Lemma 61.8.7. Let $A \to B$ be a ring map inducing algebraic extensions on residue fields. There exists a commutative diagram

\[ \xymatrix{ B \ar[r] & D \\ A \ar[r] \ar[u] & C \ar[u] } \]

with the following properties:

$A \to C$ is faithfully flat and ind-étale,

$B \to D$ is faithfully flat and ind-étale,

$\mathop{\mathrm{Spec}}(C)$ is w-local,

$\mathop{\mathrm{Spec}}(D)$ is w-local,

$\mathop{\mathrm{Spec}}(D) \to \mathop{\mathrm{Spec}}(C)$ is w-local,

the set of closed points of $\mathop{\mathrm{Spec}}(D)$ is the inverse image of the set of closed points of $\mathop{\mathrm{Spec}}(C)$,

the set of closed points of $\mathop{\mathrm{Spec}}(C)$ surjects onto $\mathop{\mathrm{Spec}}(A)$,

the set of closed points of $\mathop{\mathrm{Spec}}(D)$ surjects onto $\mathop{\mathrm{Spec}}(B)$,

for $\mathfrak m \subset C$ maximal the local ring $C_\mathfrak m$ is strictly henselian.

**Proof.**
There is a faithfully flat, ind-Zariski ring map $A \to A'$ such that $\mathop{\mathrm{Spec}}(A')$ is w-local and such that the set of closed points of $\mathop{\mathrm{Spec}}(A')$ maps onto $\mathop{\mathrm{Spec}}(A)$, see Lemma 61.5.3. Let $I \subset A'$ be the ideal such that $V(I)$ is the set of closed points of $\mathop{\mathrm{Spec}}(A')$. Choose $A' \to C'$ as in Lemma 61.8.1. Note that the local rings $C'_{\mathfrak m'}$ at maximal ideals $\mathfrak m' \subset C'$ are strictly henselian by Lemma 61.8.5. We apply Lemma 61.5.8 to $A' \to C'$ and $I \subset A'$ to get $C' \to C$ with $C'/IC' \cong C/IC$. Note that since $A' \to C'$ is faithfully flat, $\mathop{\mathrm{Spec}}(C'/IC')$ surjects onto the set of closed points of $A'$ and in particular onto $\mathop{\mathrm{Spec}}(A)$. Moreover, as $V(IC) \subset \mathop{\mathrm{Spec}}(C)$ is the set of closed points of $C$ and $C' \to C$ is ind-Zariski (and identifies local rings) we obtain properties (1), (3), (7), and (9).

Denote $J \subset C$ the ideal such that $V(J)$ is the set of closed points of $\mathop{\mathrm{Spec}}(C)$. Set $D' = B \otimes _ A C$. The ring map $C \to D'$ induces algebraic residue field extensions. Keep in mind that since $V(J) \to \mathop{\mathrm{Spec}}(A)$ is surjective the map $T = V(JD') \to \mathop{\mathrm{Spec}}(B)$ is surjective too. Apply Lemma 61.5.8 to $C \to D'$ and $J \subset C$ to get $D' \to D$ with $D'/JD' \cong D/JD$. All of the remaining properties given in the lemma are immediate from the results of Lemma 61.5.8.
$\square$

## Comments (2)

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