I'm trying to understand the proof of the following lemma in Freitag/Kiehl, "Etale Cohomology…":
1.5 Lemma. Let $A \rightarrow B$ be a finitely generated local homomorphism. We assume that it is injective and that $A$ is a normal ring. If $B$ is unramified over $A$, then $B$ is a local-etale $A$-algebra.
It had previously been shown that $B=\tilde{B}/\mathfrak a$, where $\tilde B$ is local-etale over $A$. Taking this for granted, the proof continues:
As $A$ is normal, so is $\tilde B$ … ,and thus without zero divisors. From the injectivity of $A \rightarrow B$ we conclude $\mathfrak a = 0$.
I don't understand how we conclude that $\mathfrak a =0$ based on the preceding information. (I accept that $\tilde{B}$ is normal.) Would someone be kind enough to help me understand?
Note: $f: A \rightarrow B$ local-etale means:
- $f$ is a localization of a finitely generated morphism
- $f$ is flat and unramified
Best Answer
Here's a short proof assuming the notation and content of Frietag/Kiehl before this lemma.
We know that $A\to B$ is injective and it is factored as $A\to \tilde{B}\to B$, where $A\to \tilde{B}$ is local-etale and $\tilde{B}\to B$ is surjective. Now since $A$ is normal, in particular a domain we let $K$ be its field of fractions.
Since $A$ injects into $B$ (and $K$ is flat $A$-module) so does $A\otimes_AK=K$ into $B\otimes_A K$. Thus the morphism $\tilde{B}\otimes_A K\to B\otimes_A K$ is non-zero.
Here's the crucial point. Since $A\to \tilde{B}$ is etale upon tensoring with $K$ it still remains an etale extension and etale extension where the base is a field must be a finite separable field extension (see Remark 1.2 in Frietag/Kiehl). It is not hard to see that $\tilde{B}\otimes_A K$ is the field of fractions of $\tilde{B}$.
As $\tilde{B}\otimes_A K$ is a field and since the morphism to $B\otimes_A K$ is non-zero it must be injective. Thus $\tilde{B}\otimes_A K\to B\otimes_A K$ is an isomorphism. Since $\tilde{B}$ injects into its field of fractions $\tilde{B}\otimes_A K$ we conclude $\tilde{B}\to B$ is injective and thus an isomorphism. Consequently $\mathfrak{a}=0$.