# Algebraic closure of field extension

Easy question on field theory. I imagine this has already been asked before but I couldn't find it on the search. Feel free to redirect me in case.

Let's consider a field $$K$$ and one algebraic closure $$\overline{K}$$. Then for any algebraic extension $$L$$ of $$K$$, $$\overline{K}$$ is the algebraic closure of $$L$$.

I feel this is is true. By algebraically extending $$K$$ we "add" roots of polynomials with coefficients in $$K$$ and these roots must lie in $$\overline{K}$$ by definition. However I'm not sure if by allowing coefficients in the extension $$L$$ one could get roots outside of $$\overline{K}$$. As I said, this sounds unlikely to me but I can't prove it.

Thank you for any help.

You have to be a bit careful. If $$\overline{K}$$ is an algebraic closure of $$K$$, and $$L$$ is an algebraic extension of $$K$$, it is not necessarily true that $$L$$ is contained in $$\overline{K}$$. However, $$L$$ is isomorphic to a (not necessarily unique) subfield of $$\overline{K}$$. But if I understand your question correctly, you are assuming that $$L$$ is an algebraic extension of $$K$$ which is already contained in $$\overline{K}$$, and you then want to show that $$\overline{K}$$ is an algebraic closure of $$L$$.
To say that $$\overline{K}$$ is an algebraic closure of $$L$$ is to say that $$\overline{K}$$ is algebraic over $$L$$, and if $$f \in L[X]$$ is any polynomial, then all of its roots lie in $$\overline{K}$$. Since $$\overline{K}$$ is algebraic over $$K$$, it is definitely algebraic over $$L$$. Let's show the second claim, that all the roots of $$f$$ lie in $$\overline{K}$$.
Since $$L[X] \subset \overline{K}[X]$$, we can think of $$f$$ as a polynomial in $$\overline{K}[X]$$, and let $$E$$ be its splitting field over $$\overline{K}$$. That is, $$E$$ is an algebraic extension of $$\overline{K}$$, $$f$$ has all its roots in $$E$$, and $$E = \overline{K}(\alpha_1, ... , \alpha_n)$$, where $$\alpha_1, ... , \alpha_n$$ are all the roots of $$f$$ in $$E$$.
Now each field $$\overline{K}(\alpha_i)$$ is algebraic over $$\overline{K}$$. Since $$\overline{K}$$ is also algebraic over $$K$$, the transitivity property of algebraic extensions implies that $$\overline{K}(\alpha_i)$$ is an algebraic extension of $$K$$. Therefore $$\alpha_i$$ is algebraic over the field $$K$$. Since $$\overline{K}$$ is an algebraic closure of $$K$$, this forces us to have $$\alpha_i \in \overline{K}$$. This is what we wanted to show.