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I'm reading Chapter V.2 of Hartshorne, which includes the following claim, where $\pi: X \rightarrow C$ is a ruled surface:

Note that any two fibres of $\pi$ are algebraically equivalent divisors on $X$, since they are all parametrized by the curve $C$.

While this statement is intuitively clear, I cannot work out precisely what the divisor $D$ on $X \times C$ with $D_t = \pi^{-1}(t)$ for $t \in C$ closed should be. It seems like it should be described as the locus $\{(x, c) \in X \times C \mid \pi(x) = c \}$ or something similar, but I can't see exactly what the local defining equations should be to realize this as a Cartier divisor.

I think if you take the image $f(X)$ where $f: X \rightarrow X \times C$ is $\text{id} \times \pi$, you should get the right divisor, but I'm having trouble rigorously showing that this is in fact a Cartier divisor.

I'd be interested too in exactly what conditions we need on $X$ and on $C$ - Hartshorne tends to assume everything is non-singular and proper over an algebraically closed field, but perhaps not all of these are necessary?

Dorebell
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  • $X$ and $C$ are both generally assumed to be nonsingular varieties over an algebraically closed field, I believe. This should imply smoothness, since the fibers are all smooth of dimension one (they are $\mathbb{P}^1$'s). – Dorebell Jun 23 '15 at 04:56
  • I think something like this should do it. Assuming that $\pi$ is smooth, how would we show that the graph of $\pi$ is a Cartier divisor? – Dorebell Jun 23 '15 at 06:25
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    The graph of $\pi$ is isomorphic to $X$, so it is a codimension-1 subvariety of $X \times C$, hence a Weil divisor. But on a nonsingular variety (or regular scheme) every Weil divisor is Cartier. – Relapsarian Jun 23 '15 at 09:46
  • I see - I suppose I was worried about the codimension behaving oddly, but this shouldn't be a worry with the varieties irreducible.

    Is it easy to write down local equations defining $X$ as a divisor in $X \times C$? At a point of $C$, you can take $t$ to be a uniformizing parameter of $\mathcal{O}{C,c}$ and then the ideal should be generated by $\pi^#(t) \otimes 1 - 1 \otimes t \in \mathcal{O}_X \otimes \mathcal{O}{C,c}$, but I don't know how to do this on an open set.

     – Dorebell Jun 24 '15 at 00:57
  • Eh, I take it back; it's reduced and irreducible. No issues there. – Hoot Jun 25 '15 at 21:15

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