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Let $\pi: X\to C$ be a fibration in curves where $C$ is a non-singular curve and $X$ a regular, integral surface and the generic fiber $X_\eta$ is a non-singular curve over $k(C)$ (these hypotheses might be stronger than necessary, but I just threw a bunch on to make it as nice as possible).

Now a point on $X_\eta$, say $p$ is also a point on $X$ itself. Note the generic point of $X$ is the generic point of $X_\eta$, say $\zeta$. All the other points on $X_\eta$ are closed in the curve and are height $1$ points on $X$.

I read in a paper that for any point on the generic fiber, $p$, we have $\mathcal{O}_{X,p}\simeq \mathcal{O}_{X_\eta, p}$, and at first I just thought to myself that it's obvious, but when I tried to actually think of a reason it wasn't so obvious.

If $X_\eta$ were open in $X$, then this would be clear since restricting to an open and then taking a stalk doesn't cause problems, but why should this still be true for the generic fiber which is neither open nor closed?

One noted consequence is that $k(X)=\mathcal{O}_{X,\zeta}=\mathcal{O}_{X_\eta, \zeta}=k(X_\eta)$.

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up vote 5 down vote accepted

Take affine affine open sets $U\subseteq X$ and $V\subseteq C$ such that $\pi|_U: U\rightarrow V$.Then $B:=O_X(U)$ is an $A:=O_C(V)$-algebra and $O_{X,p}=B_p$, where $p\in\mathrm{Spec} (B)$ satisfies $p\cap A=0$.

The generic fibre $X_\zeta\cap U$ equals $\mathrm{Spec}((A\setminus 0)^{-1}B)$. Hence

$ O_{X_\zeta ,p} =((A\setminus 0)^{-1}B)_{p(A\setminus 0)^{-1}B}=B_p $

since $B\setminus p$ contains $A\setminus 0$.

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Thanks. This proof actually works in much more generality than the given conditions! It seems all that was needed is that the map is dominant and of finite type between integral schemes. –  Matt Feb 24 '11 at 18:43
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