Let $X$ be a projective variety over $k$, and $\dim X \geq 2$. By a curve $C$ on $X$, I mean a proper,reduced subscheme of $X$ of dimension $1$.

(1)If $C$ is an irreducible curve on $X$, then is $C$ numerically equivalent to some $\sum n_i C_i$ with $n_i > 0$, and $C_i$ smooth curves ?

(2) If $D$ is a Cartier divisor on $X$, and $D\cdot C \geq 0$ for any smooth curve $C$ on $X$, then is $D$ necessarily to be a nef divisor?

Certainly, if (1) holds, then (2) holds.

(I ask this question because I curious why people usually call a nonconstant morphism $C \to X$ with $C$ being a smooth curve to be a curve on $X$, rather than a smooth curve $C$ exactly sitting inside $X$.)


2 Answers 2


The answer to question $(1)$ is no. In fact, take an irreducible, nodal cubic curve $A \subset \mathbb{P}^2$ and take $10$ points $p_1, \ldots, p_{10}$ on it, different from the node. Let $X$ be the blow-up of $\mathbb{P}^2$ at the points $p_i$ and $C$ the strict transform of $A$ in $X$. Then $C$ is an irreducible nodal curve isomorphic to $A$ and such that $C^2=-1$.

This implies that $C$ is isolated in its numerical equivalence class. Indeed, assume that $D$ is effective and numerically equivalent to $C$; since $CD =-1$ and $C$ is irreducible, it follows that $C$ is a component of $D$. Then $D=C+Z$, where $Z$ is effective and numerically trivial on $X$; so $Z=0$ and $C=D$. In particular, no positive linear combination of smooth curves can be numerically equivalent to $C$.

This also shows that the answer to question $(2)$ is no. In fact, the curve $C$ in the example above clearly has non-negative intersection with any irreducible curve on $X$ different from it, in particular it has non-negative intersection with all the smooth curves. However, $C$ is not nef since $C^2 =-1 <0$.

  • $\begingroup$ Thank you so much! I have two further questions regard your answer: (1) Why you claim " since $C\cdot D=-1$, ..., it follows that $C$ is a component of D" ?(2) Why $C$ has nonnegative intersection with any irreducible curve differ from $C$? I think the two questions are essentially the same: how to describe the effective divisors on $X$? $\endgroup$
    – Li Yutong
    Jul 30, 2013 at 10:50
  • 1
    $\begingroup$ If one has two irreducible curves $C$, $D$ on a smooth complex projective surface, with $C \neq D$, then $C \cdot D$ is the number of intersections of $C$ and $D$ (counted with multiplicities). In particular $C \cdot D \geq 0$. Then if $C \cdot D <0$ it follows that $C$ and $D$ necessarily have some component in common (and such component has negative self-intersection). This explains (1) and also (2). Notice that in this example one does not need to know all the effective divisors on $X$. $\endgroup$ Jul 30, 2013 at 11:46

Since Francesco answered your two questions very nicely, I thought I'd indicate some possible reasons why people call a nonconstant morphism from a smooth curve $C$ to $X$ a curve as opposed to one actually sitting on $X$.

One big reason is that one can reduce many statements to discussions about flat families over smooth curves. A well-known fact is that connected algebraic varieties (say projective for simplicity) are "path-connected" by smooth curves. Here the meaning is that any two points $p$ and $q$ can be connected by the $\textit{image}$ of a smooth curve $C$. For smooth varieties this can probably be made to work with $C$ actually sitting on $X$, but the proof of the above fact is not difficult (it's come up a few times on MO, once even by me :)). This fact is very useful because pulling back the question, whatever it is, to one over $C$ then allows us to use the result that a morphism $\pi:X\rightarrow C$ to a smooth curve $C$ is flat iff every associated point of $X$ gets sent to the generic point of $C$ (Hartshorne III.9.7). This allows us to use flatness in situations where a more general morphism might not be flat.

For example, this is used in Hartshorne III.9.13 to show that the Hilbert polynomial of an "algebraic family" of normal varieties are constant. Here the family wasn't originally flat but becomes so upon base change.

Uses like this also appear in Hilbert scheme-type arguments, for example to show that the Hilbert scheme is proper, since one often just restricts to ALL curves passing through some point to show the result in general.

I hope this indicates the uses of such terminology.

  • $\begingroup$ Great explanation! Thank you Howie! $\endgroup$
    – Li Yutong
    Aug 7, 2013 at 2:20

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