The number of possible extensions of an embedding of a field into a algebraically closed field. I've been thinking this more than a week. My problem is the second part of the
Propositon 2.7. in Algebra by Serge Lang.
In the proposition, k is a field and $\sigma:k \rightarrow L$ is an embedding of $k$ into an algebraically closed field $L$. $\alpha$ is algebraic over $k$ and $p(X) \in k[X]$ is the irreducible polynomial of $\alpha$ over $k$. $k^a$ is the algebraic closure of $k$.

Proposition 2.7.
  The number of possible extensions of $\sigma$ to $k(\alpha)$ is $\leq \deg p$, and is equal to the number of distinct roots of $p$ in $k^a$.

I understand that the number of possible extensions of $\sigma$ is equal to the number of distinct roots of $p^\sigma$ in $L$. But why is it equal to the number of distinct roots of $p$ in $k^a$?
 A: Expanding on my comments, here's an answer without Lang (which I don't have). I'll assume what you've already said, that the number of extensions of $\sigma \colon k \to L$ to $k(\alpha) \to L$ is the number of distinct roots of $p^\sigma$ in $L$ ($k$ any field, $\alpha$ algebraic over $k$, $L$ algebraically closed, all maps are embeddings of fields).
Proposition: An embedding $\sigma \colon k \to L$ into an algebraically closed field extends to an embedding $\mu \colon k^a \to L$ of the algebraic closure. $k^a/k$ is algebraic.
Proof: Standard Zorn. Let $S = \{\rho \colon K \to L : \rho|_k = \sigma,\ k \subset K \subset k^a, K/k\ {\rm algebraic}\} \neq \varnothing$. Partially order $S$ by "extension of". A non-empty totally ordered subset of $S$, $\{\rho_i \colon K_i \to L\}$, has an upper bound in $S$, as follows. Set $K = \cup_i K_i$, which is a field, and define $\rho \colon K \to L$ by $\rho|_{K_i} = \rho_i$, which is well-defined by the total order property, and $\rho \in S$. By Zorn's lemma, we have a maximal element $\mu \colon K \to L$ of $S$. If $K$ is not algebraically closed, we can find $\alpha$ algebraic over $K$, hence over $k$, not in $K$, so we can take $\alpha \in k^a - K$. But then the result you've said implies we can extend $\mu$ to $K(\alpha) \to L$ since there is at least one root of $p^\mu$, the minimal polynomial of $\alpha$ over $K$ viewed in $L$. Hence $K$ is algebraically closed, and from the minimality of the algebraic closure, $k^a = K$, and we've found our extension. $K=k^a$ was algebraic by construction, which is just a nice side note. $\Box$
Now the rest is easy. The number of distinct roots of $p^\sigma$ in $L$ is the number of distinct roots of $p$ in $k^a$ because $\mu$ gives a bijection between them.
A: Trying to address the several comments and, of course, the OP's question:
First, in prop. 2.7, the algebraic closure $\;k^a\;$ isn't even mentioned so I'm not sure what's the problem here.
Second, in prop. 2.3 was already mentioned that upon having an embedding of fields $\;\sigma: k\to L\;$ we can then consider $\;k\;$ as a subfield of $\;L\;$ so that we can drop the annoying $\;k^\sigma\;$ notation .
Third, the existence of an algebraic closure was already established in Corollary 2.6.
Fourth: with all the above, talking of a root $\;\beta\in k^\sigma\;$ of $\;p^\sigma(x)=Irr(\alpha,k,x)^\sigma\;$ or of a(nother) root $\;\beta\;$ of $\;p(x)\;$ in $\;k\;$ is just "the same" (up to the embedding, of course).
