Generalized Trace Submodules and Centers of Endomorphism Rings

It’s clear that if $R_{\mathfrak{p}}\in\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ for all $\mathfrak{p}\in\operatorname{Ass}_{R}(R)$ that $M$ must be faithful. For the converse, suppose $M$ is faithful so that there is an injection $j:R\to M^{\oplus r}$ for some $r$. As $M$ is torsionless, there is an injection $M\xrightarrow{i}R^{\oplus n}$ for some $n$. Then there is an exact sequence of the form $0\rightarrow R\xrightarrow{i^{\oplus r}\circ j}R^{\oplus nr}\to C\to 0$, so $\operatorname{pd}_{R}(C)\leq 1$. In particular, $C_{\mathfrak{p}}$ has finite projective dimension for all $\mathfrak{p}\in\operatorname{Ass}(R)$, but as $R_{\mathfrak{p}}$ has depth zero, the Auslander-Buchsbaum formula forces $C_{\mathfrak{p}}$ to be a free $R_{\mathfrak{p}}$-module. In particular $i^{\oplus r}\circ j$ is a split injection locally at every $\mathfrak{p}\in\operatorname{Ass}(R)$. Then so is $j$, and so $R_{\mathfrak{p}}\in\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ for all $\mathfrak{p}\in\operatorname{Ass}(R)$, as desired. ∎

Note $(1)$ is immediate given Remark 2.5. For $(2)$, if $L\in\operatorname{Add}_{R}(M)$, then there is a split injection $i:L\to M^{\oplus n}$ for some $n$, with retraction $p:M^{\oplus n}\to L$. There is a commutative diagram:

where $a,b$ and $\eta$ are the natural isomorphisms. In particular, $\phi^{M}_{M^{\oplus n},N}$ is an isomorphism. But there is also a commutative diagram

As $i$ is a split injection the commutativity of the top square forces $\phi^{M}_{L,N}$ to be injective, while that of the bottom square forces it to be surjective, so it is an isomorphism. The case where $N\in\operatorname{Add}_{R}(M)$ instead follows from a nearly identical argument.

For $(3)$, suppose $A$ generates $B$ covariantly with respect to $L$, so there is a map $p:A^{\oplus m}\to B$ such that $\operatorname{Hom}_{R}(L,p)$ is a surjection. Suppose we have $g\in\operatorname{tr}_{L,N}(B)$, so that there is a commutative diagram

for some $n$. Since $\operatorname{Hom}_{R}(L,p)$ is surjective, so is $\operatorname{Hom}_{R}(L,p^{\oplus n})$, and thus there is a map $q\in\operatorname{Hom}_{R}(L,A^{\oplus mn})$ so that $p^{\oplus n}\circ q=t$. Then there is a commutative diagram

which shows $g\in\operatorname{tr}_{L,N}(A)$. The case where $A$ generates $B$ contravariantly with respect to $N$ follows similarly.

For $(4)$, from part (3) we have either $\operatorname{tr}_{L,N}(L)\subseteq\operatorname{tr}_{L,N}(M)$ when $M$ generates $N$ covariantly with respect to $L$ or $\operatorname{tr}_{L,N}(N)\subseteq\operatorname{tr}_{L,N}(M)$ when $M$ generates $L$ contravariantly with respect to $N$. But from part $(2)$, we have $\operatorname{tr}_{L,N}(L)=\operatorname{tr}_{L,N}(N)=\operatorname{Hom}_{R}(L,N)$, so $\phi^{M}_{L,N}$ is surjective.

