# Question about embeddings in algebraically closed fields

I was looking at the two following well known propositions:

$1$) Let $E|F$ be a finite separable extension of degree n, and let $\sigma$ be an embedding of $F$ in $C$, where $C$ is an algebraic closure of $E$. Then $\sigma$ extends to exactly $n$ embeddings of $E$ in $C$;

$2$) The extension $E|F$ is normal if and only if every $F$-monomorphism of $E$ into an algebraic closure $C$ is actually an $F$-automorphism of $E$.

Observing the proofs, it seems to me that actually instead of $C$ we can pick any algebraically closed field $L$ containing $F$ (we don't need to assume that it contains also $E$ in these propositions, right?) $\textbf{in both propositions}$.

Is it correct? (I'm not so sure for prop 2)). What I'm thinking of, is the typical situation in number theory where you have $\mathbb C$ instead of $C$. In general $\mathbb C$ is not the algebraic closure of $E$, but just an algebraically closed field. So why in many texts the authors put algebraic closure instead of simply algebraically closed field?

• You're right that picking any algebraically closed field containing $E$ will work. Note that any such field also contains $F$ because $F$ is an algebraic extension of $E$. One reason for using $\mathbb C$ and $\mathbb R$ is because they are complete topological fields. These embeddings can be viewed as the "infinite primes" of a number field. A reason for sticking to the algebraic closure is because a lot of the theory extends to arbitrary Dedekind domains, in particular to rings like $\mathbb F_p[t]$. – Mathmo123 Dec 16 '16 at 9:09
• Actually the result you've quoted is just a result about fields. So a reason for not using $\mathbb C$ is because your field may not be contained in $\mathbb C$. – Mathmo123 Dec 16 '16 at 9:14
• sorry in my notation $F\subseteq E$, so actually you meant "picking any algebraically closed field containing $F$ will work" right? Anyway I do not understand why such an $L$ contains $E$ too. Does any algebraically closed field containing $F$ contains also any algebraic extension of $F$? – Richard Dec 16 '16 at 19:47
• Moreover, I do not understand your second comment. Do you refer to the first or second result? – Richard Dec 16 '16 at 19:55
• For the first, let $L$ algebraically closed containing $F$ and $K$ an algebraic extension of $F$. Now if $a\in K\setminus F$, then there is some non constant polynomial $f\in F[X]\subseteq L[X]$ such that $f(a)=0$. Since $L$ is algebraically closed, $f$ splits in $L[X]$, which means $a\in L$ so $K\subseteq L$. Is this correct? However, still I do not understand the second comment.Thank you – Richard Dec 16 '16 at 20:10

Because algebraic closures are intrinsic in terms of the field, and any algebraically closed field usually means that there's some topological structure. Eg. with $\Bbb C$ vs $\overline{\Bbb Q}$ the latter is purely algebraic in terms of the polynomials over $\Bbb Q$, but if you want to prove a result about embeddings like this one you prefer into $\overline{F}$ rather than $L\supseteq\overline{F}$. Because often there are many choices for such an $L$, and you don't want to write a separate proof for all cases. I would not want to prove this result two different times one for eg. $\Bbb C$ and another for $\overline{\Bbb Q_2}$. And if I'm dealing with a field of positive characteristic, obviously something like $\Bbb C$ is not even available to me.
• Thank you, but I can't appreciate the difference, since we can prove that given any field $F$, there exists an algebraically closed containing $F$ and at the same time we can prove that an algebraic closure of $F$ exists. So maybe I do not know many practical examples to appreciate the difference. The most relevant I know, is about number fields and in this case I have always found $\mathbb C$, namely an algebraically closed field, not an algebraic closure. – Richard Dec 16 '16 at 20:22