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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.

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It depends on what parts of field theory you want to assume. If you take for granted that $K=\mathbf{Q}(\alpha)$ for some $\alpha$, say with minimal polynomial $f\in\mathbf{Q}[X]$, then you can prove directly that, if $\alpha=\alpha_1,\alpha_2,\ldots,\alpha_n$ are the $n$ distinct roots of $f$ in $\mathbf{C}$ ($n=\mathrm{deg}(f)=[K:\mathbf{Q}]$), then each root gives rise to an embedding $K\rightarrow\mathbf{C}$ by sending $\alpha$ to $\alpha_i$. –  Keenan Kidwell Aug 7 '12 at 21:15
    
@KeenanKidwell Now that I finally managed to finish writing my answer I realize that you had already said the same thing 20 minutes ago. I don't know why I sometimes see notifications of new comments and other times (like this) there are no notifications at all. –  Adrián Barquero Aug 7 '12 at 21:40

2 Answers 2

up vote 3 down vote accepted

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.

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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.

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