This is a beginner's question. I don't understand the description after defining Weil divisor in Chapter 2 of Hartshorne's book 'Algebraic Geometry'.

Let $ X $ be a noetherian, integral, separated, regular in codimension one scheme. Let $ Y $ be the prime divisor of $ X $ and $ \eta $ be the generic point of $ Y $.

It is written that the local ring $ \mathcal {O} _ {\eta, X} $ in $ \eta $ becomes a DVR with the function field $ K $ of $ X $ as the field of fractions.

However, in Ex.3.6 (I'm sorry if the problem number is different because it is the Japanese version), it is written that the local ring at the generic point of the integral scheme becomes the function field of $ X $.

If I follow this, $ \mathcal {O} _ {\eta, X} $ becomes a field, and I don't think it becomes the DVR.

What am I doing wrong?

  • $\begingroup$ A simple example to keep in mind is this. Suppose $A$ is a UFD and $f\in A$ an irreducible hence prime. Then the variety cut out by $f$ has co-dimension $1$ and global sections $A/f$ which is a domain. The local ring of $V(f)$ in $\operatorname{Spec}(A)$ is a DVR. It corresponds to the localization of $A$ at $(f)$ i.e. $A_{(f)}$. This is a DVR. $\endgroup$
    – S.S.
    Jul 27 at 10:08
  • 1
    $\begingroup$ Thank you for the easy-to-understand example. I seem to need to re-study the complex definition. $\endgroup$
    – Kazsugi
    Jul 28 at 7:13

Exercise 3.6 says that if $X$ is an integral scheme and $\eta$ is the generic point of $X$, then $\mathcal{O}_{\eta,X}$ is the function field of $X$. However, in the context you are asking about, $\eta$ is not the generic point of $X$ itself, but rather of the subscheme $Y$. So, Exercise 3.6 doesn't tell you anything about $\mathcal{O}_{\eta,X}$ (instead it would tell you about $\mathcal{O}_{\eta,Y}$).

  • $\begingroup$ Thank you. I understand. For the open affine subset $U$, $ \mathcal {O}_X | _U = \mathcal {O} _U $ holds. I think my mistake was to think that $ \mathcal {O} _X | _Y = \mathcal {O} _Y $ holds in the same way. $\endgroup$
    – Kazsugi
    Jul 25 at 3:56
  • $\begingroup$ @Kazsugi that last formula is actually correct. Restriction to a subvariety (of any type) is just pullback along the inclusion $f:Y \to X$, but the formula $f^* \mathcal O_X \cong\mathcal O_Y$ holds for basically any morphism you can think of (to the extent that I don't know a counterexample even in the most pathological of cases). $\endgroup$ Jul 25 at 6:30
  • $\begingroup$ @TabesBridges the formula $f^*\mathcal{O}_X\cong\mathcal{O}_Y$ holds by definition of pullback for any morphism of ringed spaces: $f^*(-)= f^{-1}(-)\otimes_{f^{-1}\mathcal{O}_X} \mathcal{O}_Y$. (Separately, in general/for any future readers, every once in a while you'll meet a context where the $|_Y$ sort of restriction means the inverse image instead of the pullback along the inclusion and it pays to be mindful about your notation.) $\endgroup$
    – KReiser
    Jul 25 at 8:40
  • $\begingroup$ @KReiser right, right, right... $\endgroup$ Jul 25 at 14:32

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