Proof. We have an exact sequence of pointed sets

$$\begin{eqnarray}\displaystyle 1\rightarrow \ker f\rightarrow \ker h\rightarrow \ker g\rightarrow M/fL\rightarrow N/hL\rightarrow N/gM\rightarrow 1, & & \displaystyle \nonumber\end{eqnarray}$$

in which each map has the property that all its non-empty fibres have equal cardinality. Hence, any term that sits between two finite sets in the above sequence is itself finite. The first assertion of the proposition easily follows. Moreover, if all terms in the sequence are finite, then the alternating product of their cardinalities is one, which proves the second assertion. ◻

Proof. We first prove part (a). We have

$$\begin{eqnarray}\ker p_{0}=\{(1,y)\in X\times Y:h(y)=g(1)=1\}\cong \ker h,\end{eqnarray}$$

which is finite by assumption. Further, the kernel of $g:X\rightarrow M/hY$ is equal to $p_{0}W$ , so $(X:p_{0}W)\leqslant (M:hY)$ , which is also finite. So $p_{0}$ is an isogeny.

Similarly, $\ker p_{1}\cong \ker g$ , which proves part (b). Finally, part (c) is symmetric in $X$ and $Y$ , and so follows from part (a).◻

Proof. The first two assertions follow immediately from Propositions 2.4 and 2.1. The third follows from Proposition 2.1 and a routine diagram chase, which we leave to the reader. ◻

Proof. The relation is clearly symmetric. Reflexivity is also clear, since an equivalence between $(X,f,g)$ and itself is given by $(X,\operatorname{id},\operatorname{id}):X\rightleftharpoons X$ . Transitivity follows from Proposition 2.6.◻

Proof. The proof is easy, and is left to the reader. ◻

Proof. Let $c=(X,f,g)$ , $d=(Y,h,j)$ .

First, we prove the proposition in the special case that $d^{\prime }=d$ , and $c^{\prime }=(W,fp,gp)$ , where $p:W\rightarrow X$ is an isogeny. Let $(X\times _{M}Y,p_{0},p_{1})$ be the fibre product of the diagram $X\stackrel{g}{\rightarrow }M\stackrel{h}{\leftarrow }Y$ , and let $(W\times _{M}Y,p_{0}^{\prime },p_{1}^{\prime })$ be the fibre product of the diagram $W\stackrel{gp}{\rightarrow }M\stackrel{h}{\leftarrow }Y$ . Thus, $d\circ c=(X\times _{M}Y,fp_{0},jp_{1})$ , and $d\circ c^{\prime }=(W\times _{M}Y,fpp_{0}^{\prime },jp_{1}^{\prime }):L\rightleftharpoons N$ . Since $gpp_{0}^{\prime }=hp_{1}^{\prime }$ , the universal property of fibre products guarantees the existence of a unique map $i:W\times _{M}Y\rightarrow X\times _{M}Y$ with the property that $pp_{0}^{\prime }=p_{0}i$ and $p_{1}^{\prime }=p_{1}i$ :

Moreover, it is easy to see that $(W\times _{M}Y,p_{0}^{\prime },i)$ is the fibre product of the diagram $W\stackrel{p}{\rightarrow }X\stackrel{p_{0}}{\leftarrow }X\times _{M}Y$ . It follows from Proposition 2.4 that $i$ is an isogeny, which proves that $d\circ c$ is equivalent to $d\circ c^{\prime }$ .

Now, we prove the proposition in the special case that $d=d^{\prime }$ , and $c^{\prime }$ is arbitrary. Write ${\sim}$ for the equivalence relation between correspondences. Let $c^{\prime }=(X^{\prime },f^{\prime },g^{\prime })$ , and let $(W,p,q):X\rightleftharpoons X^{\prime }$ be an equivalence between $c$ and $c^{\prime }$ . Since $p$ is an isogeny, we have $c\sim (W,fp,gp)=(W,f^{\prime }q,g^{\prime }q)$ , and since $q$ is an isogeny, we have $(W,f^{\prime }q,g^{\prime }q)\sim c^{\prime }$ . We deduce from the special case of the proposition that we just proved that $d\circ c\sim d\circ (W,fp,gp)\sim d\circ c^{\prime }$ .

