Proof. Since \(M\) is faithful, \(R\) naturally embeds as a subring of both \(Z(E)\) and \(E\), so the maps \(A \otimes_R B \to A \otimes_E B\) and \(A \otimes_R B \to A \otimes_{Z(E)} B\) are surjective. As \(M\) is torsionless and faithful, it follows from [1] that \(Z(E)\) is a birational extension of \(R\). In particular, \(Z(E)_{\mathfrak{p}} \cong R_{\mathfrak{p}}\) for all \(\mathfrak{p}\in \operatorname{Ass}(R)\). Thus \(p\) is locally a surjective endomorphism and thus isomorphism at every associated prime of \(R\), from which we see that \(\operatorname{ker}(p)\) is torsion.

For \((3)\), we note as above that \(Z(E)\) is a birational extension of \(R\). Then the claim follows immediately from [8]. ◻

Proof. It follows from [7] that \(\omega_{Z(E)} \cong \operatorname{tr}_{\omega}(M)\). The claim then follows from combining Proposition 4 (2) Proposition 5 (3). ◻

Proof. Define \(\gamma:\operatorname{Hom}_{\operatorname{End}_{\Lambda}(L)}(\operatorname{End}_{\Lambda}(L),\operatorname{Hom}_{\Lambda}(L,N)) \to \operatorname{Hom}_{\Lambda}(L,N)\) by \(g \mapsto g(1_{\operatorname{End}_{\Lambda}(L)})\). We observe for any \(f \in \operatorname{Hom}_{\operatorname{End}_{\Lambda}(L)}(\operatorname{End}_{\Lambda}(L),\operatorname{Hom}_{\Lambda}(L,N))\) that \(\beta^{L,\Lambda}_{L,N}(\gamma(f))=\operatorname{Hom}_{\Lambda}(L,f(1_{\operatorname{End}_{\Lambda}(L)}))\). But if \(x \in \operatorname{End}_{\Lambda}(L)\), then \(\operatorname{Hom}_{\Lambda}(L,f(1_{\operatorname{End}_{\Lambda}(L)}))(x)=f(1_{\operatorname{End}_{\Lambda}(L)}) \circ x=f(1_{\operatorname{End}_{\Lambda}(L)} \circ x)=f(x)\). That is to say, \(\operatorname{Hom}_{\Lambda}(L,f(1_{\operatorname{End}_{\Lambda}(L)}))=f\), so \(\beta^{L,\Lambda}_{L,N} \circ \gamma=\operatorname{id}_{\operatorname{Hom}_{\operatorname{End}_{\Lambda}(L)}(\operatorname{End}_{\Lambda}(L),\operatorname{Hom}_{\Lambda}(L,N))}\).

Similarly, for any \(g \in \operatorname{Hom}_{\Lambda}(L,N)\), we have \(\gamma(\beta^{L,\Lambda}_{L,N}(g))=\gamma(\operatorname{Hom}_{\Lambda}(L,g))=\operatorname{Hom}_{\Lambda}(L,g)(1_{\operatorname{End}_{\Lambda}(L)})=g \circ 1_{\operatorname{End}_{\Lambda}(L)}=g\). Thus \(\gamma \circ \beta^{L,\Lambda}_{L,N}=\operatorname{id}_{\operatorname{Hom}_{\Lambda}(L,N)}\). So \(\beta^{L,\Lambda}_{L,N}\) and \(\gamma\) are inverses, and in particular \(\beta^{L,\Lambda}_{L,N}\) is an isomorphism. ◻

Proof. We begin with \((1)\). Suppose \(E\) and \(\operatorname{Hom}_R(M,M^{**})\) are isomorphic as right \(E\)-modules, i.e., that there is an isomorphism \(\gamma \in \operatorname{Hom}_E(E,\operatorname{Hom}_R(M,M^{**})\) By Lemma 7, \(\beta^{M,R}_{M,M^{**}}\) is an isomorphism, so there is a map of \(R\)-modules \(\alpha:M \to M^{**}\) so that \(\gamma=\operatorname{Hom}_R(M,\alpha)\).

