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Let $X$ be an integral scheme. Show that the local ring $\mathcal{O}_{\xi}$ of the generic point $\xi$ of $X$ is a field.

Proof Idea:

Let $U \subset X$ be an affine open so that $U= Spec \hspace{0.5mm} A$ where $A$ is necessarily an integral domain since $X$ is an integral scheme. Since $\overline{\xi}=X$, we must have that every open subset of $X$ contains the generic point $\xi$.

In particular, $ \xi \in U$. Furthermore, since $A$ is an integral domain $(0)$ is a prime ideal whose closure is $A$. Since irreducible schemes have unique generic points we must have that $(0)$ corresponds to our generic point $\xi$.

Thus,

$$ \mathcal{O}_{Spec \hspace{0.5mm} A, (0)} = A_{(0)} = Q(A) $$

where $Q(A)$ denotes the quotient field of $A$. Thus the stalk at $\xi$ for every open affine of $X$ is necessarily a field.

My Question:

I have show that the stalk at $\xi$ for every open affine is a field. How do I use this to show that $\mathcal{O}_{X, \xi}$ is a field? I am thinking some sort of gluing argument but I am pretty stuck.

user7090
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    If $U$ is an open subset of a locally ringed space $X$, and $\zeta\in U$, then $\mathcal{O}{X,\zeta} = \mathcal{O}{U,\zeta}$. This follows from the definition of the stalk as a set of equivalence classes. Minor nitpick: we should take $U$ to be some nonempty affine open. – Andrew Dudzik Jun 14 '16 at 22:17
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    Does this answer your question? http://math.stackexchange.com/questions/218767/relation-of-function-field-of-a-scheme-to-the-local-ring-of-its-prime-divisor?rq=1 – John M Jun 14 '16 at 22:19
  • I see. I really don't see how $\mathcal{O}{X, \xi}= \mathcal{O}{U, \xi}$ follows from the definition. Is it true that $\mathcal{O}_{U} \cong \mathcal{O}_X \vert_U $ ? – user7090 Jun 14 '16 at 22:27

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