The Stacks project

Lemma 10.153.11. Let $R \to S$ be a ring map with $S$ henselian local. Given

  1. an ├ętale ring map $R \to A$,

  2. a prime $\mathfrak q$ of $A$ lying over $\mathfrak p = R \cap \mathfrak m_ S$,

  3. a $\kappa (\mathfrak p)$-algebra map $\tau : \kappa (\mathfrak q) \to S/\mathfrak m_ S$,

then there exists a unique homomorphism of $R$-algebras $f : A \to S$ such that $\mathfrak q = f^{-1}(\mathfrak m_ S)$ and $f \bmod \mathfrak q = \tau $.

Proof. Consider $A \otimes _ R S$. This is an ├ętale algebra over $S$, see Lemma 10.143.3. Moreover, the kernel

\[ \mathfrak q' = \mathop{\mathrm{Ker}}(A \otimes _ R S \to \kappa (\mathfrak q) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak m_ S) \to \kappa (\mathfrak m_ S)) \]

of the map using the map given in (3) is a prime ideal lying over $\mathfrak m_ S$ with residue field equal to the residue field of $S$. Hence by Lemma 10.153.3 there exists a unique splitting $\tau : A \otimes _ R S \to S$ with $\tau ^{-1}(\mathfrak m_ S) = \mathfrak q'$. Set $f$ equal to the composition $A \to A \otimes _ R S \to S$. $\square$

Comments (1)

Comment #8562 by on

A few comments:

  1. is used both in the proof and the statement in different roles.
  2. Statement (3): Strictly speaking, a map mod becomes which is not of the form . So, perhaps, make this statement more precise.
  3. Because this statement is an important fact, I suggest making the proof a bit more transparent. For instance, more explanation can be added in the proof as to why reduces to , possibly highlighting the role of the uniqueness of the splitting.

There are also:

  • 6 comment(s) on Section 10.153: Henselian local rings

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