}^{n} \tau(a_i)[\tau(\alpha)]^i$$
that is my problem, because for some $i$, $\tau(a_i)$ may not be an element of $\mathbb{Q}$ ? In this case, $$\tau(f(\alpha)) \neq f(\tau(\alpha)) ?$$

Not necessarily. For example, $\mathrm{Aut}(\mathbb{C})$ is uncountable, while precisely two of these automorphisms (the identity map and complex conjugation) fix $\mathbb{R}$.
