The Stacks project

Proof. Let $B''' \subset B$ be the $A$-subalgebra generated by the images of $B' \to B$ and $B'' \to B$. As $B'$ and $B''$ are each generated by finitely many elements integral over $A$, we see that $B'''$ is generated by finitely many elements integral over $A$ and we conclude that $B'''$ is finite over $A$ (Algebra, Lemma 10.36.5). Consider the maps

The final equality holds because $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is an open immersion (and hence a monomorphism). The second arrow is injective as $B' \to B$ is flat. Hence both arrows are isomorphisms. This means that

is cartesian. Since the base change of an open immersion is an open immersion we conclude. $\square$

Proof. Let $B', B'', B'''$ be as in Lemma 49.2.1. We obtain a canonical map

and a similar one involving $B''$. If we show these maps are isomorphisms then the lemma is proved. Let $g \in B'$ be an element such that $B'_ g \to B_ g$ is an isomorphism and hence $B'_ g \to (B''')_ g \to B_ g$ are isomorphisms. It suffices to show that $(\omega ''')_ g \to \omega '_ g$ is an isomorphism. The kernel and cokernel of the ring map $B' \to B'''$ are finite $A$-modules and $g$-power torsion. Hence they are annihilated by a power of $g$. This easily implies the result. $\square$

Proof. Say $A \to B' \to B$ is a factorization with $A \to B'$ finite and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ an open immersion. In case (1) we may use the factorization $A_ f \to B'_ f \to B$ to compute $\omega _{B/A_ f}$ and use Algebra, Lemma 10.10.2. In case (2) use the factorization $A \to B' \to B_ g$ to see the result. Part (3) follows from a combination of (1) and (2). $\square$

Proof. Omitted. Hint: argue in exactly the same way as in Lemma 49.2.2 using Lemma 49.2.1. $\square$

Proof. Assume that $A \to A_1$ is flat. By construction of $\omega _{B/A}$ we may assume that $A \to B$ is finite. Then $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ and $\omega _{B_1/A_1} = \mathop{\mathrm{Hom}}\nolimits _{A_1}(B_1, A_1)$. Since $B_1 = B \otimes _ A A_1$ the result follows from More on Algebra, Lemma 15.65.4. $\square$

Proof. Choose $A \to B' \to B$ with $A \to B'$ finite such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is an open immersion. Then $B' \to C$ is quasi-finite too. Choose $B' \to C' \to C$ with $B' \to C'$ finite and $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(C')$ an open immersion. Then the source of the arrow is

which is equal to

This indeed comes with a canonical map to $\mathop{\mathrm{Hom}}\nolimits _ A(C', A) \otimes _{C'} C = \omega _{C/A}$ coming from composition $\mathop{\mathrm{Hom}}\nolimits _ A(B', A) \times \mathop{\mathrm{Hom}}\nolimits _{B'}(C', B) \to \mathop{\mathrm{Hom}}\nolimits _ A(C', A)$. $\square$

Proof. Choose factorizations $A \to B' \to B$ and $A \to C' \to C$ such that $A \to B'$ and $A \to C'$ are finite and such that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ and $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(C')$ are open immersions. Then $A \to B' \times C' \to B \times C$ is a similar factorization. Using this factorization to compute $\omega _{B \times C/A}$ gives the lemma. $\square$

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

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. $\square$

Proof. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. We will show that the localization $\omega _{B/A, \mathfrak q}$ is flat over $A_\mathfrak p$. This suffices by Algebra, Lemma 10.39.18. By Algebra, Lemma 10.145.2 we can find an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p') = \kappa (\mathfrak p)$ and such that

with $A' \to C$ finite and such that the unique prime $\mathfrak q'$ of $B \otimes _ A A'$ lying over $\mathfrak q$ and $\mathfrak p'$ corresponds to a prime of $C$. By Lemma 49.2.5 and Algebra, Lemma 10.100.1 it suffices to show $\omega _{B'/A', \mathfrak q'}$ is flat over $A'_{\mathfrak p'}$. Since $\omega _{B'/A'} = \omega _{C/A'} \times \omega _{D/A'}$ by Lemma 49.2.7 this reduces us to the case where $B$ is finite flat over $A$. In this case $B$ is finite locally free as an $A$-module and $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ is the dual finite locally free $A$-module. $\square$

Proof. If $A \to B$ is finite flat, then $B$ is finite locally free as an $A$-module. In this case $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ is the dual finite locally free $A$-module and formation of this module commutes with arbitrary base change which proves the lemma in this case. In the next paragraph we reduce the general (quasi-finite flat) case to the finite flat case just discussed.

Let $\mathfrak q_1 \subset B_1$ be a prime. We will show that the localization of the map at the prime $\mathfrak q_1$ is an isomorphism, which suffices by Algebra, Lemma 10.23.1. Let $\mathfrak q \subset B$ and $\mathfrak p \subset A$ be the prime ideals lying under $\mathfrak q_1$. By Algebra, Lemma 10.145.2 we can find an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p') = \kappa (\mathfrak p)$ and such that

with $A' \to C$ finite and such that the unique prime $\mathfrak q'$ of $B \otimes _ A A'$ lying over $\mathfrak q$ and $\mathfrak p'$ corresponds to a prime of $C$. Set $A'_1 = A' \otimes _ A A_1$ and consider the base change maps (49.2.3.1) for the ring maps $A \to A' \to A'_1$ and $A \to A_1 \to A'_1$ as in the diagram

where $B' = B \otimes _ A A'$, $B_1 = B \otimes _ A A_1$, and $B_1' = B \otimes _ A (A' \otimes _ A A_1)$. By Lemma 49.2.4 the diagram commutes. By Lemma 49.2.5 the vertical arrows are isomorphisms. As $B_1 \to B'_1$ is étale and hence flat it suffices to prove the top horizontal arrow is an isomorphism after localizing at a prime $\mathfrak q'_1$ of $B'_1$ lying over $\mathfrak q$ (there is such a prime and use Algebra, Lemma 10.39.17). Thus we may assume that $B = C \times D$ with $A \to C$ finite and $\mathfrak q$ corresponding to a prime of $C$. In this case the dualizing module $\omega _{B/A}$ decomposes in a similar fashion (Lemma 49.2.7) which reduces the question to the finite flat case $A \to C$ handled above. $\square$

Proof. Choose a factorization $A \to B' \to B$ where $A \to B'$ is finite and $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(B)$ is an open immersion. Then $\omega _{B/A}^\bullet = \omega _{B'/A}^\bullet \otimes _ B^\mathbf {L} B'$ by Dualizing Complexes, Lemmas 47.24.7 and 47.24.9 and the definition of $\omega _{B/A}^\bullet $. Hence it suffices to show there is an isomorphism when $A \to B$ is finite. In this case we can use Dualizing Complexes, Lemma 47.24.8 to see that $\omega _{B/A}^\bullet = R\mathop{\mathrm{Hom}}\nolimits (B, A)$ and hence $H^0(\omega ^\bullet _{B/A}) = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as desired. $\square$