I learned the following theorem
Let char $K \neq 2$ and let $f=a \prod_{k=1}^n (x-\alpha_k) \in K[x]$ be a non constant seperable polynomial (where $\alpha_k$ are the roots of $f$). Let further $\psi : Gal(f,K) \rightarrow S_{\alpha_1,...,\alpha_n}$ , $\sigma \mapsto \sigma_{\alpha_1,...\alpha_n}$ be the canonical embedding. Then $\psi(Gal(f,K))$ is a subgroup of $A_n$ iff the discriminant $Disc(f)$ is a square in $K$.
I wanted to try an example: Let $f(x)=x^3-2x+2 \in \mathbb{Q}[x]$. Eisenstein with $p=2$ yields that $f$ is irreducible over $\mathbb{Q}$
Since $deg(f)=3$ $\Rightarrow$ $Gal(f,\mathbb{Q})$ is a subgroup of $S_3$. I did calculate that $Disc(x^3+px+q)=-4p^3-27q^2$. Next I calculated $Disc(x^3-2x+2)=-76$ Now since $Disc(f)$ is not a square in $\mathbb{Q}$, this yields that $Gal(f,\mathbb{Q})$ is not a subgroup of $A_3$. This means that $Gal(f,\mathbb{Q}) \cong S_3$
Next thing I wanted is to have a look at the othe possible case: For that I am trying to calculate $Gal(f,\mathbb{Q}(i\sqrt19))$. One can see that $\sqrt{-76}=i\sqrt{76}=i\sqrt{2^2 19}=2i\sqrt{19}$.
In this case the Discriminant of $f$ is a square in $\mathbb{Q}(i\sqrt{19})$. so I get that $Gal(f,\mathbb{Q}(i\sqrt{19}) \leq A_3$. But how do I continue?
(Also another question is if my attempt was correct)
Hint If a polynomial $f$ of degree $n$ is irreducible, the Galois group acts transitively on the set of roots of $f$, but the only transitive subgroups of $S_3$ are $A_3$ and $S_3$ itself.
In particular, if $f$ is irreducible, then $\operatorname{Gal}(f, \Bbb Q(i \sqrt{19})) = A_3$. If $f$ isn't irreducible, the galois group is $1$ or $C_2$ according to whether $f$ factors into $3$ or $2$ factors irreducible over $\Bbb Q(i \sqrt{19})$, respectively.