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 clearly 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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