what are embeddings of $\Bbb Q(^4\sqrt 2)$ in $\overline {\Bbb Q}$ really fixing? I want to prove $\sqrt 3 \not \in \Bbb Q(\sqrt[4]2)$ using contradiction.
Suppose $\sqrt 3 \in \Bbb Q(\sqrt[4] 2)$ and set $K:=\Bbb Q(\sqrt 2,\sqrt 3)=\Bbb Q(\sqrt[4]2)$ because both fields have same degree over $\Bbb Q$.
let $σ_1 , . . . , σ_4$ be the distinct field embeddings $K \to \overline {\Bbb Q}$ that fix every element of $\Bbb Q$:
$σ_1(\sqrt[4] 2)=\sqrt[4] 2$
$σ_2(\sqrt[4]2)=-\sqrt[4] 2$
$σ_3(\sqrt[4] 2)=i\sqrt[4] 2$
$σ_4(\sqrt[4] 2)=-i\sqrt[4] 2$
Do we have $σ_1(\sqrt 3\sqrt[4]2) = \sqrt 3 σ_1(\sqrt[4] 2)$? In other words, does embedding fix $\sqrt 3$ within our hypothesis that $\sqrt 3 \in \Bbb Q(\sqrt[4] 2)$?
My understanding of embedding in Field Theory is that it is a ring morphism that permutates the conjugates of $\sqrt[4] 2$ and fixes every other element in the field.
Thank you for your help.
 A: For all $i=1,\ldots,4$ we see that $\sigma_i$ is an isomorphism $($onto a subfield of $\overline{\mathbb{Q}})$, hence must send $\sqrt{2},\sqrt{3}$ to one of their conjugates, namely, $\pm\sqrt{2}$ and $\pm\sqrt{3},$ respectively. Therefore
$$\sigma_i(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\sqrt{3})=a\pm b\sqrt{2}\pm c\sqrt{3}\pm d\sqrt{2}\sqrt{3}\in\mathbb{R}.$$
So $\sigma_i(\overline{\mathbb{Q}})\subseteq\mathbb{R}$ for all $i=1,\ldots,4.$ Then $i\sqrt[4]{2}\in\mathbb{R},$ a contradiction, therefore $\mathbb{Q}(\sqrt{2},\sqrt{3})\not=\mathbb{Q}(\sqrt[4]{2}).$
It would be hard to deduce what would happened to $\sqrt{3}$ exactly, if it were fixed or not, under these isomorphisms. To do that would require we write $\sqrt{3}$ as a linear combination of powers of $\sqrt[4]{2}$, which is impossible. So, to derive a contradiction we have to resort to other means, like what I just mentioned. 
"My understanding of embedding in Field Theory is that it is a ring morphism that permutates the conjugates of $\sqrt[4]{2}$ and fixes every other element in the field."
This is partly true, in that the embeddings do permute the conjugates of $\sqrt[4]{2},$ however, the other elements of the fields are linear combinations of powers of $\sqrt[4]{2},$ so we should expect that most are not fixed. For example, $$\sigma_2(1+\sqrt[4]{2})=1-\sqrt[4]{2}\not=1+\sqrt[4]{2}.$$
It is true that every element of the field is sent to one of its Galois Conjugates though, so while $1-\sqrt[4]{2}\not=1+\sqrt[4]{2},$ it is true that both are roots of the same irreducible polynomial over $\mathbb{Q}.$