Applying \(\operatorname{Hom}_R(M,-)\) to the exact sequence \[0 \rightarrow \operatorname{ker}(\alpha) \rightarrow M \xrightarrow{\alpha} M^{**}\] we get an exact sequence \[0 \rightarrow \operatorname{Hom}_R(M,\operatorname{ker}(\alpha)) \rightarrow E \xrightarrow{\gamma} \operatorname{Hom}_R(M,M^{**}).\] Since \(\gamma\) is an isomorphism, it follows that \(\operatorname{Hom}_R(M,\operatorname{ker}(\alpha))=0\). Thus there is a \(\operatorname{ker}(\alpha)\) regular element \(x \in \operatorname{Ann}_R(M)\). But \(\operatorname{ker}(\alpha) \hookrightarrow M\), so if \(xM=0\), then \(x\operatorname{ker}(\alpha)=0\) as well. It follows that \(\operatorname{ker}(\alpha)=0\).

Then applying \(\operatorname{Hom}_R(M,-)\) to the exact sequence \[0 \rightarrow M \xrightarrow{\alpha} M^{**} \rightarrow \operatorname{coker}(\alpha) \rightarrow 0\] we obtain, since \(\operatorname{Ext}^1_R(M,M)=0\), an exact sequence \[0 \rightarrow E \xrightarrow{\gamma} \operatorname{Hom}_R(M,M^{**}) \rightarrow \operatorname{Hom}_R(M,\operatorname{coker}(\alpha)) \rightarrow 0.\] Since \(\gamma\) is an isomorphism, we have \(\operatorname{Hom}_R(M,\operatorname{coker}(\alpha))=0\). But then \(\operatorname{Ass}_R \operatorname{Hom}_R(M,\operatorname{coker}(\alpha))=\operatorname{Supp}_R(M) \cap \operatorname{Ass}_R(\operatorname{coker}(\alpha))=\operatorname{Spec}(R) \cap \operatorname{Ass}_R(\operatorname{coker}(\alpha))=\operatorname{Ass}(\operatorname{coker}(\alpha))=\varnothing\), which implies \(\operatorname{coker}(\alpha)=0\). Therefore, \(\alpha\) is an isomorphism, so \(M\) is reflexive.

