6
$\begingroup$

I asked this on mathematics stack exchange and did not receive answer . I hope it is good manners to ask here. Thank you very much.

Let $X$ be integral scheme and $\mathcal K$ sheaf of rationnal functions on $X$. For any point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K _y / \mathcal m_y \mathcal K_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rationnal functions on $X$ can be restricted to nonzero rationnal functions on $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$?

$\endgroup$
2
  • 3
    $\begingroup$ The field $\mathbf{Q}$ of all rational numbers, when viewed as a $\mathbf{Z}$-module, vanishes when taken modulo $p$ for any prime $p$. This is not inconsistent with the fact that $1/3$ lies in the local ring $\mathbf{Z}_{(5)}$ at $5$ and has nonzero reduction in its residue field $\mathbf{F}_5$, since in $\mathbf{Q}$ we have $1/3 = 5q$ for another element $q \in \mathbf{Q}$ (albeit one not in the local ring at $5$!). It's the same phenomenon. $\endgroup$
    – BCnrd
    Oct 28, 2010 at 20:19
  • $\begingroup$ Thanks for answering, but I had computed such examples "blindly" without undersand what is behind computations. What I would really like to know is what information follow from: fiber is zero. For example in terms of classical variety $Y$ corresponding to nonclosed nongeneric point $y$ and rationnal functions on $Y$. What vectorspace over $\mathcal K (Y)$ would have clearly been zero for Italian geometer 100 years ago? Maybe this is stupid question and I should say to myself "just compute; if fiber is zero, then it is zero!" $\endgroup$ Oct 28, 2010 at 21:32

3 Answers 3

10
$\begingroup$

The non-classical aspect of this setup is that you're using a quasi-coherent sheaf that is not coherent, and beyond the coherent case one cannot expect information about a fiber (e.g., vanishing, 6 generators, etc.) to "propogate" to information in a neighborhoood (which would be the spirit behind the choice of word "coherent", I suppose). Computing the fiber of the field of all rational functions at a non-generic point likely has no classical counterpart, much as in number theory one doesn't ever try to reduce $\mathbf{Q}$ modulo 5, only $\mathbf{Z}_{(5)}$ or its subrings.

$\endgroup$
1
  • $\begingroup$ "Computing the fiber of the field of all rational functions at a non-generic point likely has no classical counterpary" is perfect answer to question and explain why I had not intuition for this concept. Thank you very much. $\endgroup$ Oct 29, 2010 at 7:24
2
$\begingroup$

Your intuition is confusing the 'fiber over a point' with `restriction to a closed subscheme'. In general these can be very different, even if they come from the same place conceptually. Rational functions give a good example of when the fiber is zero but the restriction isn't.

For an example the other way, consider the submodule $O(-y)\subset O$ consisting of regular functions which vanish at $y$. These restrict to zero at $y$, but the fiber is isomorphic to the zariski cotangent space, which at a smooth point will be a $k$-vector space of the dimension of your scheme.

Restriction of functions is the more intuitive concept, but the fiber construction is more natural from a module theoretic perspective. Consider $x\mathbb{C}[x]\subset \mathbb{C}[x]$ (a case of the previous example). The restrictions of these sets to $x=0$ differ, but they are isomorphic as modules.

There is also a module-theoretic version of 'restriction to a closed subscheme', given by the limit over all open neighborhoods of that subscheme, but this has its own counter-intuitive phenomenon. For example, it can never give different answers for $\mathbb{C}[x]$ and $x\mathbb{C}[x]$. If you take this limit for rational functions, you get all of $\mathcal{K}$, which is at least non-zero, but now it is keeping track of too much information (for example, it is distinguishing between rational functions which differ off of $y$).

$\endgroup$
3
  • $\begingroup$ Nobody uses the notation $\mathcal O(y)$ for regular functions vanishing at $y$: where have you seen that notation? If $X$ is a curve it is very confusing because one allways uses it for sheaf of rationnal functions having pole (not zero!) of order $\leq 1$ on $y$ and regular outside $y$. $\endgroup$ Oct 28, 2010 at 22:06
  • $\begingroup$ Probably this should be $\mathcal{O}(-y)$. $\endgroup$ Oct 28, 2010 at 23:51
  • $\begingroup$ Yes, corrected. $\endgroup$ Nov 2, 2010 at 22:44
0
$\begingroup$

Let me stick to the affine situation $X=Spec(R)$ for an integral domain $R$.

As you were pointing out in your question only "many rationnal functions on $X$ can be restricted to nonzero rationnal functions on $Y$", not all of them!

So In order to get a morphism from $K(X)$ to $K(Y)$ we have to restrict ourselves to the subring of functions in $K(X)$ whose pole divisor does not contain $Y$. But this is nothing but the locallization of $R_{\eta}$ where $\eta$ is the prime ideal corresponding to $Y$. Then you get a morphism "by taking the fiber"

$K(X) \supset R_{\eta} \longrightarrow R_{\eta}/\eta R_{\eta} =K(Y)$

which looks like the one you were searching for.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.