Lemma 49.2.8. Let $A \to B$ be a quasi-finite homomorphism of Noetherian rings. Then $\text{Ass}_ B(\omega _{B/A})$ is the set of primes of $B$ lying over associated primes of $A$.

Proof. Choose a factorization $A \to B' \to B$ with $A \to B'$ finite and $B' \to B$ inducing an open immersion on spectra. As $\omega _{B/A} = \omega _{B'/A} \otimes _{B'} B$ it suffices to prove the statement for $\omega _{B'/A}$. Thus we may assume $A \to B$ is finite.

Assume $\mathfrak p \in \text{Ass}(A)$ and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p$. Let $x \in A$ be an element whose annihilator is $\mathfrak p$. Choose a nonzero $\kappa (\mathfrak p)$ linear map $\lambda : \kappa (\mathfrak q) \to \kappa (\mathfrak p)$. Since $A/\mathfrak p \subset B/\mathfrak q$ is a finite extension of rings, there is an $f \in A$, $f \not\in \mathfrak p$ such that $f\lambda$ maps $B/\mathfrak q$ into $A/\mathfrak p$. Hence we obtain a nonzero $A$-linear map

$B \to B/\mathfrak q \to A/\mathfrak p \to A,\quad b \mapsto f\lambda (b)x$

An easy computation shows that this element of $\omega _{B/A}$ has annihilator $\mathfrak q$, whence $\mathfrak q \in \text{Ass}(\omega _{B/A})$.

Conversely, suppose that $\mathfrak q \subset B$ is a prime ideal lying over a prime $\mathfrak p \subset A$ which is not an associated prime of $A$. We have to show that $\mathfrak q \not\in \text{Ass}_ B(\omega _{B/A})$. After replacing $A$ by $A_\mathfrak p$ and $B$ by $B_\mathfrak p$ we may assume that $\mathfrak p$ is a maximal ideal of $A$. This is allowed by Lemma 49.2.5 and Algebra, Lemma 10.63.16. Then there exists an $f \in \mathfrak m$ which is a nonzerodivisor on $A$. Then $f$ is a nonzerodivisor on $\omega _{B/A}$ and hence $\mathfrak q$ is not an associated prime of this module. $\square$

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