Can it happen that an irreducible degree $2$ polynomial becomes a perfect square in the algebraic closure?

Question

Question. Is it possible to find a field $K$ together with a degree-$2$ polynomial $P \in K[x]$ such that $P$ is irreducible, but $P = Q^2$ for some $Q \in \overline{K}[x]$?

This can't occur if $K = \mathbb{R}$ due to the rational root theorem. I thought possibly that $x^2+x+1 \in \mathbb{F}_2[x]$ would be an example, since if $\alpha$ is a root, then so too is $-\alpha$. However, when I expand out $(x-\alpha)^2$ I don't get $x^2+x+1$. Factorizing it carefully, we find that $$x^2+x+1 = (x-\alpha)(x-(\alpha+1)),$$ so this is not an example.

Ideas, anyone?

Answer 1

The standard example is $X^2 - T$ with $K = \mathbb{F}_2(T)$.

It factors as $(X - \sqrt{T})^2$ in an extension containing $\sqrt{T}$.

In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$ (because they have characteristic $0$) or finite fields.

One of the simplest examples of a field which is not perfect is the function field in one variable $\mathbb{F}_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.

Answer 2

This can happen in characteristic two. Let $F$ be a field of characteristic two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure it equals $(t-Y)^2$ where $Y$ is a square root of $X$.

This phenomenon is called inseparability.