# Number of embeddings of a (nonsimple) extension of a field to another field

Let $$L\supset F$$ be a finite field extension of degree $$n$$ and $$K \supset F$$ be any extension. I wonder how to prove that the number of embeddings from $$L$$ to $$K$$ that restricts to identity on $$F$$ is bounded by $$n$$.

If $$F=\mathbb Q$$ and $$K=\mathbb C$$, then by primitive element theorem, we can show that the number of embedding is $$n$$. But I don't know how to deal with this general situation.

• Wlog take $K = \overline{F}$. Then $L/F$ is a tower of simple extensions $L_{i+1}/L_i$ where $L_{i+1} = L_i(\alpha_{i+1})$ and $\prod_i [L_{i+1}:L_i] = [L:F]$, at each step there are $[L_{i+1}:L_i]_s$ choices for where to send $\alpha_{i+1}$ (the separable degree, the number of $L_i$ conjugates of $\alpha_{i+1}$) – reuns Mar 16 at 21:18
• @reuns But it seems the argument shows that there are exactly $n$ embeddings... I don't quite understand the subtlety here. – No One Mar 17 at 15:34
• There are exactly $[L:F]$ embeddings $L \to \overline{F}$ leaving $F$ fixed iff $L/F$ is separable. Otherwise one of the $[L_{i+1}:L_i]_s$ is smaller than $[L_{i+1}:L_i]$ (inseparable arises in characteristic $p$ with things like $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$) – reuns Mar 17 at 17:10
• @reuns Thanks! How do we know that we have counted all the possible embeddings that fix $F$? An embedding restricted to $L_{i+1}$ does not necessarily fix $L_i$, right? – No One Mar 17 at 20:42
• To see how it works what do you obtain for $\mathbb{Q}(\sqrt{2i+1}, i)/\mathbb{Q}(i)/\mathbb{Q}$ – reuns Mar 17 at 20:45