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Let $A$ be a commutative $\mathbb k$-algebra and let $X = \mathrm{Spec} A$ be the corresponding affine scheme.

For simplicity assume $\Bbbk$ is an algebraically closed field of characteristic $0$.

The stalk of the structure sheaf $\mathcal O_X$ at a point $x \in X$ can be defined as the direct limit $$ \mathcal O_{X, x} = \varinjlim_{U \ni x} \mathcal O_X (U) $$ over all open sets $U$ containing the point $x$, which is a local ring, i.e. with exactly one maximal ideal.

Out of curiosity, it seems that we can also define the "stalk" in two points $x, y$ by $$ \mathcal O_{X, x, y} = \varinjlim_{U \supset \{ x, y \}} \mathcal O_X (U). $$

Is it true that $\mathcal O_{X, x, y}$ has exactly two maximal ideals whenever $x$ and $y$ are two distinct closed points? It seems not too difficult to show it has at least two maximal ideals, because there are two surjective maps to $\Bbbk$, but I don't know how to show that there are no more maximal ideals.

More precisely, we have an injective map $\mathcal O_{X, x, y} \to \mathcal O_{X, x} \times \mathcal O_{X, y}$ and the latter has exactly two maximal ideals. But in general, injective algebra morphisms can "lose" maximal ideals (e.g. $\Bbbk [x] \hookrightarrow \Bbbk [[x]]$)...

Earthliŋ
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    As stated, $\mathcal O_{X,x,y}$ doesn't need to have two maximal ideals. For example, if $x$ is a specialisation of $y$ (i.e. $x\in\overline{{y}}$), then all open subsets that contain $x$ also contain $y$, so $\mathcal O_{X,x,y}$ coincides with the local ring $\mathcal O_{X,x}$. Do you assume $x$ and $y$ to be closed points? – Florian Adler Mar 24 '21 at 18:00
  • @FlorianAdler Right, thank you for pointing this out. I've edited the question accordingly. – Earthliŋ Mar 24 '21 at 19:45
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    It comes from the prime avoidance lemma, see https://math.stackexchange.com/questions/159163/a-ring-of-fractions-which-has-finitely-many-maximal-ideals – Minseon Shin Mar 24 '21 at 21:25
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    @MinseonShin Is it clear that $\mathcal O_{X, (x,y)}$ is a localization of $A$? I wonder if any open $U \supset {x,y}$ contains a standard open $D(f) \supset {x,y}$. Clearly we can cover $U$ by standard opens, but why does one of them contain both $x$ and $y$? – red_trumpet Mar 25 '21 at 09:24
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    @red_trumpet That also follows from prime avoidance: We have $U=X\setminus V(I)$ for some ideal $I\subseteq A$ which is not contained in the maximal ideals corresponding to $x$ and $y$. By prime avoidance, there's an element $f\in I$ not contained in either of these prime ideals, whence $D(f)\supseteq{x,y}$. – Florian Adler Mar 25 '21 at 13:46
  • @FlorianAdler Right, thanks! – red_trumpet Mar 25 '21 at 13:50

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