Now, we prove the general case. By Lemma 2.10 and by the special case we just proved, we have

$$\begin{eqnarray}(d\circ c)^{-1}\sim c^{-1}\circ d^{-1}\sim c^{-1}\circ (d^{\prime })^{-1}\sim (d^{\prime }\circ c)^{-1}.\end{eqnarray}$$

It therefore follows from Lemma 2.10(b), that $d\circ c\sim d^{\prime }\circ c$ . By the special case of the proposition that we proved already, we also have $d^{\prime }\circ c\sim d^{\prime }\circ c^{\prime }$ , and the proposition follows.◻

Proof. This is an immediate consequence of Proposition 2.1. ◻

Proof. First, we prove the assertion on $c^{-1}\,\circ \,c$ . By definition, $c^{-1}\,\circ \,c=(X\times _{M}X,fp_{0},fp_{1}):L\rightleftharpoons L$ , where $(X\times _{M}X,p_{0},p_{1})$ is the fibre product of the diagram $X\stackrel{g}{\rightarrow }M\stackrel{g}{\leftarrow }X$ . By the universal property of the fibre product, we have a unique map $i:X\rightarrow X\times _{M}X$ with the property that $p_{0}i=p_{1}i=\operatorname{id}:X\rightarrow X$ :

Since $g$ is an isogeny, it follows from Proposition 2.4 that $p_{0}$ is an isogeny. By Proposition 2.1, the morphism $i$ is also an isogeny, so $(X,i,f):X\times _{M}X\rightleftharpoons L$ defines an equivalence between $c^{-1}\circ c$ and $(L,\operatorname{id},\operatorname{id})$ .

The claim for $c\circ c^{-1}$ follows by applying the result just proved to $c^{-1}$ in place of $c$ .◻

Proof. Let $c=(X,f,g):L\rightleftharpoons M$ be a skew correspondence, and let $d=(Y,h,j):M\rightleftharpoons L$ be a two-sided inverse in ${\mathcal{C}}_{\operatorname{skew}}$ . So $d\circ c$ is equivalent to the commensurability $(L,\operatorname{id},\operatorname{id}):L\rightleftharpoons L$ , while $c\circ d$ is equivalent to $(M,\operatorname{id},\operatorname{id}):M\rightleftharpoons M$ , and in particular both compositions are commensurabilities. We wish to prove that $g$ is then necessarily an isogeny, and for this it is enough to assume that ${\mathcal{C}}=\mathbf{Grp}$ .

Let $(Y\times _{L}X,p_{0},p_{1})$ be the fibre product of the diagram $Y\stackrel{j}{\rightarrow }L\stackrel{\,\,\,f}{\leftarrow }X$ , and let $(X\times _{M}Y,p_{0}^{\prime },p_{1}^{\prime })$ be the fibre product of the diagram $X\stackrel{g}{\rightarrow }M\stackrel{\,h}{\leftarrow }Y$ . Since $c\circ d$ is a commensurability, the morphism $gp_{1}$ is an isogeny, so $(M:gX)$ is finite. Also, since $d\circ c$ is a commensurability, the morphism $jp_{1}^{\prime }$ is an isogeny, so $\ker p_{1}^{\prime }$ is finite. But $\ker p_{1}^{\prime }=\{(x,1)\in X\times Y:g(x)=h(1)=1\}\cong \ker g$ . So $g$ is an isogeny, as claimed.◻

Proof. The first assertion follows from the fact that $G_{L}=\operatorname{Hom}_{{\mathcal{C}}_{\text{com}}}(L,L)$ . The second assertion follows from Propositions 2.6 and 2.12.◻

Proof. See [Reference LamLam01, § 4]. ◻

Proof. Let $u+I$ be a unit in $E/I$ , and let $v+I$ be its inverse. Then we have $uv$ , $vu\in 1+I\subset 1+\text{J}(E)\subset E^{\times }$ , so $u$ has both a right inverse, namely $v(uv)^{-1}$ , and a left inverse, namely $(vu)^{-1}v$ . It follows that $u$ is a unit in  $E$ .◻

Proof. First, suppose that $E$ is semisimple. Then $E$ can be written as the product of finitely many simple rings. Since the kernel of $E\rightarrow F$ is a two-sided ideal of $E$ , it must be a subproduct, and $F$ may then be identified with the complementary subproduct. Surjectivity of $E^{\times }\rightarrow F^{\times }$ is now obvious.

We pass to the general case. Denote by $I$ the image of $\text{J}(E)$ in $F$ , which is a two-sided ideal of  $F$ . The map $E\rightarrow F$ induces a surjective ring homomorphism $E/\text{J}(E)\rightarrow F/I$ , where the ring $E/\text{J}(E)$ is semisimple, so by the first part of the proof the induced map $(E/\text{J}(E))^{\times }\rightarrow (F/I)^{\times }$ is surjective. By Lemma 3.3, the map $E^{\times }\rightarrow (E/\text{J}(E))^{\times }$ is also surjective, so the map $E^{\times }\rightarrow (F/I)^{\times }$ induced by the composed ring homomorphism $E\rightarrow F/I$ is surjective as well. Now let $v\in F^{\times }$ , and choose $u\in E^{\times }$ that maps to $v+I\in (F/I)^{\times }$ . Then the image $w$ of $u$ in $F$ satisfies $w\equiv v~\text{mod}~I$ , so $w^{-1}v$ belongs to the image $1+I$ of $1+\text{J}(E)$ in $F$ . Let $x\in 1+\text{J}(E)$ map to $w^{-1}v$ . Then $ux$ maps to $ww^{-1}v=v$ , and we have $ux\in E^{\times }$ because $u\in E^{\times }$ and $x\in 1+\text{J}(E)\subset E^{\times }$ . This proves surjectivity of $E^{\times }\rightarrow F^{\times }$ , as required.◻

