**Proof.**
By Lemma 71.7.5 we see that $U \in X_{\acute{e}tale}$ affine $\mathcal{S}_ X(U) \subset \mathcal{O}_ X(U)$ is the set of nonzerodivisors in $\mathcal{O}_ X(U)$. Thus the presheaf $\mathcal{S}^{-1}\mathcal{O}_ X$ is equal to

\[ U \longmapsto Q(\mathcal{O}_ X(U)) \]

on $X_{affine, {\acute{e}tale}}$, with notation as in Algebra, Example 10.9.8. Observe that the codimension $0$ points of $X$ correspond to the generic points of $U$, see Properties of Spaces, Lemma 66.11.1. Hence if $U = \mathop{\mathrm{Spec}}(A)$, then $A$ is a ring with finitely many minimal primes such that any weakly associated prime of $A$ is minimal. The same is true for any étale extension of $A$ (because the spectrum of such is an affine scheme étale over $X$ hence can play the role of $A$ in the previous sentence). In order to show that our presheaf is a sheaf and quasi-coherent it suffices to show that

\[ Q(A) \otimes _ A B \longrightarrow Q(B) \]

is an isomorphism when $A \to B$ is an étale ring map, see Properties of Spaces, Lemma 66.29.3. (To define the displayed arrow, observe that since $A \to B$ is flat it maps nonzerodivisors to nonzerodivisors.) By Algebra, Lemmas 10.25.4 and 10.66.7. we have

\[ Q(A) = \prod \nolimits _{\mathfrak p \subset A\text{ minimal}} A_\mathfrak p \quad \text{and}\quad Q(B) = \prod \nolimits _{\mathfrak q \subset B\text{ minimal}} B_\mathfrak q \]

Since $A \to B$ is étale, the minimal primes of $B$ are exactly the primes of $B$ lying over the minimal primes of $A$ (for example by More on Algebra, Lemma 15.44.2). By Algebra, Lemmas 10.153.10, 10.153.3 (13), and 10.153.5 we see that $A_\mathfrak p \otimes _ A B$ is a finite product of local rings finite étale over $A_\mathfrak p$. This cleary implies that $A_\mathfrak p \otimes _ A B = \prod _{\mathfrak q\text{ lies over }\mathfrak p} B_\mathfrak q$ as desired.

At this point we know that (1) and (2) hold. Proof of (3). Let $s \in \mathcal{O}_{X, \overline{x}}$ be a nonzerodivisor. Then we can find an étale neighbourhood $(U, \overline{u}) \to (X, \overline{x})$ and $f \in \mathcal{O}_ X(U)$ mapping to $s$. Let $u \in U$ be the point determined by $\overline{u}$. Since $\mathcal{O}_{U, u} \to \mathcal{O}_{X, \overline{x}}$ is faithfully flat (as a strict henselization), we see that $f$ maps to a nonzerodivisor in $\mathcal{O}_{U, u}$. By Divisors, Lemma 31.23.6 after shrinking $U$ we find that $f$ is a nonzerodivisor and hence a section of $\mathcal{S}_ X(U)$. Part (4) follows from (3) by computing stalks.
$\square$

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