How do I axiomatize algebraically closed fields? The goal is to write a set of axioms that formalize the property of a field to be "algebraically closed". Of all the possible equivalent definitions, I thought I should start by "a field $F$ is algebraically closed iff every non constant polynomial has a root in $F$". 
I am given a signature $\Sigma$ consisting of a set of two constants, $0$ and $1$, and two functions $+$ and $*$
The problem I encountered is that I cannot list a set of axioms without listing all possible polynomials with coefficients in the field $F$, that is, start by 
$$(A1)\quad  \forall a_0 \exists a ( a + a_0 = 0)$$
Now I should list the polynomials of degree $2$ over $F$, so I need to add the axiom
$$(A2)\quad \forall a_0 \forall a_1 \exists a (a^2+a_1a + a_0 = 0)$$
and so on
$$(An)\quad \forall a_0 \dots \forall a_n \exists a (a^n + a_{n-1}a^{n-1}+\dots a a_1a + a_0 = 0)$$
so how should I proceed? Is there away to formalizing the "algebraically closed" property for a filed with only one axiom, or at least a finite set of axioms? Thanks in advance!
 A: First of all, this isn't a "problem": "axiomatize" doesn't mean you have to give a finite set of axioms.  It's totally fine to use an infinite set of axioms.
In fact, it is impossible to axiomatize algebraically closed fields using only finitely many axioms.  For instance, if you take $\mathbb{Q}$ and repeatedly adjoin roots to all polynomials of degree $\leq n$ to obtain a new field $K_n$, then you can show that for each $\alpha\in K_n$, the degree of the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has no prime factors greater than $n$.  It follows that $K_n$ is not algebraically closed (it has no roots of irreducible polynomials over $\mathbb{Q}$ of degree $p$ whenever $p>n$ is prime), but it satisfies the first $n$ of your axioms.
Now suppose there were a finite set $S$ of axioms that characterized algebraically closed fields.  Each of those axioms would follow from your set of axioms, and by compactness each one would follow from only finitely many of your axioms.  So only finitely many of your axioms would imply every axiom in $S$, and thus imply the field is algebraically closed.  But cannot be the case, since for any finite subset of your axioms, the field $K_n$ for $n$ sufficiently large satisfies all of them but is not algebraically closed.