For $(5)$, suppose the pair $(M,N)$ is covariantly reflexive with respect to $L$. Let $t_{1},\dots,t_{n}$ be a minimal generating set for $\operatorname{Hom}_{R}(M,N)$ as a right $\operatorname{End}_{R}(M)$-module and let $p:(\operatorname{End}_{R}(M))^{\oplus n}\to\operatorname{Hom}_{R}(M,N)$ be the surjection mapping each standard basis element $e_{i}$ to $t_{i}$. For each $i=1,\dots,n$, let $j_{i}:\operatorname{Hom}_{R}(L,M)\to(\operatorname{End}_{R}(M))^{\oplus n}\otimes_{\operatorname{End}_{R}(M)}\operatorname{Hom}_{R}(L,M)$ be given by $j_{i}(a)=e_{i}\otimes a$. Now pick $f\in\operatorname{Hom}_{\operatorname{End}_{R}(L)}(\operatorname{tr}_{L,N}(M),\operatorname{Hom}_{R}(L,N))$. We claim $f$ factors through the map $i$, equivalently, that $\operatorname{im}f\subseteq\operatorname{tr}_{L,N}(M)$. For this, it suffices to show the images under $f$ of the generators $f_{i}\circ a$ for $a\in\operatorname{Hom}_{R}(L,M)$ are contained in $\operatorname{tr}_{L,N}(M)$. For each $i=1,\dots,n$, set $w_{i}=f\circ\phi^{M}_{L,N}\circ(p\otimes\operatorname{id}_{\operatorname{Hom}_{R}(L,M)})\circ j_{i}$. Then each $w_{i}\in\operatorname{Hom}_{\operatorname{End}_{R}(L)}(\operatorname{Hom}_{R}(L,M),\operatorname{Hom}_{R}(L,N))$, and as $(M,N)$ is covariantly reflexive with respect to $L$ there are unique maps $s_{i}\in\operatorname{Hom}_{R}(M,N)$ so that $w_{i}=\operatorname{Hom}_{R}(L,s_{i})$. Now given a generator $f_{i}\circ a$ of $\operatorname{tr}_{L,N}(M)$, we have $f_{i}\circ a=\phi^{M}_{L,N}((p\otimes\operatorname{id}_{\operatorname{Hom}_{R}(L,M)})(j_{i}(a)))$, so $f(f_{i}\circ a)=w_{i}(a)=\operatorname{Hom}_{R}(L,s_{i})(a)=s_{i}\circ a\in\operatorname{tr}_{L,N}(M)$, and the claim follows. The case where $(L,M)$ is contravariantly reflexive with respect to $N$ instead follows from a similar argument.

The claims of $(6)$ follow from the commutative diagram

whose vertical arrows are the natural isomorphisms owed to the flatness of $S$.

∎

There is a short exact sequence of $\operatorname{End}_{R}(N)-\operatorname{End}_{R}(L)$ bimodules

By Proposition 3.1 (2) and (6), the hypotheses force $\phi^{M}_{L,N}$ to be an isomorphism locally at every $\mathfrak{p}\in\operatorname{Supp}_{R}(L)\cap\operatorname{Ass}_{R}(N)=\operatorname{Ass}_{R}(\operatorname{Hom}_{R}(L,N))$. In particular, $K_{\mathfrak{p}}=0$ for all $\mathfrak{p}\in\operatorname{Ass}_{R}(\operatorname{Hom}_{R}(L,N))$, and it follows that $\operatorname{Hom}_{R}(K,\operatorname{Hom}_{R}(L,N))=0$. As the $R$-modules $\operatorname{Hom}_{\operatorname{End}_{R}(N)}(K,\operatorname{Hom}_{R}(L,N))$ and $\operatorname{Hom}_{\operatorname{End}_{R}(L)}(K,\operatorname{Hom}_{R}(M,N))$ are submodules of $\operatorname{Hom}_{R}(K,\operatorname{Hom}_{R}(L,N))=0$, it follows they are $0$ as well.

For $(1)$, we apply $\operatorname{Hom}_{\operatorname{End}_{R}(L)}(-,\operatorname{Hom}_{R}(L,N))$ to $\epsilon$ to see that $\operatorname{Hom}_{\operatorname{End}_{R}(L)}(\phi^{M}_{L,N},\operatorname{Hom}_{R}(L,N))$ gives an isomorphism $\operatorname{Hom}_{\operatorname{End}_{R}(L)}(\operatorname{tr}_{L,N}(M),\operatorname{Hom}_{R}(L,N))\to\operatorname{Hom}_{\operatorname{End}_{R}(L)}(\operatorname{Hom}_{R}(M,N)\otimes_{\operatorname{End}_{R}(M)}\operatorname{Hom}_{R}(L,M),\operatorname{Hom}_{R}(L,N))$. By Hom-tensor adjointness, the right hand term is naturally isomorphic to

But as the pair $(M,N)$ is covariantly reflexive with respect to $L$, this term is naturally isomorphic to $\operatorname{End}_{\operatorname{End}_{R}(M)}(\operatorname{Hom}_{R}(M,N))$. The claim then follows from Proposition 3.1 (5), noting the composition $\operatorname{End}_{\operatorname{End}_{R}(L)}(\operatorname{tr}_{L,N}(M))\to\operatorname{End}_{\operatorname{End}_{R}(M)}(\operatorname{Hom}_{R}(M,N))$ is a map of rings.

