Why is a normal scheme of finite type over a field of characteristic zero which is geometrically connected automatically geometrically irreducible?

Question

I know the basic fact that a connected normal scheme of finite type over $k$ must be irreducible since the irreducible components cannot intersect.

If $X$ is a normal integral separated scheme of finite type over a field $k$ of characteristic 0 and such that $X(k)$ is non empty, then $X$ is geometrically connected. The geometrical reducedness is automatic since $k$ is perfect but I do not understand $X_{k'}$ should be normal (in order to be irreducible) for $k'/k$ a field extension and I have read geometrical connectedness should imply geometrical integrality in my case.

Answer 1

The trick is that one only has to check geometric normality over purely inseparable extensions - since you're in characteristic zero, the only purely inseparable extension is the trivial one, and as soon as you have normality you have geometric normality. See for instance Stacks 038O for a reference.

Intuitively, what's going on here is that a point on an integral scheme of finite type over a field which is normal but not geometrically normal must either be on multiple irreducible components after some finite separable extension, or acquire some nilpotents after some finite purely inseparable extension. The first case leads to non-normality, so it's out, and all you have to check is the second case.