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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 $y$?

Thanks for answering. But what I would 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!"

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 $y$?

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 $y$?

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 $y$?

Thanks for answering. But what I would 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!"