The Stacks project

Proof. Since $A \to B$ is flat we have $H^ i(K) \otimes _ A B = H^ i(K \otimes _ A^\mathbf {L} B)$. Since $K \otimes _ A^\mathbf {L} B$ is in $D^ b_{\textit{Coh}}(B)$ we first find that $K$ is in $D^ b(A)$ and then we see that $H^ i(K)$ is a finite $A$-module by Algebra, Lemma 10.83.2. Let $M$ be a finite $A$-module. Then

by More on Algebra, Lemma 15.99.2. Since $K \otimes _ A^\mathbf {L} B$ has finite injective dimension, say injective-amplitude in $[a, b]$, we see that the right hand side has vanishing cohomology in degrees $> b$. Since $A \to B$ is faithfully flat, we find that $R\mathop{\mathrm{Hom}}\nolimits _ A(M, K)$ has vanishing cohomology in degrees $> b$. Thus $K$ has finite injective dimension by More on Algebra, Lemma 15.69.2. To finish the proof we have to show that the map $A \to R\mathop{\mathrm{Hom}}\nolimits _ A(K, K)$ is an isomorphism. For this we again use More on Algebra, Lemma 15.99.2 and the fact that $B \to R\mathop{\mathrm{Hom}}\nolimits _ B(K \otimes _ A^\mathbf {L} B, K \otimes _ A^\mathbf {L} B)$ is an isomorphism. $\square$

Proof. Observe that $A \to B$ satisfies (1) if and only if $A \to B$ satisfies (2) by More on Algebra, Lemma 15.33.5. Observe that in both (2) and (3) the relative dualzing complex $\varphi ^!(A) = \omega _{B/A}^\bullet$ is an invertible object of $D(B)$, see Lemmas 47.25.4 and 47.25.5. Moreover we have $\varphi ^!(K) = K \otimes _ A^\mathbf {L} \omega _{B/A}^\bullet$ in both cases, see Lemma 47.24.10 for case (3). Thus $\varphi ^!(K)$ is the same as $K \otimes _ A^\mathbf {L} B$ up to tensoring with an invertible object of $D(B)$. Hence $\varphi ^!(K)$ is a dualizing complex for $B$ if and only if $K \otimes _ A^\mathbf {L} B$ is (as being a dualizing complex is local and invariant under shifts). Thus we see that if $K$ is dualizing for $A$, then $K \otimes _ A^\mathbf {L} B$ is dualizing for $B$ by Lemma 47.24.3. To descend the property, see Lemma 47.26.1. $\square$

Proof. Write $E = \bigcup E_ n$ as in Lemma 47.7.3. It suffices to show that $E_ n \otimes _{A/\mathfrak m^ n} B/\mathfrak n^ n$ is the injective hull of $l$ over $B/\mathfrak n$. This reduces us to the case where $A$ and $B$ are Artinian local. Observe that $\text{length}_ A(A) = \text{length}_ B(B)$ and $\text{length}_ A(E) = \text{length}_ B(E \otimes _ A B)$ by Algebra, Lemma 10.52.13. By Lemma 47.6.1 we have $\text{length}_ A(E) = \text{length}_ A(A)$ and $\text{length}_ B(E') = \text{length}_ B(B)$ where $E'$ is the injective hull of $l$ over $B$. We conclude $\text{length}_ B(E') = \text{length}_ B(E \otimes _ A B)$. Observe that

where we have used flatness of $A \to B$ and $\mathfrak n = \mathfrak mB$. Thus there is an injective $B$-module map $E \otimes _ A B \to E'$ by Lemma 47.7.2. By equality of lengths shown above this is an isomorphism. $\square$

Proof. Étale maps satisfy the assumption by Algebra, Lemma 10.143.5. By Lemma 47.3.7 and Proposition 47.5.9 we may assume $I$ is the injective hull of $\kappa (\mathfrak p)$ for some prime $\mathfrak p \subset A$. Then $I$ is a module over $A_\mathfrak p$. It suffices to prove $I \otimes _ A B = I \otimes _{A_\mathfrak p} B_\mathfrak p$ is injective as a $B_\mathfrak p$-module, see Lemma 47.3.2. Thus we may assume $(A, \mathfrak m, \kappa )$ is local Noetherian and $I = E$ is the injective hull of the residue field $\kappa$. Our assumption implies that the Noetherian ring $B/\mathfrak m B$ is a product of fields (details omitted). Thus there are finitely many prime ideals $\mathfrak m_1, \ldots , \mathfrak m_ n$ in $B$ lying over $\mathfrak m$ and they are all maximal ideals. Write $E = \bigcup E_ n$ as in Lemma 47.7.3. Then $E \otimes _ A B = \bigcup E_ n \otimes _ A B$ and $E_ n \otimes _ A B$ is a finite $B$-module with support $\{ \mathfrak m_1, \ldots , \mathfrak m_ n\}$ hence decomposes as a product over the localizations at $\mathfrak m_ i$. Thus $E \otimes _ A B = \prod (E \otimes _ A B)_{\mathfrak m_ i}$. Since $(E \otimes _ A B)_{\mathfrak m_ i} = E \otimes _ A B_{\mathfrak m_ i}$ is the injective hull of the residue field of $\mathfrak m_ i$ by Lemma 47.26.3 we conclude. $\square$