# Number of embeddings $\tilde{\theta} : K(\alpha) \to L$ extending $\theta : K \to L$

Proposition

Let $K(\alpha)/K$ be a finite simple extension, with $f \in K[X]$ the minimal polynomial for $\alpha$. Given a field extension $\theta : K \to L$, the number of embeddings $\tilde{\theta} : K(\alpha) \to L$ extending $\theta$ is precisely the number of distinct roots of $\theta(f)$ in $L$

The proof I have of this proceeds by stating that "An embedding $K(\alpha) \to L$ extending $\theta$ must send $\alpha$ to a zero of $\theta(f)$"

I just can't see why this is true. Any help would be greatly appreciated.

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If $L$ is a $K$-algebra, the morphisms of $K$-algebras $f: K[X]/(f(X))\to L$ are in bijective correspondence with the elements $l\in L$ such that $f(l) =0$.
To be precise, the correspondence is the following bijection of sets (see d) below) :
$$Hom_{K-Alg} (K[X]/(f(X)),L)\stackrel {\simeq}{\to}Zero_L(f(X)):\overline {P(X)}\mapsto P(l) \quad (\ast)$$

Notice carefully that:

a) The above is true if $K$ is any ring, not necessarily a field.

b) The statement above does not require $L$ to be a field either, but only that $L$ be a $K$-algebra.
Recall that a polynomial $P(X)\in K[X]$ can be evaluated at any element $l$ of any $K$-algebra $L$.

c) The set of algebra morphisms $Hom_{K-Alg} (A,L)$ between two $K$- algebras has no algebraic structure and is often empty.
This fact seems not to be as well-known as it deserves.
In particular, the bijection above $(\ast)$ may very well be the one between the empty set and itself, even if $K$ and $L$ are fields. Can you provide an example?

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$\theta(f)\big(\tilde{\theta}(\alpha)\big)=\tilde{\theta}(f)\big(\tilde{\theta}(\alpha)\big)=\tilde{\theta}\big(f(\alpha)\big)=\dots$

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Why is the second equality true? –  Anon Dec 3 '11 at 20:34
Okay, I can see why it's true on a computational level (i.e. by just looking at such an $f$ and evaluating both sides). But how would one arrive at this fact? I'm expecting there to be a more algebraic reason for it (with some kind of commutative diagram) –  Anon Dec 3 '11 at 20:40
@Anon: You could draw a commutative diagram involving $K,K(\alpha)$, and $L$, but it seems quite unnecessary. Unless I’m missing something, this just uses the definition of an embedding in a straightforward way. I’m no algebraist, but I find it hard to see how one wouldn’t arrive at it, to be honest. –  Brian M. Scott Dec 3 '11 at 21:03
Anon, the point is that $\tilde{\theta}$ preserves addition and multiplication. So if $f=\sum a_ix^i$, then $\tilde{\theta}(f)(\tilde{\theta}(\alpha))=\sum \tilde{\theta}(a_i)\tilde{\theta}(\alpha)^i=\sum\tilde{\theta}(a_i\alpha^i)= \tilde{\theta} (\sum a_i\alpha^i)= \tilde{\theta}(f(\alpha))$. –  Kevin Dec 3 '11 at 21:06