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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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    Regularity is fpqc-local, so there won't be any counterexamples. Perhaps I haven't understood your question. Which $L$ to you take in your counterexample?2011-12-15
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    In the down-to-earth situation of algebraic-varieties-as-zero-sets-of-families-of-polynomials, a singular point is a point which solves a certain system of polynomial equations: extending scalars cannot make it disappear!2011-12-15
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    @MarianoSuárez-Alvarez: Sorry for the confusion. I changed the title to make more sense.2011-12-15
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    @MartinBrandenburg: In my example I claim that $L[x,y]/(y^2-x^3)$ is singular for all $L/k$. Therefore, no matter which base change you make, the origin will remain singular.2011-12-15
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    Yes I just wanted to add: In the affine case (which you always have locally) the Jacobi criterion tells you if the point is nonsingular, namely that the rank of the Jacobian is maximal. But the rank does not change after some field extension. For example, Neil's parabel becomes after base change the "same" parabel $L[x,y]/(y^2-x^3)$ and the point $(0,0)$ is still singular.2011-12-15
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    @MartinBrandenburg: Could you expand a bit on what you mean by regularity is fpqc-local, and therefore there won't be any counterexamples?2011-12-15
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    @Haki: I don't know what you understood in my comment: I was not saying there is any confusion: I was answering your question (Indeed, my point was exactly what Martin wrote)2011-12-15
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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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    Replace $y+\bar{x}^2$ with $(y+\bar{x})^2$ in the displayed formula. The problem is "nonsingular" means smooth for some people. This is regrettable.2011-12-16
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    Thanks for pointing out my typo.2011-12-16
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    @QiL: The adjective *nonsingular* is somewhat ambiguous. Personally, I wouldn't use it myself over a non-perfect field, but if I saw someone else using it, I would presume it meant *smooth*, not just *regular*. Am I in a minority? Regards,2011-12-16
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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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    No, it is false! See my answer.2011-12-16
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    @Andrea: Dear Andrea, Presumably Hannah means *smooth*, not merely regular. Regards,2011-12-16