Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 $X\to S$ be a scheme and let $D'$ and $D$ be relative effective Cartier divisors on $X$ satisfying $D' \subset D$ and let $D''$ satisfy $D = D' + D''$. Let $$0 \to \mathscr{O}_X(D'') \to \mathscr{O}_X(D) \to \mathscr{L} \to 0$$ be the induced exact sequence of sheaves on $X$. Of course $\mathscr{O}_X(D')$, $\mathscr{O}_X(D'')$ and $\mathscr{O}_X(D)$ are flat, but

Question: Is $\mathscr{L}$ necessarily flat? If not, when does flatness fail.

A colleague of a colleague has said that a positive answer (i.e. that $\mathscr{L}$ is indeed flat) is "at the beginning of Katz-Mazur", and I presume he is referring to Section 1.3, but I don't see how to derive the result from the material there. Section 1.3 is where the existence of the relative effective Cartier divisor $D''$ such that $D = D' + D''$ is stated, from which we can derive the exact sequence $$0 \to \mathscr{O}_X(D'') \to \mathscr{O}_X(D) \to \mathscr{O}_X(D) \otimes \iota_*\mathscr{O}_{D'} \to 0$$ by tensoring $$0 \to \mathscr{I}_{D'} \to \mathscr{O}_X \to \iota_*\mathscr{O}_{D'} \to 0$$ with $\mathscr{O}_X(D)$. So it would suffice to show that $\mathscr{O}_X(D) \otimes \iota_*\mathscr{O}_{D'}$ is flat, for which it would suffice to show that $\iota_*\mathscr{O}_{D'}$ is flat. Any ideas?

share|cite|improve this question
up vote 2 down vote accepted

This is in the definition of a relative effective Cartier divisor: $O_{D'}$ is flat over $S$.

share|cite|improve this answer
Of course! Silly me. – Hamish Oct 26 '12 at 21:51

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.