This is a very basic question, but I can't find a clean answer anywhere.

In introductory algebraic geometry books working over the complex numbers, it's usual to use these three words interchangeably. A point on a variety $X$ is smooth/regular/nonsingular if the dimension of the tangent space at the point is equal to the dimension of the variety.

On the other hand, I know that people sometimes find it important to distinguish between these terms, maybe when defining smoothness of morphisms, or working over non-closed fields,...

I want to make sure I know the right definitions of these terms in current use. In what contexts should each be defined? What implies what? How should I think of them?

Edit: See for example https://en.wikipedia.org/wiki/Regular_scheme , which says there are regular schemes that aren't smooth. There are also these notes of Vakil, where he has crossed out "smooth" and replaced it with "nonsingular": https://math.stanford.edu/~vakil/0708-216/216class21.pdf The notes seem to suggest it's because "smooth" is reserved as a property of morphisms. Is there a reason Wiki is happy to say "smooth scheme" but Ravi isn't? Is "nonsingular" the same as "regular"?

  • $\begingroup$ Have you considered searching in some books or internet. I did not see any case where these are used in different meanings $\endgroup$ Commented Oct 28, 2019 at 13:54
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    $\begingroup$ I clarified the question. That page appears to use them interchangeably, which as the examples I've added show is not always done. I am trying to understand the distinction. $\endgroup$
    – Lester
    Commented Oct 28, 2019 at 14:15
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    $\begingroup$ I tend to use the terms interchangeably when describing a variety and when working over a base field (especially an algebraically closed field). But given a map of varieties $f:X\to Y$ over such a field, the map $f$ is smooth, or "$X$ is smooth over $Y$", means something much stronger than simply that $X$ is a non-singular variety. It means that every fiber of $f$ is non-singular. (This isn't the technical definition of smooth morphism, but it conveys what's going on in this setting.) $\endgroup$ Commented Oct 28, 2019 at 14:16
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    $\begingroup$ For a variety over an algebraically closed field, all 3 are equivalent. For schemes, I would avoid the classical term "nonsingular", which seems ambigous. Smooth implies regular in general; it is equivalent to "geometrically regular". $\endgroup$ Commented Oct 28, 2019 at 14:16
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    $\begingroup$ (2) Smoothness is related to number theory: an extension $L/K$ of number fields is everywhere unramified if and only if $\mathrm{Spec} \mathcal{O}_L \to \mathrm{Spec} \mathcal{O}_K$ is smooth. $\endgroup$ Commented Oct 28, 2019 at 18:45

1 Answer 1


In the general context, "regular" is a property of a scheme (or a ring, or local ring), and "smooth" is a property of a morphism of schemes.

"Regular" means exactly that at every point, the dimension of the (Zariski) tangent space is equal to the (Krull) dimension (of the local ring at that point).

A map $f: X \to Y$ is smooth if the fibers over geometric points of $Y$ are regular, and $f$ is locally of finite presentation and flat.

We also use the relative point of view, so a scheme $X$ over $S$ is a smooth scheme over $S$ if the map $X \to S$ is smooth.

The first potential source of confusion is that a map between two regular schemes can fail to be smooth. This is not hard to see once you realize that there's no need for the fibers to be regular - for instance $xy$ defines a map $\mathbb A^2 \to \mathbb A^1$.

The second potential source of confusion is that a regular scheme over a perfect field is necessarily smooth over that field, but for an imperfect field this fails. See some examples. Usually over an imperfect field you want to consider smooth schemes and not regular ones as they are better-behaved.

Going in the other direction, smooth schemes over non-regular bases can fail to be regular, but smooth schemes over regular bases will be regular.

I think people rarely use "nonsingular" when they are trying to be careful about this distinction, but I think it's more likely to mean "regular".

  • $\begingroup$ Thank you good sir. $\endgroup$
    – Lester
    Commented Oct 28, 2019 at 14:16
  • $\begingroup$ @Bort Indeed, thanks. $\endgroup$
    – Will Sawin
    Commented Oct 28, 2019 at 14:31
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    $\begingroup$ A third important difference: When one says that a ring $R$ is smooth, one generally means that $R$ is smooth over some subfield $k$ which is clear from context. But a ring can be regular without containing ANY subfield, like the ring of integers. $\endgroup$ Commented Oct 28, 2019 at 15:00
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    $\begingroup$ @DavidESpeyer That's a good point, although a scheme can also be smooth without lying over any subfield, like a smooth scheme over $\mathbb Z$. $\endgroup$
    – Will Sawin
    Commented Oct 28, 2019 at 17:01
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    $\begingroup$ Right, although I think most algebraic geometers are unlikely to say "smooth" when they mean "smooth over $\mathbb{Z}$". To me, my comment is the main response to "why define regular at all"? The situation with imperfect fields suggests that regular is a worse behaved notion than smooth. But what is good about regularity is that I can say that $\mathbb{Z}[x,y]/(xy-p)$ is regular, whereas it is not smooth in any useful sense. $\endgroup$ Commented Oct 28, 2019 at 17:17

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