For \((2)\), we follow a similar approach as for \((1\)). We first note that is follows from [1] that \(Z(\operatorname{End}_R(M^*)) \cong \operatorname{End}_R(\operatorname{tr}_R(M))\), while [7] implies that \(\operatorname{End}_R(\operatorname{tr}_R(M)) \cong \operatorname{End}_{E}(M^*)\). From Hom-tensor adjointness we see that \(\operatorname{End}_{E}(M^*) \cong \operatorname{Hom}_E(M,M^{**})\). So there is an isomorphism of \(Z(E)\)-modules, \(\eta:Z(E) \to \operatorname{Hom}_E(M,M^{**})\). By Lemma 7, \(\beta^{M,E}_{M,N}\) is an isomorphism. Thus there is a map \(\theta:M \to M^{**}\) of left \(E\)-modules so that \(\eta=\operatorname{Hom}_E(M,\theta)\). Applying \(\operatorname{Hom}_E(M,-)\) to the exact sequence \[0 \rightarrow \operatorname{ker}(\theta) \rightarrow M \xrightarrow{\theta} M^{**},\] we get an exact sequence \[0 \rightarrow \operatorname{Hom}_E(M,\operatorname{ker}(\theta)) \rightarrow Z(E) \xrightarrow{\eta} \operatorname{Hom}_E(M,M^{**}).\] As \(\eta\) is an isomorphism, it follows that \(\operatorname{Hom}_E(M,\operatorname{ker}(\theta))=0\). As \(M\) is free on the punctured spectrum of \(R\), localizing at any \(\mathfrak{p}\in \operatorname{Spec}(R)-\{\mathfrak{m}\}\), we have \(M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{\oplus r_{\mathfrak{p}}}\), and we see that \[\operatorname{Hom}_E(M,\operatorname{ker}(\theta))_{\mathfrak{p}} \cong \operatorname{Hom}_{E_{\mathfrak{p}}}(M_{\mathfrak{p}},\operatorname{ker}(\theta)_{\mathfrak{p}}) \cong \operatorname{Hom}_{M_{r_{\mathfrak{p}}}(R_{\mathfrak{p}})}(R_{\mathfrak{p}}^{\oplus r_{\mathfrak{p}}},\operatorname{ker}(\theta)_{\mathfrak{p}})=0.\] As there is an isomorphism of left \(M_{r_{\mathfrak{p}}}(R_{\mathfrak{p}})\)-modules \(M_{r_{\mathfrak{p}}}(R_{\mathfrak{p}}) \cong (R_{\mathfrak{p}}^{\oplus r_{\mathfrak{p}}})^{\oplus r_{\mathfrak{p}}}\), it follows that \(\operatorname{ker}(\theta)_{\mathfrak{p}}=0\). It follows that \(\operatorname{ker}(\theta)\) is an \(R\)-module of finite length, but as \(M\) is torsion-free, it must be that \(\operatorname{ker}(\theta)=0\). Applying \(\operatorname{Hom}_E(M,-)\) to the exact sequence \[0 \rightarrow M \xrightarrow{\theta} M^{**} \rightarrow \operatorname{coker}(\theta) \rightarrow 0\] we get an exact sequence \[0 \rightarrow Z(E) \xrightarrow{\eta} \operatorname{Hom}_E(M,M^{**}) \rightarrow \operatorname{Hom}_E(M,\operatorname{coker}(\theta)) \rightarrow 0.\] As \(\eta\) is an isomorphism, \(\operatorname{Hom}_E(M,\operatorname{coker}(\theta))=0\). By a similar argument as above, we obtain that \(\operatorname{coker}(\theta)\) is an \(R\)-module of finite length. Any ascending chain \(C_1 \subseteq C_2 \subseteq \cdots \subseteq C_n\) of \(E\)-submodules \(\operatorname{coker}(\theta)\) is also an ascending chain of \(R\)-modules, and as \(\operatorname{coker}(\theta)\) has finite length as an \(R\)-modules, it follows that \(\operatorname{coker}(\theta)\) also has finite length as a left \(E\)-module. Suppose \(\operatorname{coker}(\alpha) \ne 0\) so that its \(E\)-module length is nonzero. Then as \(\operatorname{coker}(\theta)\) has a finite composition series, it must contain a simple left \(E\)-module. But as \(E\) is local, the only simple left \(E\)-module is \(E/J(E)\), so there is an embedding \(i:E/J(E) \hookrightarrow \operatorname{coker}(\theta)\). Moreover, as \(E/J(E)\) is a division ring, \(M/J(E)E\) is a free \(E/J(E)\)-module and so we take an \(E\)-linear projection \(q:M/J(E)E \twoheadrightarrow E/J(E)\). Letting \(p:M \to M/J(E)M\) be the natural projection of left \(E\)-modules, we see that \(i \circ q \circ p\) is a nonzero element of \(\operatorname{Hom}_E(M,\operatorname{coker}(\alpha))\). It follows that \(\operatorname{coker}(\alpha)=0\). Therefore, \(\theta\) is an isomorphism so \(M\) is reflexive. ◻

