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If $X$ is a normal variety, and $p \in X$, is it true that there is a curve $C \subset X$ with $p \in C$ a smooth point (on C)?

It is obviously false if normality is dropped - take $X$ to be a singular curve. I have no specific reason beyond this to request normality, though the examples I am familiar with pass this test. I somewhat doubt that this could be true, so I am asking for a counter example.

(The vague motivation is that I am thinking about the valuative criterion for separatedness recently, and would like to understand the intuition that there are no curves $C$ with double points on a separated scheme - i.e with two centers for the same valuation on $C$. And I like DVRs, though I guess one can just take an arbitrary curve passing through the point and take its normalization to get a discrete valuation on the curves function field with some prescribed center. I am still curious about the geometric question anyway.)

The other side of this question:

Is there a (normal) variety $X$ with a singularity so "bad" that all curves passing through that point acquire that singularity?

Elle Najt
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1 Answers1

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Here's an example to show the answer to the first question is "no".

Let $S$ be a surface which is factorial but not smooth. These singularities do exist, but they are rather special --- see the answer of Victor Protsak to this MO question for some details. By definition of "factorial", every Weil divisor, in particular every curve, on $S$ is a Cartier divisor.

Now, it is a general fact that if $X$ is a variety with a singular point $p$, and $D$ an effective Cartier divisor on $X$ passing through $p$, then $D$ is also singular at $p$. The proof is easy: take an affine open set containing $p$ in which $D$ is principal, defined by some regular function $f$ say. Then the Jacobian matrix for the ideal defining $D$ is obtained from the matrix for the ideal defining $X$ by adding one row, so its rank is at most one more.

So let $p$ be a singular point of $S$. By factoriality, any curve $C$ on $S$ is a Cartier divisor, so if $C$ is a curve passing through $p$, then $C$ must be singular at $p$.

Note that this example works not because the singularity is especially "bad", but rather the opposite --- factorial surface singularities are mild. By contrast the surface ordinary double point $xy=z^2$ is not factorial, but here there are smooth curves passing through the singularity. The point is precisely that these smooth curves are Weil divisors that fail to be Cartier.

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Nefertiti
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  • Thanks. And the "general fact" about Cartier divisors is quite easy to prove, no? It just follows from the observing that the differential of a local equation cannot drop the dimension of the tangent space by more than one, so the point remains smooth. – Elle Najt Mar 28 '16 at 19:24
  • @AreaMan: Right. – Nefertiti Mar 28 '16 at 20:09
  • Could you expand a little on what makes a singularity "mild"? – Elle Najt Mar 29 '16 at 04:07
  • Dear Pooh Bear, I would be quite grateful to you if you wrote down a reasonably complete proof that a (presumably effective) Cartier divisor passing through a singular point of a scheme is itself singular. – Georges Elencwajg Mar 29 '16 at 19:25
  • @GeorgesElencwajg: yes, effective, thanks! The proof is just what AreaMan said above. I added some words of explanation in the answer. – Nefertiti Mar 30 '16 at 06:43
  • @AreaMan: I was not being very precise when I used the term "mild". All I had in mind was that smooth points are factorial, whereas some singularities aren't. – Nefertiti Mar 30 '16 at 06:57