Number of field homomorphisms from an extension field of $\mathbb Q$ to $\mathbb C$ 
Take $\mathbb{Q}$ $\subset$ $K$ $\subset$ $\mathbb{C}$ with $[K:\mathbb{Q}]$ finite. How would you show that the number of field homomorphisms from $K$ to $\mathbb{C}$ is equal to $[K: \mathbb{Q}]$? 

I guess it is clear that any element of  $\mathbb{Q}$ would map to itself, so you would need to look at the generators of $K$ (i.e. write elements of $K$ as linear combinations of elements of $\mathbb{C}$ with coefficients in $\mathbb{Q}$). 
Can you give me some suggestions on where to start? Thank you.
 A: One way to prove this is by first using the primitive element theorem to show that $K = \mathbb{Q}(\alpha)$ for some algebraic number $\alpha$, whose minimum polynomial over $\mathbb{Q}$ has degree $n = [K:\mathbb{Q}]$.
Then you can in fact describe the $n$ monomorphisms $\phi_i : K \hookrightarrow \mathbb{C}$ for $i = 1, \dots, n$ by the effect they have on $\alpha$. 
In particular if the minimal polynomial of $\alpha$ over $\mathbb{Q}$, say $p_\alpha(x)$ factorices over $\mathbb{C}$ as
$$
p_\alpha(x) = (x - \alpha_1)\cdots (x - \alpha_n)
$$
with $\alpha_1 = \alpha$, then the monomorphisms $\phi_i$ are given by $\phi_i(\alpha) = \alpha_i$.
For example if you have the extension $K = \mathbb{Q}(\sqrt{5})$ then the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ is $p(x) = x^2 - 5 = (x- \sqrt{5})(x + \sqrt{5})$. Then there are only two monomorphisms given by $\phi_1(\sqrt{5}) = \sqrt{5}$ and $\phi_2(\sqrt{5}) = -\sqrt{5}$. Then since the elements of $K = \mathbb{Q}(\sqrt{5})$ are of the form $a + b\sqrt{5}$ they are given by
$$
\phi_1(a + b\sqrt{5}) = a + b\sqrt{5} \quad \quad \phi_2(a + b\sqrt{5}) = a - b\sqrt{5}
$$
Now, if you already now the description of the monomorphisms, the proof that indeed these are monomorphisms and that every monomorphism $K \hookrightarrow \mathbb{C}$ is of this form is not difficult and you should try to complete it on your own.
A: Maybe this is overkill, but this follows from the fact that the extension $K/\mathbb Q$ is separable. For example, you can look at Theorem 3.8 in this note. If you look at the proof, you will get an idea of what is involved in proving your statement.
