Intuition for rational functions

Question

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 $y$?

Answer 1

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.

Answer 2

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 $y$).

Answer 3

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.