5
$\begingroup$

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?

$\endgroup$
6
  • 2
    $\begingroup$ 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 $\mathbb F_p[t]$. $\endgroup$ – Mathmo123 Dec 16 '16 at 9:09
  • $\begingroup$ 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 $\mathbb C$. $\endgroup$ – Mathmo123 Dec 16 '16 at 9:14
  • $\begingroup$ 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 $F$? $\endgroup$ – Richard Dec 16 '16 at 19:47
  • $\begingroup$ Moreover, I do not understand your second comment. Do you refer to the first or second result? $\endgroup$ – Richard Dec 16 '16 at 19:55
  • $\begingroup$ 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 $\endgroup$ – Richard Dec 16 '16 at 20:10
1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ – Richard Dec 16 '16 at 20:22
  • $\begingroup$ @Richard because the embeddings go into the algebraic closure. A minimal algebraically closed field is useful because anything you prove about things into it extend automatically to arbitrary algebraically closed fields. Proving weaker theorems when the premises allow for strong conclusions is just a waste of effort, it's being lazy without any real purpose, you don't save any effort, you just end up with a weak theorem. It's like using a smartphone to just make phone calls, you spent a lot of money on something you don't even use to it's full potential. $\endgroup$ – Adam Hughes Dec 16 '16 at 20:40
  • $\begingroup$ It seems the opposite to me, if you prove something for any algebraically closed field, you prove the theorem also for an algebraic closure, since it is a particular kind of algebraically closed field $\endgroup$ – Richard Dec 16 '16 at 22:30
  • $\begingroup$ If I prove there is an embedding or some other fact for some algebraically closed field, it does not mean there is true for every algebraically closed field unless the "some" field is the algebraic closure. I can see why you are thinking the way you are, but because of the up-to-isomorphism-uniqueness feature of the algebraic closure, this particular theorem is basically either or proves the other. If you're just asking if both are true if you replace with any algebraically closed field, yes, but more generally useful theorems are better stated with the closure in general. $\endgroup$ – Adam Hughes Dec 16 '16 at 22:35
  • $\begingroup$ I can trust you that in general useful theorems are better stated with the closure, but it looks very weird. I mean, if you use in your proof that the algebraic closure is an algebraic extension we are ok, but if, like in these cases, you use just the fact that polynomials split, then I can't see why stating these theorems with the closure. It's just a restriction since you are using only one hypothesis regarding closure (namely that they are algebraically closed, but not using the algebraic extension issue). $\endgroup$ – Richard Dec 16 '16 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.