Questions about flat limits and associate points, Vakil's section 24.4.12 Suppose $(A,m)$ is a discrete valuation ring and $[m]$ is the closed point of $\operatorname{Spec}A$ and $\eta$ is its generic point, which is also the only nontrivial open set. If we have a morphism $\pi:X \rightarrow\operatorname{Spec}A$, and $Y$ is a closed subscheme of the fiber over $\eta$ $X|_{\eta}$, which is naturally the open subscheme $\pi^{-1}(\eta)$. The scheme theoretic closure of $Y$ in $X$ is denoted by $Y'$, which is the smallest closed subscheme contains $Y$.
In that section, it claims that the induced map of $Y' \rightarrow\operatorname{Spec}A$ is flat. From exercise 24.4.K, we need to show that all the associate points of $Y'$ map to the generic point $\eta$. Since $Y$ is mapped to $\eta$ by $\pi$, we only need to show that $Y'$ does not contain other associate points, i.e. the scheme-theoretic closure process does not introduce new associate points. This is intuitively very reasonable since $Y'$ is the smallest closed subscheme contains $Y$. But I do not know how to rigourously prove this.
Ex 24.4.L is an inverse of this, i.e. if $Y'$ is a closed subscheme that its fiber over $\eta$, $Y'|_{\eta}$ is $Y$ and $Y'$ is flat over $\operatorname{Spec}A$, then $Y'$ is the scheme theoretic closure of $Y$. This is also intuitively very reasonable, i.e. $Y'$ is flat over $\operatorname{Spec}A$ so its fiber $Y'|_{[m]}$ contains no associate points of $Y'$, and no associate points means smaller! But I do not understand associate points well and I do not now how to show this.
 A: I'll generalize this as follows:
Suppose a locally Noetherian scheme $X$ has a global section $f$, and $\mathscr{F}$ is a coherent sheaf on $X$. Let $U=D(f)$, and $\mathscr{G}$ be a quasi-coherent quotient of $\left.\mathscr{F}\right|_U$ (so $$\left.\mathscr{F}\right|_U\to \mathscr{G}\to 0$$ is exact). Then there is a unique quasi-coherent quotient $\mathscr{G}'$ of $\mathscr{F}$, such that

*

*$\mathscr{G}\cong \mathscr{G}'\mid_U$ as quotients of $\mathscr{F}|_U$.

*All of the associated points of $\mathscr{G}'$ lie in $U$.

In your case $f$ is the pullback of the uniformizor in $A$, $\mathscr{F}=\mathscr{O}_X$, and the last condition is equivalent to $\mathscr{G}'$ being flat over $A$.
Proof:
Define $\mathscr{G}'$ as the image of $\mathscr{F}\to i_*\mathscr{G}$, the adjunction of $\mathscr{F}\mid_U\to \mathscr{G}$. To show this works and is unique, we may work affine-locally on $X$. Obviously 1 holds ($\mathscr{F}\to \mathscr{G}'$ pulls back to $\mathscr{F}\mid_U\to \mathscr{G}$).
Suppose $X=\text{Spec}\ A$, $\mathscr{F}=\tilde{M}$, where $M$ is a finite $A$-module, and $\mathscr{G}=\tilde{N}$ where $N$ is a quotient of $M_f$.
Consider $\mathscr{G}'=\tilde{N'}$. The condition imply $M\to N'\to 0$ is exact, and $M_f\to N$ factors through an isomorphism $N'_f\to N$. Condition 2 is equivalent to $f$ not being a zero-divisor, or that that natural map $N'\to N'_f$ is injective,  equivalently $N'\to N$ is injective. This is equivalent to $N'$ being the image of the composition $M\to N'\to N$, which is the same as $M\to M_f\to N$. This implies there is only one option for $N'$, namely the image of
$$M\to M_f\to N$$
In this case, one sees that indeed $N'_f\cong N$ (obviously $N'_f\to N$ is injective, but also subjective because $M_f\to N$ is). In addition, $f$ is not a zero-divisor on $N'$ because it's not on $N$. This completes the proof.
