Definition:
Let $R$ be a commutative ring with unity. An ideal $I\triangleleft R$ is said to be pure if the map $R\rightarrow R/I$ is flat.
Question:
In the proof of Stacks Project Lemma 10.107.3 (04PT), it is stated that, if $I$ is a pure ideal of $R$, then $$I=\operatorname{Ker}(R\rightarrow\prod_{\mathfrak p\in V(I)}R_{\mathfrak p}),$$ and says this follows from the $7$th point of Stacks Project Lemma 10.107.2 (04PS). But from the cited lemma I can only see that $I\subseteq\operatorname{Ker}(R\rightarrow\prod\limits_{\mathfrak p\in V(I)}R_{\mathfrak p})$, and I am confused why the converse inclusion holds as well.
Thoughts:
Let $x\in\operatorname{Ker}(R\rightarrow\prod\limits_{\mathfrak p\in V(I)}R_{\mathfrak p})$. Then, for all $\mathfrak p\in V(I)$, we have $xy=0$ for some $y\not\in\mathfrak p$. Thus $x\in\sqrt I$, so $x^n\in I$ for some $n$.
Then from Lemma 10.107.2, we have $x^n=x^nz$ for some $z\in I$. But I don't see any way to continue showing that $x\in I$.
I also tried to show that the annihilator of $x$ has a non-empty intersection with $1+I$, but I am not even sure it has a non-empty intersection with $1+\mathfrak p$ for $\mathfrak p\in V(I)$.
Any help is sincerely appreciated. Thanks in advance.
It turns out that I might not need the equality $I=\operatorname{Ker}(R\rightarrow\prod\limits_{\mathfrak p\in V(I)}R_{\mathfrak p})$ to show that $I$ is determined by $V(I)$.
Define $K=\operatorname{Ker}(R\rightarrow\prod\limits_{\mathfrak p\in V(I)}R_{\mathfrak p})$, and $J=\left\{x\in R\mid x\in xK\right\}$. We would like to show that $I=J$, and hence $I$ is indeed determined by $V(I)$ alone.
As stated in the question, if $x\in K$, then $\forall\mathfrak p\in V(I)$, we know $\exists y\not\in\mathfrak p$ such that $xy=0$. Hence $x\in\bigcap\limits_{\mathfrak p\in V(I)}\mathfrak p=\sqrt I$, and it follows that $K\subseteq\sqrt I$.
Now we show that $J$ is an ideal containing $I$. From Stacks Project Lemma 10.107.2, we know that $\forall x\in I$, $x\in xI\subseteq xK$, thus $I\subseteq J$ follows from the definition.
Next let $x,y\in J$. Take $f,g\in K$ such that $x=xf,\,y=yg$. Then we see that $(x+y)=(x+y)(f+g-fg)$ by a simple calculation, so $x+y\in J$ as well. And for any $r\in R$, it is obvious that $rx\in J$. Thus $J$ is an ideal of $R$ containing $I$.
Finally we show that $I=J$.
Take $x\in J$. By definition, $x=xk$ for some $k\in K$. Hence $x=xk=xk^2=\cdots=xk^n,\,\forall n\in\mathbb N$. As we observed in the beginning, $K\subseteq\sqrt I$, so $\exists n\in\mathbb N$ such that $k^n\in I$. It follows that $x=xk^n\in I$. Therefore $I=J$ as we wanted to show.
Hope this helps other confused people.