Proof. Let \(x_1,\dots,x_n\) be a minimal \(R\)-generating set for \(M\) and let \(p:R^{\oplus n} \to M\) be given on standard basis vectors by \(p(e_i)=x_i\). Applying \(\operatorname{Hom}_R(-,M)\) to the short exact sequence \[0 \rightarrow \Omega^1_R(M) \rightarrow R^{\oplus n} \xrightarrow{p} M \rightarrow 0\] we obtain a short exact sequence \[0 \rightarrow E \xrightarrow{\operatorname{Hom}_R(p,M)} \operatorname{Hom}_R(R^{\oplus n},M) \rightarrow \operatorname{Hom}_R(\Omega^1_R(M),M) \rightarrow 0\] Applying \(\operatorname{Hom}_E(-,M)\) to this exact sequence, we obtain an exact sequence \[0 \rightarrow (\Omega^1_R(M))^{\dagger \dagger} \rightarrow (R^{\oplus n})^{\dagger \dagger} \xrightarrow{p^{\dagger \dagger}} \operatorname{Hom}_E(E,M).\] Noting that \(x_1,\dots,x_n\) is a \(Z(E)\)-generating set for \(M\), we have a surjection \(q:Z(E)^{\oplus n} \to M\) given on standard basis vectors by \(q(e_i)=x_i\). Moreover, there is a commutative diagram:

\[\begin{tikzcd}[cramped,row sep=3.15em,yshift=.3ex] 0 & {\operatorname{ker}(q)} & {Z(E)^{\oplus n}} & M & 0 \\ 0 & {(\Omega^1_R(M))^{\dagger \dagger}} & {(R^{\oplus n})^{\dagger \dagger}} & {\operatorname{Hom}_E(E,M)} \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-4, to=1-5] \arrow["q", from=1-3, to=1-4] \arrow["{p^{\dagger \dagger}}", from=2-3, to=2-4] \arrow["{i^{\dagger \dagger}}", from=2-2, to=2-3] \arrow["{\gamma^M_M}", from=1-4, to=2-4] \arrow["{\beta_M^{R^{\oplus n}}}", from=1-3, to=2-3] \arrow["w", dashed, from=1-2, to=2-2] \arrow[from=2-1, to=2-2] \end{tikzcd}\]

As the maps \(\beta^{R^{\oplus n}}_M\) and \(\beta^M_M\) are isomorphisms, it follows that \(p^{\dagger \dagger}\) is surjective and then that \(w\) is an isomorphism. So we have established \((1)\). For \((2)\), as \(\operatorname{Ext}^1_R(M,M)=0\), it follows from [9] that \(M^{\vee} \otimes_R M\) is torsion-free. Since \(R\) is a subring of \(Z(E)\), there is a surjection \(\beta:M^{\vee} \otimes_R M \twoheadrightarrow M^{\vee} \otimes_{Z(E)} M\) whose kernel is torsion since \(Z(E)_{\mathfrak{p}} \cong R_{\mathfrak{p}}\) for all \(\mathfrak{p}\in \operatorname{Ass}(R)\). As \(M^{\vee} \otimes_R M\) is torsion-free, it follows \(\beta\) is an isomorphism. Then \[M^{\vee} \otimes_R M \cong M^{\vee} \otimes_{Z(E)} M \cong \operatorname{Hom}_R(M,\operatorname{tr}_{\omega}(M)) \otimes_{Z(E)} M\] \[\cong \operatorname{Hom}_{Z(E)}(M,\operatorname{tr}_{\omega}(M)) \otimes_{Z(E)} M \cong \operatorname{Hom}_{Z(E)}(M,\omega_{Z(E)}) \otimes_{Z(E)} M.\] As \(\operatorname{Hom}_{Z(E)}(M,\omega_{Z(E)}) \otimes_{Z(E)} M\) is torsion-free, [9] forces \(\operatorname{Ext}^1_{Z(E)}(M,M)=0\) and we have \((2)\). ◻

