Identifying the stalk of an integral scheme at the generic point

Question

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 $X$?

Answer 1

Let me compile the discussion from the comments to mark this as answered.

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 $\mathcal{O}_{X,\xi} \cong \mathcal{O}_{\operatorname{Spec} A, (0)} \cong A_{(0)} \cong K(A)$.