Lemma 15.44.12. Let $A$ be a ring. Let $B$ be a filtered colimit of étale $A$-algebras. Let $\mathfrak p$ be a prime of $A$. If $B$ is Noetherian, then there are finitely many primes $\mathfrak q_1, \ldots , \mathfrak q_ r$ lying over $\mathfrak p$, we have $B \otimes _ A \kappa (\mathfrak p) = \prod \kappa (\mathfrak q_ i)$, and each of the field extensions $\kappa (\mathfrak p) \subset \kappa (\mathfrak q_ i)$ is separable algebraic.

**Proof.**
Write $B$ as a filtered colimit $B = \mathop{\mathrm{colim}}\nolimits B_ i$ with $A \to B_ i$ étale. Then on the one hand $B \otimes _ A \kappa (\mathfrak p) = \mathop{\mathrm{colim}}\nolimits B_ i \otimes _ A \kappa (\mathfrak p)$ is a filtered colimit of étale $\kappa (\mathfrak p)$-algebras, and on the other hand it is Noetherian. An étale $\kappa (\mathfrak p)$-algebra is a finite product of finite separable field extensions (Algebra, Lemma 10.141.4). Hence there are no nontrivial specializations between the primes (which are all maximal and minimal primes) of the algebras $B_ i \otimes _ A \kappa (\mathfrak p)$ and hence there are no nontrivial specializations between the primes of $B \otimes _ A \kappa (\mathfrak p)$. Thus $B \otimes _ A \kappa (\mathfrak p)$ is reduced and has finitely many primes which all minimal. Thus it is a finite product of fields (use Algebra, Lemma 10.24.4 or Algebra, Proposition 10.59.6). Each of these fields is a colimit of finite separable extensions and hence the final statement of the lemma follows.
$\square$

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