Lemma 10.145.3. 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
an étale ring map R \to R',
a prime \mathfrak p' \subset R' lying over \mathfrak p,
a product decomposition
R' \otimes _ R S = A_1 \times \ldots \times A_ n \times B
with the following properties
we have \kappa (\mathfrak p) = \kappa (\mathfrak p'),
each A_ i is finite over R',
each A_ i has exactly one prime \mathfrak r_ i lying over \mathfrak p', and
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 10.145.2. 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_1 \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
Comments (2)
Comment #3966 by Manuel Hoff on
Comment #4101 by Johan on