Proof. We first note that is it well-known that \(E:=\operatorname{End}_R(I)\) is commutative in this situation; see e.g. [8]. From [10] (cf. [11]), the condition that \(\operatorname{Ext}^1_R(I,I)=0\) forces \(I^{\vee} \otimes_R I\) to be torsion-free. Then from Proposition 5 (2) it follows that \(I^{\vee} \otimes_R I \cong I^{\vee} \otimes_{E} I\). Moreover, the trace map \(I^{\vee} \otimes_{E} I \twoheadrightarrow \operatorname{tr}_{\omega_R}(I)\) also has a torsion kernel, and so \(I^{\vee} \otimes_E I \cong \operatorname{tr}_{\omega_R}(I)\). From [7], we have \(\omega_E \cong \operatorname{tr}_{\omega_R}(I)\), and then, similar to above, the condition \(\operatorname{Ext}^1_{E}(I,E)=0\) forces \(I \otimes_E \operatorname{tr}_{\omega_R}(I)\) to be torsion-free via [10]. But then \[I \otimes_E \operatorname{tr}_{\omega_R}(I) \cong I \otimes_E (I^{\vee} \otimes_R I) \cong (I \otimes_E I^{\vee}) \otimes_R I \cong (I^{\vee} \otimes_E I) \otimes_R I \cong (I^{\vee} \otimes_R I) \otimes_R I \cong I^{\vee} \otimes_R (I \otimes_R I).\] Applying \(I^{\vee} \otimes_R -\) to the natural exact sequence \[0 \rightarrow \operatorname{Tor}^R_1(I,R/I) \rightarrow I \otimes_R I \xrightarrow{p} I^2 \rightarrow 0,\] we have an exact sequence \[I^{\vee} \otimes_R \operatorname{Tor}^R_1(I,R/I) \rightarrow I^{\vee} \otimes_R (I \otimes_R I) \xrightarrow{I^{\vee} \otimes p} I^{\vee} \otimes_R I^2 \rightarrow 0.\] As \(I^{\vee} \otimes_R \operatorname{Tor}^R_1(I,R/I)\) is torsion, so is \(\operatorname{ker}(I^{\vee} \otimes p)\), and as \(I^{\vee} \otimes_R (I \otimes I)\) is torsion-free from above, it follows that \(I^{\vee} \otimes p\) is an isomorphism. In particular, comparing minimal number of generators on both sides, we have \[\mu_R(I^{\vee})\mu_R(I)^2=\mu_R(I^{\vee})\mu_R(I^2).\] Since \(I^{\vee}\) is nonzero, this forces \(\mu_R(I)^2=\mu_R(I^2)\). But the map \(p\) induces a surjection \(S^2_R(I) \twoheadrightarrow I^2\), so in particular \(\mu_R(I)^2=\mu_R(I^2) \le \dfrac{\mu_R(I)(\mu_R(I)+1)}{2}\). But this occurs if and only if \(\mu_R(I) \le 1\), so \(I\) is principal, as desired. ◻

Proof. As \(I\) is two-generated, there is an exact sequence \[0 \to I^* \to R^{\oplus 2} \to I \to 0,\] see [12]. Applying \(- \otimes_R E\), we have, since \(\operatorname{Tor}^R_1(I,E)=0\), an exact sequence of the form \[0 \to I^* \otimes_R E \to E^{\oplus 2} \to I \otimes_R E \to 0.\] In particular, the depth lemma forces \(I^* \otimes_R E\) to be torsion-free. But \(I^*\) is naturally an \(E\)-module, and as \(I\) is faithful, \(R\) is naturally a subring of \(E\). There is thus a surjection \(I^* \otimes_R E \to I^* \otimes_E E \cong I^*\) of \(R\)-modules whose kernel is torsion by rank considerations. But as \(I^* \otimes_R E\) is torsion-free, we have \(I^* \otimes_R E \cong I^*\), which forces \(\mu_R(E)=1\). As \(E\) is a faithful \(R\)-module, it follows that \(E \cong R\). If \(I\) is in addition reflexive, then by Theorem 1 and Proposition 4 we have \(\operatorname{End}_R(\operatorname{tr}_R(I)) \cong (\operatorname{tr}_R(I))^* \cong R\), and it follows from [9] that \(\operatorname{tr}_R(I) \cong R\), so that \(\operatorname{tr}_R(I)\) being a trace ideal must equal \(R\). But then by Proposition 4 (3), it can only be that \(I \cong R\), so \(I\) is principal. ◻

