The statement of exercise 17 in section 14.2 of "Abstract Algebra" by Dummit & Foote says the following:
The statement assumes that there is a finite number of embeddings of $K$ into the algebraic closure of $F$, but I cannot see why this is true. There is a similar question on this site (here), but the answers didn't help me.
Let $K/F$ be a finite extension. Then $K$ is finitely generated by algebraic elements over $F$, say $K=F(\alpha_1,...,\alpha_n)$. Let $\varphi$ be an embedding of $F$ into its algebraic closure $F^a$. There is a finite number of extensions $\varphi_1:F(\alpha_1)\mapsto F^a$ of $\varphi$, because each extension must map $\alpha_1$ to a root of the minimal polynomial of $\varphi(\alpha_1)$ over $F'$, hence there are only finitely many options for $\varphi_1(\alpha)$ and, consequently, finitely many options for $\varphi_1$. Following this argument by induction, we see that there is a finite number of embeddings of $K$ into $F^a$ extending $\varphi$. The way I see it (and I might be mistaken), this is essentially the argument of the first answer to the question I mentioned earlier, but it does not show that the number of embeddings of $K$ into $F^a$ is finite, as there may be infinitely many embeddings $\varphi:F\rightarrow F^a$, and each one can be extended to an embedding of $K$ into the algebraic closure. The second answer seems to assume that the embedding fixes the elements of $F$ (in which case the result easily follows by the inductive argument above). My question then is as follows:
- Can we assume that, in the exercise statement, by "embeddings," the authors implicitly mean "embeddings fixing $F$"?
If this is the case, then everything becomes clearer to me. Otherwise, how can I justify that there is a finite number of embeddings of $K$ into the closure? And what is the mistake in my reasoning above?