This question shows that if every polynomial of degree 2 and odd degree has a root in $K[X]$, then $K$ is algebraically closed.
I wonder what examples can I find to see that the condition "odd degree" is necessary so that I cannot prove the statement:
Let $K$ be a field, if every polynomial of degree two is reducible then $K$ is algebraically closed.
This other question seems to give a procedure.
If I start with $\mathbb{F}_p$ then I know that adding a root of an irreducible of degree two over $\mathbb{F}_p$ will give me:
- $\mathbb{F}_{p^2}$ by Moore's theorem there is only one field of cardinality $p^2$.
- Any other root of an irreducible of degree 2 over $\mathbb{F}_p$ must be here .
The problem is that I do not get the roots of irreducible over $\mathbb{F}_{p^2}$ and therefore this process has to be iterated obtaining at the end the union $\cup_{i \ge 1} \mathbb{F}_{p^{2^i}}$. This does not need to be a fieldso that I shoud take the minimal field over it $\langle \cup_{i \ge 1} \mathbb{F}_{p^{2^i}} \rangle$.
In $\cup_{i \ge 1} \mathbb{F}_{p^{2^i}}$, it is clear that there cannot be a root of an irreducible of degree $3$. However, in principle I could find such a root in $\langle \cup_{i \ge 1} \mathbb{F}_{p^{2^i}} \rangle$. Why is it not the case?
Actually, $\bigcup_{i \ge 1} \mathbb{F}_{p^{2^i}}$ is a field. More generally, the union of any nested sequence of subfields $F_0\subseteq F_1\subseteq F_2\subseteq\dots$ is a field. This is because every field axiom involves only finitely many elements at a time, and any finite set of elements will be contained in $F_n$ for some $n$.