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 $y$? 



Intuition for rational functions
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. 


