# Schemes with isomorphic stalks

Fact: If $X$ and $Y$ are varieties and we have $\mathcal{O}_{X,q} \cong \mathcal{O}_{Y,q}$ then there are neighborhoods $U$ of $p$ and $V$ of $q$ which are isomorphic.

I understand the intuition behind this result: Since the open sets are so big, the scheme $$\varprojlim_{p \in U\subseteq X \; \text{open}} U$$ captures a lot of infomation about $X$ at the point $p$.

I can imagine a proof which goes like this: We can assume without loss of generality that the varieties in question are affine. Therefore we are reduced to proving that an isomorphism $R_{\mathfrak{p}} \to S_{\mathfrak{q}}$ of rings induces an isomorphism $R_f \to S_g$ for suitable elements $f \notin \mathfrak{p}$ and $g \notin \mathfrak{q}$.

This approach is a bit disappointing in my opinion because the result is so geometric and we reduce it to a formal proof about localization.

Question: Can anyone explain a geometric proof of this result to me?

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If you're looking for a better intuition, I think what you want is not a different proof, but to recognize that the proof you've given (via localizations) <b>is</b> geometric. – Steven Landsburg Oct 24 '12 at 16:07
there's a reason why it's not "geometric geometry" ;) – Mattia Talpo Oct 24 '12 at 16:12
The intuition and the approach for a proof are quite vague. After all, the result fails for general schemes. We need some finiteness condition, for example being of finite presentation over a given base. The proof then proceeds by looking at the denominators of the images of chosen generators. If you state this in geometric terms, it just becomes more complicated. – Martin Brandenburg Oct 24 '12 at 17:50
@Martin: Ahhh the fact that the result fails for general schemes is good! Is there a standard counterexample? – Daniel Barter Oct 24 '12 at 18:58
@Daniel: Take for $X$ the affine line over $\mathbb C$ and $Y$ the spectrum of $\mathbb C(x)$. – Qing Liu Oct 24 '12 at 20:26