Difficulty with basic Galois Theory

Question

The automorphisms in a Galois group are always described as a mapping from some value to another e.g. $i \to i \sqrt2$. My basic understanding is that these are permutations of roots of a polynomial, and this is what my question concerns.

I understand that for $L/K$ a Galois extension, $L$ has to be the splitting field of some $F \in K[X]$. Is this the polynomial which has its roots permuted?

Secondly, why do the permutations of the roots solely define the automorphisms of $L$ over $K$?

Answer 1

If $L/K$ is finite field extensions and $\text{Aut}(L/K)$ is group of all field automorphism of $L$ which fixes $F$ pointwise, then we say extension is galois whenever $[L:K]=|\text{Aut}(L/K)|$, generally we have $[L:K] \geq \text{Aut}(L/K)$.

$L/K$ is galois iff $L/K$ is splitting field for some separable $f(x)\in K[x]$.

Here the very simple example:

Consider $L=\mathbb Q(\sqrt{2})$ and $K=\mathbb Q$ Note that $L=\{ a+b\sqrt2 | a,b \in \mathbb Q \}$ and Corresponding polynomial for which $L/K$ is splitting field is $f(x)=x^2-2$.

Let, $\sigma \in \text{Aut}(L/K)$ then $\sigma(a+b\sqrt2)=a+b\sigma(\sqrt2)$ ,since $\sigma$ fixes $K=\mathbb Q$. So each $\sigma$ is only determined by where its maps the $\sqrt2$,which is one of roots of $f$.

Also note that, $\sigma(\sqrt2) \in L$ and $(\sigma(\sqrt2))^2-2=0$ ,whih means $\sigma(\sqrt2)$ mus be one of roots of $f(x)$. [This is crucial point]

So, $\sigma(\sqrt2) \in \{\sqrt2,-\sqrt2\}$

So,$\text{Auto}(L/K)=\{ \text{id}: (a+b\sqrt2) \mapsto (a+b\sqrt2), \sigma:(a+b\sqrt2) \mapsto (a-b\sqrt2) \} = \mathbb Z/2\mathbb Z$

Answer 2

By definition, the elements of the Galois group of $L/K$ are the field automorphisms of $L$ that fix $K$ pointwise. These maps are in particular also vector space automorphisms of $L$ seen as a finite dimensional vector space over $K$. Hence, to write down what a map does, it is enough to indicate the image of a basis of $L/K$. Depending on how the extension $L/K$ is defined, you can find a basis consisting of products of roots of some polynomial (for instance, that's certainly the case if $L$ is a splitting field. Do you see why?). But since the image of a product of roots $\alpha \cdot \beta$ is already determined by the image of $\alpha$ and the image of $\beta$ (because the elements of the Galois group are also ring automorphisms of $L$, hence compatible with multiplication), it is enough to write down the image of the roots of the polynomial.

Note that, regardless of how $L$ is defined, for each element $\alpha \in L$, and for each $\phi$ in the Galois group, $\phi(\alpha)$ is still a root of the minimal polynomial of $\alpha$ over $K$ (try to show this, it is a nice exercise). Hence $\phi$, being bijective, induces a permutation of the roots of the minimal polynomial of every element of $L$.