# Identifying the stalk of an integral scheme at the generic point

Let $$X$$ be an integral scheme. Then since $$X$$ is irreducible it has a generic point $$\eta$$. Suppose $$\operatorname{Spec}A$$ is an nonempty affine open subset of $$X$$. Then $$\eta$$ is also the unique generic point of $$\operatorname{Spec}A$$. I would like to understand in what way the stalk at $$\eta$$, $$\mathcal{O}_{X,\eta}$$, is identified with $$K(A)$$, the fraction field of $$A$$.

Since $$\operatorname{Spec}A$$ is nonempty, I know that $$A\neq (0)$$. Further, since $$X$$ is integral, $$\operatorname{Spec}A$$ is also integral, and it follows that $$A$$ is an integral domain. So $$(0)$$ is a prime ideal, and is contained in every other prime ideal of $$A$$. I am not really sure where to go from here. What am I missing?

I also know that the unique generic point of $$\operatorname{Spec}A$$ is the nilradical of $$A$$, the intersection of all prime ideals of $$A$$, of which $$(0)$$ is an element. Further, the nilradical is prime since $$\operatorname{Spec}A$$ is irreducible.

Do we get that $$(0)$$ is the generic point of $$\operatorname{Spec}A$$ since $$V((0))=\operatorname{Spec}A$$, and hence $$(0)$$ is the generic point of $$X$$?

• Do you know that stalks of a sheaf can be calculated by first restricting to any open set? Do you know the stalks of the structure sheaf for an affine scheme are given by localization? (Which localization?) Sep 22, 2020 at 22:35
• @KReiser For 1 I know that stalks are given by sections locally up to equivalence, but I'm not sure if this is the same thing you're saying. For 2, I know that the stalk at of the structure sheaf of $\operatorname{Spec}A$ at a point $P$ is $A_P$. Sep 22, 2020 at 22:41
• Yes, this is the same. Next, what's $A_{(0)}$? Sep 22, 2020 at 22:53
• Well, you're inverting all non-zero elements of $A$, right? Which is exactly what you do to get $K(A)$ from $A$. Sep 22, 2020 at 23:08
• Yes, if $X$ is an irreducible scheme then for any open subscheme $U$, we have that the generic points of $U$ and $X$ are the same. It's a straightforward topological exercise: suppose not, then we get all sorts of contradictions. Sep 22, 2020 at 23:13

If $$X$$ is a topological space, $$x\in X$$, $$U\subset X$$ an open subset containing $$x$$, and $$\mathcal{F}$$ a sheaf on $$X$$, then the stalk of $$\mathcal{F}$$ at $$x$$ is the same as the stalk of $$\mathcal{F}|_U$$ at $$x$$. Using your definition of the stalk as equivalence classes that agree on an open neighborhood of $$x$$, we can prove this by restricting a representative $$(f,V)$$ to $$(f|_U,V\cap U)$$. (With other definitions, there are other proofs - categorically, open neighborhoods of $$x$$ inside $$U$$ form a cofinal subset of the open neighborhoods of $$x$$, which means that calculating the limit along them gives the same value.)
Next, the generic point of $$X$$ is the generic point of the affine open subscheme $$\operatorname{Spec} A\subset X$$ as well (exercise: prove this, perhaps by assuming it's not and seeing what contradiction you can get, like disjoint open subsets or a decomposition of $$X$$ as a union of two proper closed subsets). Since $$(0)$$ is the generic point of $$\operatorname{Spec} A$$, we have that $$\mathcal{O}_{X,\xi} \cong \mathcal{O}_{\operatorname{Spec} A, (0)} \cong A_{(0)} \cong K(A)$$.
• The following seems to be a straightforward way of showing that $\xi$ is a generic point of any nonempty open $U \subset X$. First, $\xi \in U$ is indeed true since $\{\xi\}$ is dense in $X$. Secondly, the closure of $\{\xi\}$ in $U$ is precisely $\overline{\{\xi\}} \cap U = X \cap U = U$ and we are done. Am I missing something or is this fine? Jun 6, 2021 at 12:15