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. 
 A: 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?
A: $\theta(f)\big(\tilde{\theta}(\alpha)\big)=\tilde{\theta}(f)\big(\tilde{\theta}(\alpha)\big)=\tilde{\theta}\big(f(\alpha)\big)=\dots$
