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Let $X$ be a scheme and $\operatorname{Spec} A\subset X$ be an affine with $x\in\operatorname{Spec} A$,

Is it true that $\mathscr{O}_{X,x}=\mathscr{O}_{\operatorname{Spec} A,x}$?

I ask this question because I feel that many "obvious" remarks made in Gortz-Wedhorn and Vakil FOAG seem to rely on this or a similar fact: such as in Vakil

"We say a ringed space is a locally ringed space if its stalks are local rings. Thus Exercise 4.3.F shows that schemes are locally ringed spaces." (Where 4.3.F is Show that the stalk of $\mathscr{O}_{\operatorname{Spec} A}$ at the point $[\mathfrak{p}]$ is the local ring $A_{\mathfrak{p}}$).

Is this the case? If not, why is the above statement immediate from 4.3.F?

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2 Answers 2

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Yes that's true, the statement you want to prove is that if $U$ is any open subset of any ringed space $X$ then for $x\in U$ the natural map $\mathcal O_{U,x}\to\mathcal O_{X,x}$ is an isomorphism. It's pretty straightforward, for instance for surjectivity if you have an element $f_x\in\mathcal O_{X,x}$ then it is represented by a function $f\in\mathcal O_X(V)$ for some $V$ containing $x$, and then the stalk of $f|_{U\cap V}$ in $\mathcal O_{U,x}$ is the element you want mapping to $f_x$.

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    $\begingroup$ Nice answer. The slogan here is that a stalk of the structure sheaf is a "purely local" piece of information which isn't sensitive to the open subset you compute it in. As an aside, both $\mathcal{O}_{U, x}$ and $\mathcal{O}_{X, x}$ are defined as colimits of certain rings coming from the structure sheaf. Using the universal property of colimits, you can define maps from each stalk to the other in the obvious way, and the usual uniqueness argument tells you that these maps are inverse to one another. $\endgroup$ Nov 16, 2020 at 1:24
  • $\begingroup$ Thank you for the answer, this makes me curious about the resultant isomorphism $A_{\mathfrak{p}}\simeq B_{\mathfrak{p}}$ for two different affines $\operatorname{Spec} A$ and $\operatorname{Spec} B$ containing $[\mathfrak{p}]$. $\endgroup$
    – Tejas Rao
    Nov 16, 2020 at 1:26
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Yes, as long as $\operatorname{Spec} A$ is an open affine set in $X$.

More generally if $U$ is an open subset of a ringed space $(X,\mathcal O_X)$, and $\mathcal O_U$ is the sheaf of rings on $U$ defined by $\mathcal O_U(V) = \mathcal O_X(V)$ for all open sets $V \subset U \subset X$, then $\mathcal O_{X,x} = \mathcal O_{U,x}$ for all $x \in U$. This is because

$$\mathcal O_{X,x} = \varinjlim_V \mathcal O_X(V)$$ as $V \subset X$ runs through the open neighborhoods of $x$ ordered by reverse inclusion, and $$\mathcal O_{U,x} = \varinjlim_V \mathcal O_U(V) = \varinjlim_V \mathcal O_X(V)$$

as $V \subset U$ runs through the open neighborhoods of $x$. Although the direct system of rings defining $\mathcal O_{U,x}$ is a subsystem of the one defining $\mathcal O_{X,x}$, it is easy to see that they define the same direct limit. It is the same principle by which the infinite polynomial ring $\mathbb C[x_1, x_2, x_3, ...]$ is the direct limit of the direct system

$$\mathbb C[x_1] \subset \mathbb C[x_1, x_2] \subset \mathbb C[x_1, x_2,x_3] \subset \cdots$$

as well as the subsystem

$$\mathbb C[x_1] \subset \mathbb C[x_1,x_2,x_3] \subset \mathbb C[x_1,x_2,x_3,x_4,x_5] \subset \cdots.$$

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