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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
rj7k8
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2 Answers2

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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 $\mathfrak{a}=0$.

rj7k8
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user2902293
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  • Sometimes 'normal ring' means a ring which is reduced, that is integrally closed in its 'Total ring of fractions'. Not sure if Frietag/Kiehl, means by convention that 'normal ring' is a domain. – Evans Gambit Jul 30 '21 at 05:30
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Despite of great answer by user2902293, I would like to present another solution which looks like such one Freitag/Kiehl might had in mind in their book:

$A \to \tilde B$ is local-etale, therefore unramified, therefore quasi-finite. In general $A \to \tilde B$ is not finite, but on page 7 the authors remarked that by Zariski Main Theorem there exist a finite $A$-algebra $C$, such that $\tilde B$ arises as localization of $C$. In summary we have factorization $A \to C \to \tilde B$ where the first one is finite and second is localization. Now we consider the preimage of $\mathfrak{a}$ with respect these morphisms.

Assume $\mathfrak{a} \neq 0$. It should be clear that its preimage with resp a localization
$A \cap \mathfrak{a}$ stays non-zero (why?) and the preimage of non-zero ideal with respect integral map (finite is integral) is also non-zero (Finite extension of domains : preimage of non-zero ideal is non-zero?)

Therefore $A \cap \mathfrak{a} \neq 0$ and is mapped by $A \to B$ to zero, but it was assumed to be injective, a contradiction!

user267839
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