Consider the field morphism $\sigma: \mathbb{Q}(\sqrt2) \rightarrow \mathbb{Q}(\sqrt2), \sqrt2 \mapsto - \sqrt2$. I am trying to find all extensions $\overline{\sigma}: L \rightarrow L$ of $\sigma$, where $L=\mathbb{Q}(\sqrt2,\sqrt3)$.
(Note: I mean "extensions" as in extension of a function (field morphism), not "field extension". Those are defined as: Let $K(a)/K$ be a field extension and $\sigma: K \rightarrow L$ be a field morphism. A field morphism $\overline{\sigma}: K(a) \rightarrow L$ is called "an extension of $\sigma$ onto $L$" if $\overline{\sigma}_{|_K}=\sigma$ )
The first step is to calculate the minimal polynomial of $\sqrt3$, $m_{\alpha, \mathbb{Q}(\sqrt2)}$
I assume that $m_{\alpha, \mathbb{Q}(\sqrt2)}=x^2-3$, and because $[\mathbb{Q}(\sqrt2,\sqrt3) : \mathbb{Q}(\sqrt2)]=2$, I know that $m_{\alpha, \mathbb{Q}(\sqrt2)}$ has to have degree $2$.
The next thing is to calculate $\sigma(x^2-3)$ $\sigma(x^2-3)=\sigma(1)x^2-\sigma(3)=x^2-3$
I know that for every root r of $\sigma(x^2-3)=x^2-3$, there exists a extension $\overline{\sigma}: (\mathbb{Q}(\sqrt2))(\sqrt3 ) \rightarrow (\mathbb{Q}(\sqrt2))(\sqrt3 )$, such that $\overline{\sigma}(\alpha)=r$, in my case this means:
$\overline{\sigma_1}(\sqrt3)=\sqrt3$ and
$\overline{\sigma_2}(\sqrt3)=-\sqrt3$
Question Nr.1 Is the calculation of $\sigma(x^2-3)$ correct? Since there where no $\sqrt2$ or multiples of it in the coefficients, I get the results I got.
Question Nr.2 Are my calculations of the extensions of $\sigma$ correct? (If the awnser to Nr.1 is no, then if the approach for the extensions is correct.)
Extend $\sigma$ and define $\sigma(x)=x$. Then $\sigma(x^2-3)=x^2-3.$ and the calculation is correct.
Remark: Now, extend $\sigma$ and define $\sigma(x)=-x$ or $\sigma(x)=x+1$.
Then $\sigma(x^2-3)=\sigma(x)^2-3 \neq x^2-3$ for the latter case.
Only focus on the case extending $\sigma$ to $\hat{\sigma}(x)=x+1$.
Let $f(x)=x^2-3$, $\alpha = \sqrt 3$.
Then $\hat{\sigma}(f(x))=(x+1)^2-3$ and $\gamma := -1\pm \sqrt 3$ (pick any one) is a root of $\hat{\sigma}(f(x))$.
Here I use the same symbol $\hat{\sigma}$, to map $\alpha$ to $\gamma +1$
Note that $\hat{\sigma}(\alpha)=\gamma +1$ and it is a root of $f(x)$.
$\hat{\sigma}(\alpha)=\gamma +1=\pm\sqrt 3$ so the result is the same.
Remark: $\gamma +1$ is not a root of $\hat{\sigma}(f(x))$ and it is different for the case if we choose $\hat{\sigma}(x)=x$.
For the general case, the result is the same if $\hat{\sigma}$ is an isomorphism... and there is not a "must" to fix this mapping.