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

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 10.122.1. 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 10.21.2. 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 10.60.7.) We conclude $\mathfrak q'$ is isolated in its fibre as desired. $\square$

Comment #3960 by Manuel Hoff on

Typo: In the disjoint union decomposition in the proof, it says two times $S_1'$.

Comment #3964 by Manuel Hoff on

One more thing, that confuses me about the proof: The $S_ i \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$ aren't subrings of $(S \otimes _ R \kappa (\mathfrak p)) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$,  so it doesn't really make sense for me to consider their images in $S' \otimes _{R'} \kappa (\mathfrak p')$ (however, they can be considered as sub-vectorspaces of, which loses their structure of being rings).

The main point of the argument is (or so I think) that the decomposition $S \otimes _ R \kappa (\mathfrak p) = S_1 \times S_2$ induces a decomposition $S' \otimes _{R'} \kappa (\mathfrak p') =S'_1 \times S'_2$ with ring maps $S_i \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S_i'$ that fit into a commutative diagram.

Comment #4098 by on

OK, if $R = R_1 \times R_2$ is a product of rings over a field k, then $R_1$ is a non-unital-sub-k-algebra of $R$. Thinking of it that way, I think the original proof works fine. But the way you say it is clearer, so I've made the changes you suggested. I've left the "induced decomposition" to the reader to figure out... we could add this earlier in the chapter. See this commit.

Comment #4321 by on

Greetings, i think clarifying what the induced decomposition is would be helpful.

Comment #4322 by on

@#4321: Suppose we have a ring map $A \to B$ and a product decomposition $A = A_1 \times A_2$ of rings. Then we get an induced product decomposition $B = B_1 \times B_2$ by setting $B_i$ equal to $e_i B$ where $e_i \in A$ is the idempotent corresponding to the factor $A_i$. So $e_1 = (1, 0) \in A = A_1 \times A_2$ and $e_2 = (0, 1) \in A = A_1 \times A_2$. Another way to say this is that it is the unique product decomposition $B = B_1 \times B_2$ of $B$ such that $A \to B$ sends $A_i$ into $B_i$.

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