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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! :)

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

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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 $K$.

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  • $\begingroup$ Ah yes! That's very nice and elegantly simple! Thanks a lot! :) $\endgroup$ – C. Brendel Jan 23 at 22:17
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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 $b\in A$.

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  • $\begingroup$ Ah! That's also a very nice way of solving the problem! :) Thanks! $\endgroup$ – C. Brendel Jan 23 at 22:19

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