# Intuition for fiber of rational function sheaf

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 $y$?

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@evgeniamerkulova: If you want your accounts merged you should ask for it on Meta. – Nuno Oct 27 '10 at 21:12
Nuno,thank you for kind advice. – evgeniamerkulova Oct 27 '10 at 21:30
@Evgenia: I am sorry, I know algebraic geometry only in very informal terms. Please excuse my question. I take the example of $\mathbb Z$, and the sheaf of rational functions would be the constant sheaf $\mathbb Q$. Then the maximal ideal $m_y$ would be the zero ideal, and the fiber would be $\mathbb Q$. Right? – user1119 Nov 13 '10 at 11:35
You need not ask excuse or be sorry : algebraic geometry is not so easy ! If $y$ correspond to prime number $p$, then $\mathcal K _y = \mathbb Q$ but $m _y \mathcal K _y=p \mathbb Q = \mathbb Q$ so fiber at $y$ is $\mathbb Q / \mathbb Q =0$. So you should be carefull that maximal ideal $m_y$ is not zero: it contain $p$ for example. – evgeniamerkulova Nov 13 '10 at 14:32
@Evgenia: Ah so you seem to be taking $m_y$ to be the max. ideal inside the local ring at $p$ for the structure sheaf. In my mind I had replaced the structure sheaf with the sheaf of rational functions. Sorry about the mistake. To get the fiber, you seem to be tensoring the stalk at $y$ with the residue field at $y$. I hope I have understood the definition correctly now.If so, taking the affine line over a field $k$, the fiber of the field of rational functions at $y$ seems to be the function field in one variable over $k$. Am I right? – user1119 Nov 13 '10 at 14:54