Find all distinct isomorphisms $\sigma$ of $\mathbb Q(\sqrt{1+\sqrt{2}})$ in a subfield of $\mathbb {\bar Q}$ such that $\sigma(a) = a$ for all $a \in \mathbb Q$.
I was only able to find the irreducible polynomial (I don't even know if it's necessary) very lost here.
We know that $\alpha =\sqrt{1+\sqrt{2}} \to \alpha^2 = 1 + \sqrt{2}$.
Then.
$(\alpha^2 -1)^2 = 2$.
$\alpha^4 – 2\alpha^2 – 1 \to x^4 – 2x^2 – 1$.
I also know that.
A field $F$ is algebraically closed if every nonconstant polynomial in $F[x]$ has a zero in $F$.
But I'm confused about the isomorphism part.
Thanks for any help.
Let's say you're right about $x^4-2x^2-1$ being irreducible. (If it were reducible, we would get fewer isomorphisms, because there'd be, in this case only $2$ (we can see there's no rational root), roots to permute.)
Let the zeros be $\alpha,\beta, \gamma ,\delta $.
There's an isomorphism between $\Bbb Q(\alpha)$ and $\Bbb Q(\beta)$ sending $\alpha \to\beta$.
And there are two more. Plus the identity, making $4$.