Automorphisms (in the context of Galois Theory) Can someone give an explanation of what an automorphism is, in the context of Galois Theory? I keep thinking it is the set of maps which send roots of polynomials to their conjugates but I feel that this not the proper explanation of an automorphism. If someone could provide an explanation, and maybe a simple example it would be greatly appreciated.
 A: Let $L/K$ be an extension of fields (that is $K$ is a field included in $L$) an automorphism $\phi$ of $L/K$ is by definition, a ring automorphism of $L$ (that is $\phi(l_1+l_2)=\phi(l_1)+\phi(l_2)$, $\phi(l_1.l_2)=\phi(l_1).\phi(l_2)$, $\phi(0)=0$ and $\phi(1)=1$) with the condition that for any $k\in K$, $\phi(k)=k$. That is, an automorphism of this extension $L/K$ is an automorphism of $L$ which fixes $K$ pointwise, in particular one sees that this is a $K$-linear isomorphism of $L$ which respects the multiplication. It is clearly a group usually denoted $Gal(L/K)$. 
The link between roots of polynomials and $Gal(L/K)$ goes as follows. If $P\in K[X]$ then write :
$$P(X)=\sum_{n=0}^dk_nX^n $$
Then for any $l\in L$ :
$$\phi(P(l))=\sum_{n=0}^d\phi(k_n)\phi(l)^n=\sum_{n=0}^dk_n\phi(l)^n=P(\phi(l)) $$
That is for any $P\in K[X]$ the set of roots of $P$ in $L$ is globally fixed by $Gal(L/K)$. In particular, if $L$ is the decomposition field of some square-free polynomial $P$ (whose roots are noted $Roots(P):=\{x_1...x_d\}$) then this means that $L=K[x_1,...,x_d]$. And then the action of $Gal(L/K)$ on $Roots(P)$ is faithfull. That is how you identify (in that case wich is actually quite general) the automorphism group of $L/K$ with a permutation group of $dim_K(L)$ elements. The induced permutation group is transitive if and only if $P$ is irreducible.
Now for some examples. 

$L:=\mathbb{Q}[\sqrt{2}]$ and $K:=\mathbb{Q}$. 

Then I claim that $L$ is the decomposition field of $X^2-2$ over $K$
 so the Galois group of $L/K$ is a subgroup of $\mathfrak{S}_2$ hence
 it has $1$ or $2$ elements. The first automorphism is $Id_L$, the
 second sends $a+b\sqrt{2}$ to $a-b\sqrt{2}$. This gives two explicit elements of $Gal(L/K)$.

$L:=\mathbb{Q}[^3\sqrt{2}]$ and $K:=\mathbb{Q}$. 

We know that $X^3-2$ is a minimal polynomial for $^3\sqrt{2}$, furthermore if $j$ is a primitive $3$-root of unity then the three roots of $X^3-2$ will be :
$$^3\sqrt{2}\text{, } ^3\sqrt{2}j\text{ and } ^3\sqrt{2}j^2$$
Now if $\phi$ is a non-trivial automorphism of $L/K$ then it should move $^3\sqrt{2}$ and it should move it to $^3\sqrt{2}j^k$ for $k=1,2$ because it still has to be a root of $X^3-2$. But this is impossible because $L\subseteq \mathbb{R}$ and $j,j^2\notin \mathbb{R}$ hence you have no non-trivial automorphism of $L/K$ here.
