
Lemma 10.141.22. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Assume $R \to S$ finite type. Then there exists

1. an étale ring map $R \to R'$,

2. a prime $\mathfrak p' \subset R'$ lying over $\mathfrak p$,

3. a product decomposition

$R' \otimes _ R S = A_1 \times \ldots \times A_ n \times B$

with the following properties

1. we have $\kappa (\mathfrak p) = \kappa (\mathfrak p')$,

2. each $A_ i$ is finite over $R'$,

3. each $A_ i$ has exactly one prime $\mathfrak r_ i$ lying over $\mathfrak p'$, and

4. $R' \to B$ not quasi-finite at any prime lying over $\mathfrak p'$.

Proof. Denote $F = S \otimes _ R \kappa (\mathfrak p)$ the fibre ring of $S/R$ at the prime $\mathfrak p$. As $F$ is of finite type over $\kappa (\mathfrak p)$ it is Noetherian and hence $\mathop{\mathrm{Spec}}(F)$ has finitely many isolated closed points. If there are no isolated closed points, i.e., no primes $\mathfrak q$ of $S$ over $\mathfrak p$ such that $S/R$ is quasi-finite at $\mathfrak q$, then the lemma holds. If there exists at least one such prime $\mathfrak q$, then we may apply Lemma 10.141.21. This gives a diagram

$\xymatrix{ S \ar[r] & R'\otimes _ R S \ar@{=}[r] & A_1 \times B' \\ R \ar[r] \ar[u] & R' \ar[u] \ar[ru] }$

as in said lemma. Since the residue fields at $\mathfrak p$ and $\mathfrak p'$ are the same, the fibre rings of $S/R$ and $(A \times B)/R'$ are the same. Hence, by induction on the number of isolated closed points of the fibre we may assume that the lemma holds for $R' \to B$ and $\mathfrak p'$. Thus we get an étale ring map $R' \to R''$, a prime $\mathfrak p'' \subset R''$ and a decomposition

$R'' \otimes _{R'} B' = A_2 \times \ldots \times A_ n \times B$

We omit the verification that the ring map $R \to R''$, the prime $\mathfrak p''$ and the resulting decomposition

$R'' \otimes _ R S = (R'' \otimes _{R'} A_1) \times A_2 \times \ldots \times A_ n \times B$

is a solution to the problem posed in the lemma. $\square$

Comment #3966 by Manuel Hoff on

Typo: In tge sentence after the diagram, one $A$ lacks a $_1$.

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