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.
 A: 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.
