# Is the intersection of all algebraically closed subextensions an algebraic closure?

Let E be an extension of k which is algebraically closed and let A be the intersection of all the subextensions of E which are algebraically closed. I guess that A is actually the algebraic closure of k but I'm not quite sure. Is it correct?

Edit: I should apologize for the misunderstanding caused by the inexact description of my question. What I actually want to ask is how we can prove A is an algebraic closure of k, if we haven't proved the existence of an algebraic closure of k? In particular, I have just seen a proof which shows that A, as the intersection of all algebraically closed subextensions of E, is an algebraically closed field. So I am hoping for a proof that A is algebraic over k, given that we haven't known anything about the algebraic closure. Thanks!

• I don't think the proof you're hoping for exists. The definition of $A$ really doesn't help to prove it is algebraic over $k$ unless you already know there exists an algebraically closed subextension consisting of only elements that are algebraic over $k$ (i.e., an algebraic closure of $k$ in $E$). Jan 28, 2022 at 4:14
• @EricWofsey Thanks. I tend to agree with you now, because all proofs I have found out use the union of all subextensions of an algebraically closed extension E, which are algebraic over k, so that one an use both property to prove this fact. So it is highly possible that your suggestion reaches the essence of this problem.
– zyy
Jan 29, 2022 at 5:45

The algebraic closure of $$k$$ inside $$E$$ (i.e. elements of $$E$$ which are algebraic over $$k$$), which is also an algebraic closure of $$k$$ (in the abstract sense) is an algebraically closed subextension of $$E$$ and is contained in every algebraically closed subextension of $$E$$, so the answer is yes.
• @zyy The set $F$ of elements of $E$ which are algebraic over $k$ is well-defined and is an algebraic closure of $k$. It exists once you assume that $E$ algebraically closed containing $k$ exists. Evidently $A=F$. Jan 28, 2022 at 11:20