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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?

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    Ok I get it now. Thnx a lot.2011-12-15

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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!

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    Dear @MattE, my feeling is a lot of people use nonsingular for varieties over characteristic $0$ or algebraically closed fields. In any case, I agree that it is preferable to use smooth to avoid ambiguity.2011-12-16
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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.)

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    @Andrea: Dear Andrea, Presumably Hannah means *smooth*, not merely regular. Regards,2011-12-16