Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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? – Martin Brandenburg Dec 15 '11 at 21:21
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! – Mariano Suárez-Alvarez Dec 15 '11 at 21:27
@MarianoSuárez-Alvarez: Sorry for the confusion. I changed the title to make more sense. – Haki Dec 15 '11 at 21:33
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. – Martin Brandenburg Dec 15 '11 at 21:34
@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) – Mariano Suárez-Alvarez Dec 15 '11 at 21:45

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!

share|cite|improve this answer
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. – user18119 Dec 16 '11 at 12:24
Thanks for pointing out my typo. – Andrea Dec 16 '11 at 13:08
@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, – Matt E Dec 16 '11 at 16:27
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. – user18119 Dec 16 '11 at 17:37

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.)

share|cite|improve this answer
No, it is false! See my answer. – Andrea Dec 16 '11 at 10:33
@Andrea: Dear Andrea, Presumably Hannah means smooth, not merely regular. Regards, – Matt E Dec 16 '11 at 16:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.