Let $X$ be integral scheme and $\mathcal K$ sheaf of rational 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 rational functions on $X$ can be restricted to nonzero rational functions on $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$?
Intuition for fiber of rational function sheaf
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algebraic-geometry
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0@Evgenia. Ah I see what you mean. Sorry for the error I made. Thanks a lot for explaining everything. – 2010-11-14