Proof. First suppose \(M\) is a a reflexive \(Z(E)\)-module. By Theorem 1, Proposition 5 (3), and Proposition 4 (2), we have \[Z(\operatorname{End}_{Z(E)}(M))=Z(E) \cong \operatorname{End}_{Z(E)}(\operatorname{tr}_{Z(E)}(M)) \cong \operatorname{Hom}_{Z(E)}(\operatorname{tr}_{Z(E)}(M),Z(E)).\] Then [9] forces \(\operatorname{tr}_{Z(E)}(M) \cong Z(E)\), and being a trace ideal, we must have \(\operatorname{tr}_{Z(E)}(M)=Z(E)\). It follows from Proposition 4 (3) that \(M\) is a \(Z(E)\)-generator. But as \(\operatorname{Ext}^1_R(M,M)=0\), this forces \(\operatorname{Ext}^1_R(Z(E),Z(E))=0\), and then [10] forces \(Z(E) \otimes_R Z(E)^{\vee}\) to be torsion-free. But as \(M\) is faithful, it follows from Proposition 5 (2) that there is a natural surjection \(Z(E) \otimes_R Z(E)^{\vee} \twoheadrightarrow Z(E) \otimes_{Z(E)} Z(E)^{\vee} \cong Z(E)^{\vee}\) whose kernel is torsion. Since \(Z(E) \otimes_R Z(E)^{\vee}\) is torsion-free, it follows \(Z(E) \otimes_R Z(E)^{\vee} \cong Z(E)^{\vee}\), so that \(\mu_R(Z(E))=1\). As \(Z(E)\) is a faithful \(R\)-module, it follows that \(Z(E) \cong R\).

If additionally, \(M\) is a reflexive \(R\)-module, then we have from Theorem 1, that \(\operatorname{End}_R(\operatorname{tr}_R(M)) \cong Z(E) \cong R\), and then Proposition 4 (2) gives that \((\operatorname{tr}_R(M))^* \cong R\). Applying [9] once more, we see that \(\operatorname{tr}_R(M) \cong R\), and being a trace ideal, we have \(\operatorname{tr}_R(M)=R\). That \(M\) is a generator then follows from Proposition 4 (3), completing the proof of (1).

For (2), suppose \(\operatorname{Ext}^2_{Z(E)}(M,M)=0\). Let \(x_1,x_2,\dots,x_n\) be a minimal generating set for \(M\) and take a short exact sequence \[0 \rightarrow \Omega^1_R(M) \xrightarrow{i} R^{\oplus n} \xrightarrow{p} M \rightarrow 0\] with \(p\) given on standard basis vectors by \(p(e_i)=x_i\). As in the proof of Proposition 10, there is an exact sequence

\[0 \rightarrow (\Omega^1_R(M))^{\dagger \dagger} \xrightarrow{i^{\dagger \dagger}} (R^{\oplus n})^{\dagger \dagger} \xrightarrow{p^{\dagger \dagger}} M^{\dagger \dagger} \rightarrow 0\] fitting in a commutative diagram:

\[\begin{tikzcd}[cramped,row sep=3.15em,yshift=.3ex] & {\Omega^1_R(M) \otimes_R Z(E)} & {R^{\oplus n} \otimes_R Z(E)} & {M \otimes_R Z(E)} & 0 \\ 0 & {(\Omega^1_R(M))^{\dagger \dagger}} & {(R^{\oplus n})^{\dagger \dagger}} & {M^{\dagger \dagger}} & 0 \arrow["{i \otimes Z(E)}", from=1-2, to=1-3] \arrow[from=1-4, to=1-5] \arrow["{q \otimes Z(E)}", from=1-3, to=1-4] \arrow["{p^{\dagger \dagger}}", from=2-3, to=2-4] \arrow["{i^{\dagger \dagger}}", from=2-2, to=2-3] \arrow["{\beta^M_M}", from=1-4, to=2-4] \arrow["{\beta_M^{R^{\oplus n}}}", from=1-3, to=2-3] \arrow["{\beta_{M}^{\Omega^1_R(M)}}", from=1-2, to=2-2] \arrow[from=2-1, to=2-2] \arrow[from=2-4, to=2-5] \end{tikzcd}\]

