# Constructive proof of the existence of an algebraic closure

It is well-known that, assuming the axiom of the choice (in the form of Zorn’s lemma), one can prove that any field $F$ has an algebraic closure. One proof roughly goes as follows: consider the partially-ordered set of algebraic extensions $K/F$, ordered by inclusion; show that it satisfies the hypotheses of Zorn’s lemma; then Zorn’s lemma implies the existence of a maximal element; show that this maximal element is algebraically closed.

Now, in my Algebraic Number theory class, the professor gave a “constructive” proof of this fact: let $S$ be the set consisting of all non constant polynomials in $F[X]$, and construct the ring $R=F[X_f:f\in S]$ (that is, a polynomial ring with one variable for each element of $S$); let $A_0$ be the ideal of $R$ generated by polynomial $\{f(X_f):f\in S\}$; show that it is a proper ideal, and so it is contained in a maximal ideal $A$; the quotient $R/A$ is thus a field, which contains $F$ as a subfield; iterate, and take the union of all these fields; then, this union is the algebraic closure of $F$. He also mentionned (if I recall correctly) that one can prove that after only one iteration, you get the algebraic closure. Now, I put quotation marks around “constructive”, because I suspect that one still needs the axiom of choice at some point to prove that, indeed, you have the algebraic closure (although I haven’t checked the details).

So, here’s my question:

Is it the case that this proof is constructive?

Another question:

Can one prove the existence of an algebraic closure within ZF (without the axiom of choice)?

“…is a proper ideal, and so is contained in a maximal ideal of $A$.”