Lemma 10.121.6. Let

\[ \xymatrix{ S \ar[r] & S' & & \mathfrak q \ar@{-}[r] & \mathfrak q' \\ R \ar[u] \ar[r] & R' \ar[u] & & \mathfrak p \ar@{-}[r] \ar@{-}[u] & \mathfrak p' \ar@{-}[u] } \]

be a commutative diagram of rings with primes as indicated. Assume $R \to S$ of finite type, and $S \otimes _ R R' \to S'$ surjective. If $R \to S$ is quasi-finite at $\mathfrak q$, then $R' \to S'$ is quasi-finite at $\mathfrak q'$.

**Proof.**
Write $S \otimes _ R \kappa (\mathfrak p) = S_1 \times S_2$ with $S_1$ finite over $\kappa (\mathfrak p)$ and such that $\mathfrak q$ corresponds to a point of $S_1$ as in Lemma 10.121.1. This product decomposition induces a corresponding product decomposition for any $S \otimes _ R \kappa (\mathfrak p)$-algebra. In particular, we obtain $S' \otimes _{R'} \kappa (\mathfrak p') = S'_1 \times S'_2$. Because $S \otimes _ R R' \to S'$ is surjective the canonical map $(S \otimes _ R \kappa (\mathfrak p)) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S' \otimes _{R'} \kappa (\mathfrak p')$ is surjective and hence $S_ i \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S'_ i$ is surjective. It follows that $S'_1$ is finite over $\kappa (\mathfrak p')$. The map $S' \otimes _{R'} \kappa (\mathfrak p') \to \kappa (\mathfrak q')$ factors through $S_1'$ (i.e. it annihilates the factor $S_2'$) because the map $S \otimes _ R \kappa (\mathfrak p) \to \kappa (\mathfrak q)$ factors through $S_1$ (i.e. it annihilates the factor $S_2$). Thus $\mathfrak q'$ corresponds to a point of $\mathop{\mathrm{Spec}}(S_1')$ in the disjoint union decomposition of the fibre: $\mathop{\mathrm{Spec}}(S' \otimes _{R'} \kappa (\mathfrak p')) = \mathop{\mathrm{Spec}}(S_1') \amalg \mathop{\mathrm{Spec}}(S_2')$, see Lemma 10.20.2. Since $S_1'$ is finite over a field, it is Artinian ring, and hence $\mathop{\mathrm{Spec}}(S_1')$ is a finite discrete set. (See Proposition 10.59.6.) We conclude $\mathfrak q'$ is isolated in its fibre as desired.
$\square$

## Comments (3)

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