can singular points become nonsingular after a base change

Question

Let $X$ be a normal surface over a field $k$. Assume that $X$ is singular.

Does there exist a field extension $L/k$ (finite or infinite) such that $X_L$ is nonsingular?

The answer is no in general. Here's an example: $k[x,y]/(y^2-x^3)$. (To get a surface consider the example $k[x,y,z]/(y^2-x^3)$.)

But still, does there exist an $X$ such that the answer is yes?

Answer 1

The converse is also true. That is, let $X$ be a nonsingular variety over $k$. Then $X_L$ is still nonsingular. In conclusion, a variety $X$ is nonsingular if and only if $X_L$ is nonsingular. (This works for all $k$, perfect or imperfect.)

Answer 2

This is a counterexample to Hannah's answer.

Consider a field $k$ of characteristic $2$ and an element $\alpha \in k$ such that the square root of $\alpha$ is not in $k$. (For example you can pick $k=\mathbb{F}_2(t)$ and $\alpha = t$.) The polynomial $x^2 - \alpha$ is irreducible over $k$, hence the ring $L = k[x] / (x^2 - \alpha)$ is a field. The ring $$ L \otimes_k L = L[y] / (y^2 - \alpha) = L[y] / (y^2 - \bar{x}^2) = L[y] / (y + \bar{x})^2 $$ is local, but is non reduced. Therefore the variety $\mathrm{Spec} L$ is regular, but $(\mathrm{Spec} L)_L$ is not regular.

We should distinguish between regular and smooth!