I was trying to solve this:
Let $E$ be a extension field of $K$. If $K$ is a algebraically closed field, then the algebraic closure $A$ of $K$ on $E$ is a algebraically closed field.
But for this I need:
Every element in $E$ which is algebraic element on $A$ belongs to $A$
How can I show the second affirmation?
Thanks!
Clearly, the extension $\;E/K\;$ is either trivial or transcendental (why?) . Now, supose $\;\ell\in E\;$ is algebraic over $\;A\;,\;\;K\le A\le E\;$ , which means there exists non zero $\;p(x)\in A[x]\;\;s.t.\;\;p(\ell)=0\;$ .
But if $\;p(x)=a_0+a_1x+\ldots+a_nx^n\;$ , then $\;a_i\in A\;\;\forall\;i\;$ , and this means that $\;a_i\;$ is algebraic over $\;K\;$ (why?) , so that $\;K(a_0,...,a_n)/K\;$ is a finite extension, and thus also $\;K(a_0,...,a_n,\ell)=K(a_0,...,a_n)(\ell)\;$ is finite over $\;K\;$ , so $\;\ell\;$ is algebraic over $\;K\;$ and thus $\;\ell\in A\;$