Lemma 10.124.3. Let $R \to S$ be a ring map, $\mathfrak q$ a prime of $S$ lying over $\mathfrak p$ in $R$. If

1. $R$ is Noetherian,

2. $R \to S$ is of finite type, and

3. $R \to S$ is quasi-finite at $\mathfrak q$,

then $R_\mathfrak p^\wedge \otimes _ R S = S_\mathfrak q^\wedge \times B$ for some $R_\mathfrak p^\wedge$-algebra $B$.

Proof. There exists a finite $R$-algebra $S' \subset S$ and an element $g \in S'$, $g \not\in \mathfrak q' = S' \cap \mathfrak q$ such that $S'_ g = S_ g$ and in particular $S'_{\mathfrak q'} = S_\mathfrak q$, see Lemma 10.123.14. We have

$R_\mathfrak p^\wedge \otimes _ R S' = (S'_{\mathfrak q'})^\wedge \times B'$

by Lemma 10.97.8. Observe that under this product decomposition $g$ maps to a pair $(u, b')$ with $u \in (S'_{\mathfrak q'})^\wedge$ a unit because $g \not\in \mathfrak q'$. The product decomposition for $R_\mathfrak p^\wedge \otimes _ R S'$ induces a product decomposition

$R_\mathfrak p^\wedge \otimes _ R S = A \times B$

Since $S'_ g = S_ g$ we also have $(R_\mathfrak p^\wedge \otimes _ R S')_ g = (R_\mathfrak p^\wedge \otimes _ R S)_ g$ and since $g \mapsto (u, b')$ where $u$ is a unit we see that $(S'_{\mathfrak q'})^\wedge = A$. Since the isomorphism $S'_{\mathfrak q'} = S_\mathfrak q$ determines an isomorphism on completions this also tells us that $A = S_\mathfrak q^\wedge$. This finishes the proof, except that we should perform the sanity check that the induced map $\phi : R_\mathfrak p^\wedge \otimes _ R S \to A = S_\mathfrak q^\wedge$ is the natural one. For elements of the form $x \otimes 1$ with $x \in R_\mathfrak p^\wedge$ this is clear as the natural map $R_\mathfrak p^\wedge \to S_\mathfrak q^\wedge$ factors through $(S'_{\mathfrak q'})^\wedge$. For elements of the form $1 \otimes y$ with $y \in S$ we can argue that for some $n \geq 1$ the element $g^ ny$ is the image of some $y' \in S'$. Thus $\phi (1 \otimes g^ ny)$ is the image of $y'$ by the composition $S' \to (S'_{\mathfrak q'})^\wedge \to S_\mathfrak q^\wedge$ which is equal to the image of $g^ ny$ by the map $S \to S_\mathfrak q^\wedge$. Since $g$ maps to a unit this also implies that $\phi (1 \otimes y)$ has the correct value, i.e., the image of $y$ by $S \to S_\mathfrak q^\wedge$. $\square$

Comment #5389 by Bjorn Poonen on

The logic of this proof is not clear to me - how does "the lemma follow" from the previous sentences of the proof?

Also, in the penultimate sentence, "right vertical" should be "right vertical arrow".

Comment #5623 by on

OK, this was a bit of a puzzler. I think you were right that the proof wasn't complete. Please see the corresponding changes here or just wait till I update the site in a few days.

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