Let $k$ be a field. There exists an extension $k^a$ which is algebraic over $k$ and algebraically closed.
Proof: Let $E$ be an extension of $k$ which is algebraically closed and let $$k^a:=\bigcup N$$ where the union taken over $N$ such that $k\subset N\subset E$, $N$ is algebraic over $k$. Note that this union is not empty because $k$ belongs to this family.
1) Let's show that $k^a$ is algebraic over $k$. Take $\alpha\in k^a$ then $\alpha\in N$ for some $N$ which is algebraic over $k$. Hence $\alpha$ is algebraic over $k$. Right?
2) Let's show that $k^a$ is algebraically closed. Take any $f\in k^a[X]$ with degree $\geq 1$ then we can consider it as polynomial over $E$, since $E$ is algebraically closed then $f$ has root $\beta\in E$. Hence $\beta$ is algebraic over $k^a$.
Question 1: How to show that $\beta\in k^a$?
Question 2: Also it's not obvious to me but why $k^a$ is a field?
Would be very grateful for detailed answer, please.
EDIT (This is the proof for the theorem which professor Lubin gave): We want to show that $k^a$ is algebraically closed. Take any monic polynomial $f(X)\in k^a[X]$, namely $f(X)=X^m+z_{m-1}X^{m-1}+\dots+z_1X+z_0$ then we can consider $f$ as an element of $E[X]$ but $E$ being algebraically closed means that $f(X)$ has root $\beta\in E$. Also $f(X)\in k(z_0,z_1,\dots,z_{m-1})$. Since $\beta$ is algebraic over $k(z_0,z_1,\dots,z_{m-1})$ then $[k(z_0,z_1,\dots,z_{m-1},\beta):k(z_0,z_1,\dots,z_{m-1})]<\infty$. Since each $z_i\in k^a$ then each $z_i$ is algebraic over $k$ and hence $[k(z_0,z_1,\dots,z_{m-1}):k]<\infty$.
Thus $[k(z_0,z_1,\dots,z_{m-1},\beta):k]<\infty$. Note that $$[k(z_0,z_1,\dots,z_{m-1},\beta):k]=[k(z_0,z_1,\dots,z_{m-1},\beta):k(\beta)][k(\beta):k].$$ It follows that $[k(\beta):k]<\infty$ and it follows that $\beta\in k^a$. We are done.
Best Answer
I would use a slightly different strategy, defining $k^a$ to be the union of all finite extensions of $k$ in the universe $E$. (And you know that finite implies algebraic.) Now, the union of two fields $F_1$ and $F_2$ is not ordinarily a field, but this set is contained the compositum $F_1\vee F_2$, which you may have seen defined as the smallest field containing both $F_1$ and $F_2$, or equivalently as $F_1(F_2)$. In any case, $F_1\vee F_2$ is obviously finite over $k$.
Under this definition, $k^a$ is a field, ’cause if you have elements $z,w\in k^a$, each is in a finite extension, and $z+w, zw\in k(z,w)$ (you can fill in the rest), and every element is algebraic over $k$, so the field itself is algebraic over $k$.
Algebraically closed? Take a monic polynomial $f(X)=X^m+z_{m-1}X^{m-1}+\cdots+z_1X+z_0\in k^a$, in fact $f\in k(z_0,\cdots,z_{m-1})[X]$, so that every root $\beta$ of $f$ has $[k(\beta):k]<\infty$, and thus $\beta\in k^a$.