How many homomorphisms of fields $\mathbb{Q}(\sqrt[4]2)\to\mathbb R$ are there?
I think the answer is 2 because $$\mathbb{Q}(\sqrt[4]2) \simeq\frac{\mathbb Q[x]}{(x^4-2)}$$ and only two root of $x^4-2$ are real (that means that I think also that the number of homomorphisme to $\mathbb C$ is 4).
I'm wrong? I'm not sure to had well wrapped my head around this.
PS: also sorry for my bad English, I'm doing what I can.
You're right on both counts.
$ \dfrac{\mathbb Q[x]}{(x^4-2)} $ can be seen as an abstract field containing a root of $x^4-2$, a construction due to Kronecker.
This field can be written as $\mathbb Q[u]$ with $u^4=2$. It is abstract in the sense that this $u$ does not want to be any of the complex roots of $x^4-2$. They are all the same from the algebraic point of view.
$\mathbb{Q}(\sqrt[4]2)$ can be seen as a concrete realization of $\mathbb Q[u]$ in $\mathbb R$ for instance, in the sense that there is a homomorphism $\mathbb Q[u] \to \mathbb{Q}(\sqrt[4]2)$ that sends $u$ to $\sqrt[4]2$.
If $K$ is a field, then the number of homomorphism $\mathbb Q[u] \to K$ is the number of roots of $x^4-2$ in $K$.
So there are $2$ homomorphisms when $K=\mathbb R$ and $4$ when $K=\mathbb C$.
Note that the number of homomorphism $\mathbb Q[u] \to K$ is not necessarily the number of subfields of $K$ that are isomorphic to $\mathbb Q[u]$.
All this generalizes to $u$ satisfying a polynomial equation $p(x)=0$ with $p \in \mathbb Q[x]$ irreducible.