# If a field extension is separable, then $[E:F]=[E:F]_s$?

Lang's definition

Let $E/F$ be an algebraic field extension and $\bar{F}$ be an algebraic closure of $F$. Define $[E:F]_s$ as the cardinal of field monomorphisms $\sigma:E\rightarrow \bar{F}$ fixing $F$.

Let $E/F$ be a separable extension. I know that if the extension is finite, $[E:F]=[E:F]_s$. I'm curious about the case extension is infinite. Is it still $[E:F]=[E:F]_s$?

There is an example that a separable extension possessing uncountable degree, so it does not seem easy to prove this.

• Actually, no. In infinite case, there can be much more embeddings than the degree of extension. See this answer for example. – Ennar Oct 19 '16 at 21:07

No, it's not true. Consider the separable algebraic extension $K = \mathbb Q (\sqrt p : p \mbox{ prime})$ of $\mathbb Q$. You can show that $[K : \mathbb Q] = |\mathbb N|$, but the extensions of an embedding $\mathbb{Q} \to \overline{\mathbb{Q}}$ to $K$ are in bijection with infinite binary strings: they are all possible combinations of choosing either $\sqrt p \mapsto \sqrt p$ or $\sqrt p \mapsto -\sqrt p$ for each prime $p$, so $[K : \mathbb Q]_s = |\mathbb R|$.