For (2), we follow a similar approach; applying $\operatorname{Hom}_{\operatorname{End}_{R}(N)}(-,\operatorname{Hom}_{R}(L,N))$ we observe that $\operatorname{Hom}_{\operatorname{End}_{R}(N)}(\phi^{M}_{L,N},\operatorname{Hom}_{R}(L,N))$ gives an isomorphism $\operatorname{Hom}_{\operatorname{End}_{R}(N)}(\operatorname{tr}_{L,N}(M),\operatorname{Hom}_{R}(L,N))\to\operatorname{Hom}_{\operatorname{End}_{R}(N)}(\operatorname{Hom}_{R}(M,N)\otimes_{\operatorname{End}_{R}(M)}\operatorname{Hom}_{R}(L,M),\operatorname{Hom}_{R}(L,N))$ and Hom-tensor adjointness shows the right hand term is isomorphic to

Since the pair $(L,M)$ is contravariantly reflexive with respect to, this term is naturally isomorphic to $\operatorname{End}_{\operatorname{End}_{R}(M)}(\operatorname{Hom}_{R}(L,M))$, and we have the claim.

∎

The claim follows immediately from Theorem 3.3 (2) with $L=R$.

∎

It follows from Proposition 2.8 that $R_{\mathfrak{p}}\in\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ for all $\mathfrak{p}\in\operatorname{Ass}(R)$. The claim thus follows from Corollary 3.4 with $N=R$.

∎

For the first claim, it suffices from Theorem 3.3 to show, for every $\mathfrak{p}\in\operatorname{Ass}_{R}(R)$, that one of $R_{\mathfrak{p}}$ or $\omega_{\mathfrak{p}}$ is in $\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$. But this follows immediately from noting that $R_{\mathfrak{p}}\in\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ when $M_{\mathfrak{p}}$ is a generator and that $w_{\mathfrak{p}}\in\operatorname{Add}_{R_{\mathfrak{p}}}(M_{\mathfrak{p}})$ when $M^{\vee}_{\mathfrak{p}}$ is a generator, since we may dualize a splitting $M^{\vee}_{\mathfrak{p}}\to R_{\mathfrak{p}}$. So we have $\operatorname{End}_{R}(\operatorname{tr}_{\omega}(M))\cong Z(\operatorname{End}_{R}(M))$. If $R$ has dimension $1$, then we note $Z(\operatorname{End}_{R}(M))$ embeds in a direct sum of copies of $M$, so is maximal Cohen-Macualay. Note that $\operatorname{End}_{R}(\operatorname{tr}_{\omega}(M))\cong\operatorname{Hom}_{R}(\operatorname{tr}_{\omega}(M),\omega)$ from Proposition 3.1 (5), since any pair $(M,N)$ is covariantly reflexive with respect to $R$. So from [BH93, Theorem 3.3.7 (b)], we have that a canonical module for $Z(\operatorname{End}_{R}(M))$ is $\operatorname{Hom}_{R}(Z(\operatorname{End}_{R}(M)),\omega)\cong\operatorname{Hom}_{R}(\operatorname{End}_{R}(\operatorname{tr}_{\omega}(M),\omega)\cong\operatorname{Hom}_{R}(\operatorname{Hom}_{R}(\operatorname{tr}_{\omega}(M),w),\omega)\cong\operatorname{tr}_{\omega}(M)$, with the last isomorphism owed to the fact that $\operatorname{tr}_{\omega}(M)$ embeds in $\omega$, and is thus maximal Cohen-Macaulay when $R$ has dimension $1$. ∎