Lemma 10.156.3. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p$ in $R$. Assume $R \to S$ is quasi-finite at $\mathfrak q$. Let $\kappa _2^{sep}/\kappa (\mathfrak q)$ be a separable algebraic closure and denote $\kappa _1^{sep} \subset \kappa _2^{sep}$ the subfield of elements separable algebraic over $\kappa (\mathfrak q)$ (Fields, Lemma 9.14.6). The commutative diagram

$\xymatrix{ R_{\mathfrak p}^{sh} \ar[r] & S_{\mathfrak q}^{sh} \\ R_{\mathfrak p} \ar[u] \ar[r] & S_{\mathfrak q} \ar[u] }$

of Lemma 10.155.10 identifies $S_{\mathfrak q}^{sh}$ with the localization of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at the prime ideal which is the kernel of the map

$R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q} \longrightarrow \kappa _1^{sep} \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak q) \longrightarrow \kappa _2^{sep}$

Moreover, the ring map $R_{\mathfrak p}^{sh} \to S_{\mathfrak q}^{sh}$ is a finite local homomorphism of local rings whose residue field extension is the extension $\kappa _2^{sep}/\kappa _1^{sep}$ which is both finite and purely inseparable.

Proof. Since $R \to S$ is quasi-finite at $\mathfrak q$ we see that the extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is finite, see Definition 10.122.3 and Lemma 10.122.2. Hence $\kappa _1^{sep}$ is a separable algebraic closure of $\kappa (\mathfrak p)$ (small detail omitted). In particular Lemma 10.155.10 does really apply. Next, the compositum of $\kappa (\mathfrak p)$ and $\kappa _1^{sep}$ in $\kappa _2^{sep}$ is separably algebraically closed and hence equal to $\kappa _2^{sep}$. We conclude that $\kappa _2^{sep}/\kappa _1^{sep}$ is finite. By construction the extension $\kappa _2^{sep}/\kappa _1^{sep}$ is purely inseparable. The ring map $R_{\mathfrak p}^{sh} \to S_{\mathfrak q}^{sh}$ is indeed local and induces the residue field extension $\kappa _2^{sep}/\kappa _1^{sep}$ which is indeed finite purely inseparable.

Note that $R_{\mathfrak p}^{sh} \otimes _ R S$ is quasi-finite over $R_{\mathfrak p}^{sh}$ at the prime ideal $\mathfrak q'$ given in the statement of the lemma, see Lemma 10.122.6. Hence the localization $S'$ of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at $\mathfrak q'$ is henselian and finite over $R_{\mathfrak p}^{sh}$, see Lemma 10.153.4. Note that the residue field of $S'$ is $\kappa _2^{sep}$ as the map $\kappa _1^{sep} \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak q) \to \kappa _2^{sep}$ is surjective by the discussion in the previous paragraph. Furthermore, as a localization $S'$ is a filtered colimit of étale $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$-algebras. By Lemma 10.155.12 we see that $S_{\mathfrak q}^{sh}$ is the strict henselization of $R_{\mathfrak p}^{sh} \otimes _{R_{\mathfrak p}} S_{\mathfrak q}$ at $\mathfrak q'$. Thus $S' = S_\mathfrak q^{sh}$ by the uniqueness result of Lemma 10.154.7. $\square$

## Comments (2)

Comment #4679 by Peng DU on

In the last formula in the Proof, it should be $S'=S_{\mathfrak{q}}^{sh}$.

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