$\newcommand\frp{\mathfrak{p}}$I am trying to understand 034M of the Stacks Project, whose statement is the title of this post. The proof seems to implicitly assume that $R$ is a subring of $R_\frp$ (in the sentence “as $R \subset R_{\frp}$ is flat we see that $R_{\frp} \subset Q(R) \otimes_R R_{\frp}$”). However, this is not the case for a non-domain $R$ (the preimage of the prime $(0)\subset R_\frp$ along $R\to R_\frp$ is prime and thus different to $(0)\subset R$). So what is going on?
After looking for a while on MSE, this is the only related thing I've found and it seems not to be of help.
This is what I've tried so far: Let $\frp\subset R$ be any prime and consider this commutative diagram:
where the first (resp., the second) bottom arrow is an inclusion because localization is exact (resp., because the map $H$ here is always injective). Suppose $x\in Q(R)$ satisfies an integral equation over $R$. Then we can map this equation to $Q(R_\frp)$ to deduce that the image of $x$ in $Q(R_\frp)$ is integral over $R$ and hence over $R_\mathfrak{p}$. Since we are assuming that $R_\mathfrak{p}$ is a normal domain, it follows that the image $y$ of $x$ in $Q(R)_\mathfrak{p}$ is actually in $R_\mathfrak{p}$.
Define the ideal $I=\{f\in R\mid fx\in R\}$ of $R$. We want to show that $I$ is not contained in any prime ideal of $R$. To look for a contradiction, suppose $I\subset\frp$ for some prime $\frp\subset R$. Write $x=g/s$ inside $Q(R)$. In particular, $s\in I$. We can write $y=a/t$ for some $a\in\frp$ and $t\in R\setminus\frp$. We have $g=sy=sa/t$ in $R_\frp$, so there is $u\in R\setminus\frp$ such that $gtu=sau$ in $R$. Hence, $g\in\frp$.
And now, what? I don't know what else to deduce, or if this is the right way at all.
$\def\frp{\mathfrak{p}}$EDIT (24/8/23): Johan corrected the typo: the map $R\to R_\frp$ is always flat (since it is a localization map), regardless of whether it is injective or not. From here, the rest of the SP's proof is sound. (I guess I had a compiling error and my mind halted the first time I read the proof and saw “$R\subset R_\frp$.”)
Here's a little bit more detail in the SP's proof: flatness tells us that $R_\frp\cong R\otimes_RR_\frp\to Q(R)\otimes_RR_\frp\cong S^{-1}(R_\frp)$ is injective, where $S\subset R_\frp$ is the image in $R_\frp$ of the set of nonzerodivisors of $R$. Note that $S^{-1}(R_\frp)\subset Q(R)$ (this is because a localization map always preserve nonzerodivisors). These identifications is what allows to write in the proof “as $R_\frp$ is normal we see that $x\otimes 1\in Q(R)\otimes_R R_\frp$ is an element of $R_\frp$.”
Btw, note that the converse is false in general: there are rings that are integrally closed in its total field of fractions while not being normal (the ring $R=\mathbb{Z}/4\mathbb{Z}$ satisfies $Q(R)=R$; on the other hand, $R$ has a unique prime $\mathfrak{p}=(2)\subset R$, but $R_\mathfrak{p}=R$ is not a domain).
However, the converse turns out to be true if we add the hypotheses of $R$ reduced and finitely many minimal primes, see 030C.