# $K \subset L$ a field extension and $L$ algebraically closed implies that $A$ is the algebraic closure of $K$ [duplicate]

Let $$K \subset L$$ be a field extension and $$A:=\{l \in L\,:\,a\text{ algebraic over }K\}$$. I already have shown that $$A$$ again is a field and $$K \subset A$$ is algebraic. Also: If $$l \in L$$ is algebraic over $$A$$ it's also algebraic over $$K$$.

I now have to prove the following statement:

Suppose $$L$$ is algebraically closed. Show that $$A$$ is algebraically closed and that $$K \subset A$$ is algebraic.

As mentioned the latter I already have proven. Any hints how to proceed? I thought about considering a polynomial $$a(x) \in A[x] \setminus A$$ which decomposes into linear factors of $$L[x]$$ since $$L$$ is closed. From there on it would've been quite nice to show that these linear factors indeed are contained in $$A[x]$$ but I unfortunately do not know how to proceed.

Thanks for checking in! :)

## marked as duplicate by Dietrich Burde, Paul Frost, Cesareo, darij grinberg, clathratusJan 24 at 3:29

Suppose $$L$$ is algebraically closed. Then if $$f(x) \in A[x]$$ is a monic polynomial, we may view it as a polynomial $$f(x) \in L[x]$$. Since $$L$$ is algebraically closed, this has a root $$\alpha \in L$$. We want to show that actually $$\alpha \in A \subset L$$. But since $$\alpha$$ is the root of a polynomial with coefficients in $$A$$, we immediately have that $$\alpha$$ is algebraic over $$A$$, and you said you've already shown that this implies that $$\alpha$$ is algebraic over $$K$$. But then by definition of $$A$$, we have that $$\alpha \in A$$. Thus $$A$$ is algebraically closed.
The reason that $$K \subset A$$ is algebraic is simply by definition: any element $$\alpha \in A$$ is by definition algebraic over $$K$$.
Let $$f(x)=a_0+a_1x+\dots+a_nx^n\in A[x]$$ have positive degree. Then $$f(x)$$ has a root $$b$$ in $$L$$.
Now note that $$b$$ has finite degree over $$K[a_0,a_1,\dots,a_n]$$, which in turn has finite degree over $$K$$. Hence $$K[b]$$ has finite degree over $$K$$ and so $$b\in A$$.