a faithfully flat $A$-algebra. As $B$ is a flat $A$-module, it follows from [@BourbakiAC §IV.2.6 Lem.1] applied with $E=A$ and $F=B$ that $\mathfrak{p}\in\mathrm{Ass}(B)$ implies $\mathfrak{p}\cap A=0$ ($A$ is a domain, so $\mathrm{Ass}(A)=\{0\}$). It then follows from [@BourbakiAC §V.2.1 Cor.1] that if $\mathfrak{p}\in\mathrm{Ass}(B)$, then $\mathfrak{p}$ is a minimal prime of $B$. Indeed, $A$ being noetherian and $B$ a finite $A$-module, $B$ is an integral extension of $A$. We can apply [*loc. cit.*]{} to the inclusion $\mathfrak{q}\subseteq\mathfrak{p}$ where $\mathfrak{q}$ is a minimal prime ideal of $B$ (both ideals $\mathfrak{q}$ and $\mathfrak{p}$ being above the prime ideal $(0)$ of $A$ since $\mathfrak{p}\cap A=\mathfrak{q}\cap A=0$).

Let $\mathfrak{d}_{B/A}$ be the discriminant of $B/A$ (its existence comes from the fact that $B$ is a finite faithfully flat $A$-algebra, hence a finite projective $A$-module). As the extension is generically étale, we can find $f\in\mathfrak{d}_{B/A}$ such that $B_f$ is étale over $A_f$. As $A_f$ is a domain, $B_f$ is then reduced. Thus the nilradical $\mathfrak{n}$ of $B_f$ is killed by some power of $f$. Replacing $f$ by this power, we can assume that the vanishing ideal of $\mathfrak{n}$ contains $f$. Assume that $\mathfrak{n}$ is nonzero and let $\mathfrak{p}$ be a prime ideal of $B$ minimal among prime ideals 