Skip to content
Permalink
Browse files

Strengthen a lemma

  • Loading branch information...
aisejohan committed Jan 25, 2019
1 parent 6bbb4fb commit 69b19486e0ac418a81261da197298c33651b3e05
Showing with 9 additions and 5 deletions.
  1. +9 −5 algebra.tex

\begin{lemma}
\label{lemma-quasi-finite-permanence}
Let $A \to B$ and $B \to C$ be finite type ring homomorphisms.
Let $\mathfrak r$ be a prime of $C$ lying over
Let $A \to B$ and $B \to C$ be ring homomorphisms such that $A \to C$
is of finite type. Let $\mathfrak r$ be a prime of $C$ lying over
$\mathfrak q \subset B$ and $\mathfrak p \subset A$.
If $A \to C$ is quasi-finite at $\mathfrak r$, then
$B \to C$ is quasi-finite at $\mathfrak r$.
\end{lemma}

\begin{proof}
Using property (3) of Lemma \ref{lemma-isolated-point-fibre}:
By assumption there exists some $c \in C$ such that
Observe that $B \to C$ is of finite type
(Lemma \ref{lemma-compose-finite-type})
so that the statement makes sense.
Let us use characterization (3) of Lemma \ref{lemma-isolated-point-fibre}.
If $A \to C$ is quasi-finite at $\mathfrak r$, then
there exists some $c \in C$ such that
$$
\{\mathfrak r' \subset C \text{ lying over }\mathfrak p\} \cap D(c) =
\{\mathfrak{r}\}.
$$
Since the primes $\mathfrak r' \subset C$ lying over $\mathfrak q$
form a subset of the primes $\mathfrak r' \subset C$ lying over
$\mathfrak p$ we conclude.
$\mathfrak p$ we conclude $B \to C$ is quasi-finite at $\mathfrak r$.
\end{proof}

\noindent

0 comments on commit 69b1948

Please sign in to comment.
You can’t perform that action at this time.