Lemma 15.44.13. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Then

\[ R^ h \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i) \quad \text{resp.}\quad R^{sh} \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , s} \kappa (\mathfrak r_ i) \]

where $\mathfrak q_1, \ldots , \mathfrak q_ t$, resp. $\mathfrak r_1, \ldots , \mathfrak r_ s$ are the prime of $R^ h$, resp. $R^{sh}$ lying over $\mathfrak p$. Moreover, the field extensions $\kappa (\mathfrak p) \subset \kappa (\mathfrak q_ i)$ resp. $\kappa (\mathfrak p) \subset \kappa (\mathfrak r_ i)$ are separable algebraic.

**Proof.**
This can be deduced from the more general Lemma 15.44.12 using that the henselization and strict henselization are Noetherian (as we've seen above). But we also give a direct proof as follows.

We will use without further mention the results of Lemmas 15.44.1 and 15.44.3. Note that $R^ h/\mathfrak pR^ h$, resp. $R^{sh}/\mathfrak pR^{sh}$ is the henselization, resp. strict henselization of $R/\mathfrak p$, see Algebra, Lemma 10.150.10 resp. Algebra, Lemma 10.150.16. Hence we may replace $R$ by $R/\mathfrak p$ and assume that $R$ is a Noetherian local domain and that $\mathfrak p = (0)$. Since $R^ h$, resp. $R^{sh}$ is Noetherian, it has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, resp. $\mathfrak r_1, \ldots , \mathfrak r_ s$. Since $R \to R^ h$, resp. $R \to R^{sh}$ is flat these are exactly the primes lying over $\mathfrak p = (0)$ (by going down). Finally, as $R$ is a domain, we see that $R^ h$, resp. $R^{sh}$ is reduced, see Lemma 15.44.4. Thus we see that $R^ h \otimes _ R \kappa (\mathfrak p)$ resp. $R^{sh} \otimes _ R \kappa (\mathfrak p)$ is a reduced Noetherian ring with finitely many primes, all of which are minimal (and hence maximal). Thus these rings are Artinian and are products of their localizations at maximal ideals, each necessarily a field (see Algebra, Proposition 10.59.6 and Algebra, Lemma 10.24.1).

The final statement follows from the fact that $R \to R^ h$, resp. $R \to R^{sh}$ is a colimit of étale ring maps and hence the induced residue field extensions are colimits of finite separable extensions, see Algebra, Lemma 10.141.5.
$\square$

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