By the snake lemma, there is an exact sequence of the form \[0 \rightarrow \operatorname{Tor}^R_1(M,Z(E)) \rightarrow \Omega^1_R(M) \otimes_R Z(E) \xrightarrow{\beta_M^{\Omega^1_R(M)}} (\Omega^1_R(M))^{\dagger \dagger} \rightarrow T \rightarrow 0\] where \(T:=\operatorname{ker}(\beta^M_M)\). Splitting this exact sequence into short exact sequences \[\clubsuit: 0 \rightarrow \operatorname{Tor}^R_1(M,Z(E)) \rightarrow \Omega^1_R(M) \otimes_R Z(E) \xrightarrow{b} C \rightarrow 0\]

\[\spadesuit: 0 \rightarrow C \xrightarrow{a} (\Omega^1_R(M))^{\dagger \dagger} \rightarrow T \rightarrow 0\]

we see as \(\operatorname{Tor}^R_1(M,Z(E))\) is torsion, that \(\operatorname{Hom}_{Z(E)}(b,M)\) is an isomorphism. By Hom-tensor adjointness, we may identify \(\operatorname{Hom}_{Z(E)}(q \otimes Z(E),M)\) and \(\operatorname{Hom}_{Z(E)}(i \otimes Z(E),M)\) with \(\operatorname{Hom}_R(p,M)\) and \(\operatorname{Hom}_R(i,M)\) respectively. In particular, \(\operatorname{Hom}_{Z(E)}(i \otimes Z(E),M)\) is surjective, and then so is \(\operatorname{Hom}_{Z(E)}(\beta_M^{\Omega^1_R(M)},M)\) by commutativity of the diagram. As \(\operatorname{Hom}_{Z(E)}(b,M)\) is an isomorphism, it follows that \(\operatorname{Hom}_{Z(E)}(a,M)\) is surjective.

Now, as \(\operatorname{Hom}_E(\operatorname{Hom}_R(\Omega^1_R(M),M),M) \cong \Omega^1_{Z(E)}(M)\) uo to free \(Z(E)\)-summands, we have \[\operatorname{Ext}^1_{Z(E)}(\operatorname{Hom}_E(\operatorname{Hom}_R(\Omega^1_R(M),M),M),M) \cong \operatorname{Ext}^1_{Z(E)}(\Omega^1_{Z(E)}(M),M) \cong \operatorname{Ext}^2_{Z(E)}(M,M)=0.\]

Thus, applying \(\operatorname{Hom}_{Z(E)}(-,M)\) to \(\spadesuit\) and considering the long exact sequence in \(\operatorname{Ext}\), we see that \(\operatorname{Ext}^1_{Z(E)}(T,M)=0\). But \(T\) is torsion while \(M\) is torsion-free, and as \(Z(E)\) has dimension \(1\), it can only be that \(T=0\). Then \(\beta^M_M\) is an isomorphism which forces \(Z(E)\) to be a cyclic \(R\)-module. But as it is faithful, we have \(Z(E) \cong R\). If \(M\) is reflexive, then by [1], we have \(\operatorname{End}_R(\operatorname{tr}_R(M)) \cong \operatorname{Hom}_R(\operatorname{tr}_R(M),R) \cong R\), so \(\operatorname{tr}_R(M) \cong R\) by [13], which implies that \(M\) is a generator. ◻