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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.

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    $\begingroup$ 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 $\alpha_{i+1}$) $\endgroup$ – reuns Mar 16 at 21:18
  • $\begingroup$ @reuns But it seems the argument shows that there are exactly $n$ embeddings... I don't quite understand the subtlety here. $\endgroup$ – No One Mar 17 at 15:34
  • $\begingroup$ 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 $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$) $\endgroup$ – reuns Mar 17 at 17:10
  • $\begingroup$ @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? $\endgroup$ – No One Mar 17 at 20:42
  • $\begingroup$ To see how it works what do you obtain for $\mathbb{Q}(\sqrt{2i+1}, i)/\mathbb{Q}(i)/\mathbb{Q}$ $\endgroup$ – reuns Mar 17 at 20:45

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