Proof. Let $I$ be the kernel of the isogeny $E\rightarrow F$ . Then $I$ is finite, and we may identify $F$ with $E/I$ . Write $\operatorname{End}I$ for the endomorphism ring of the additive group of $I$ . Let $J$ and $R$ , respectively, be the kernel and the image of the ring homomorphism $E\rightarrow \operatorname{End}I$ sending $a\in E$ to the map $x\mapsto ax$ . Then $R$ , being a subring of $\operatorname{End}I$ , is a finite ring, $J$ is a two-sided ideal of $E$ , and we have a ring isomorphism $E/J\rightarrow R$ . By Lemma 3.5, the combined map $E\rightarrow F\times R$ induces a ring isomorphism $\unicode[STIX]{x1D711}:E/(I\cap J)\rightarrow F\times _{E/(I+J)}R$ . Now we first prove that the map $(E/(I\cap J))^{\times }\rightarrow F^{\times }$ is surjective. Let $u\in F^{\times }$ . Write $v$ for the image of $u$ in $(E/(I+J))^{\times }$ . Since $R$ is finite and hence left-artinian, by Lemma 3.4 we can choose $w\in R^{\times }$ mapping to $v\in (E/(I+J))^{\times }$ . Then $(u,w)$ belongs to $F^{\times }\times _{(E/(I+J))^{\times }}R^{\times }=(F\times _{E/(I+J)}R)^{\times }$ , so $\unicode[STIX]{x1D711}^{-1}(u,w)$ is a unit of $E/(I\cap J)$ that maps to $u\in F^{\times }$ . This proves that $(E/(I\cap J))^{\times }\rightarrow F^{\times }$ is surjective. From $(I\cap J)\cdot (I\cap J)\subset JI=0$ it follows that for each $x\in I\cap J$ the element $1+x$ has inverse $1-x$ and therefore belongs to  $E^{\times }$ ; this implies $I\cap J\subset \text{J}(E)$ , so by Lemma 3.3 the map $E^{\times }\rightarrow (E/(I\cap J))^{\times }$ is surjective. The composed map $E^{\times }\rightarrow F^{\times }$ is then surjective as well.◻

Proof. (a) Put $I=\{x\in F:FxF\subset E\}$ . Then $I$ is a two-sided ideal of $F$ that is contained in  $E$ , and we proceed to show that $I$ has finite index in  $F$ . Put $J=\{x\in F:Fx\subset E~\text{and}~xF\subset E\}$ . Then we have $I\subset J\subset E\subset F$ . Denote by $D$ the finite abelian group $F/E$ , by $\operatorname{End}D$ its endomorphism ring, and by $(\operatorname{End}D)^{\operatorname{opp}}$ the ring opposite to $\operatorname{End}D$ . Both of these rings are finite. The natural left and right $E$ -module structures on $D$ induce a ring homomorphism $E\rightarrow (\operatorname{End}D)\times (\operatorname{End}D)^{\operatorname{opp}}$ of which $J$ is the kernel. It follows that $J$ is a two-sided ideal of $E$ of finite index in  $E$ . There is a well-defined group homomorphism

$$\begin{eqnarray}\displaystyle J & \rightarrow & \displaystyle \operatorname{Hom}(D\otimes _{\mathbb{Z}}D,D)\nonumber\\ \displaystyle x & \mapsto & \displaystyle ((y+E)\otimes (z+E)\mapsto yxz+E)\nonumber\end{eqnarray}$$

for $x\in J$ , $y,z\in F$ . Its kernel is $I$ , and since $D\otimes _{\mathbb{Z}}D$ is finite, the group $J/I$ is finite. Because each of $F/E$ , $E/J$ , $J/I$ is finite, the ring $F/I$ is finite. This proves part (a).

(b) Let $I$ be as in part (a). Then $F/I$ and $(F/I)^{\times }$ are finite, so the kernel $K$ (say) of the natural group homomorphism $F^{\times }\rightarrow (F/I)^{\times }$ has finite index in $F^{\times }$ . If $x\in K$ , then $x^{-1}\in K$ , so both $x$ and $x^{-1}$ are in $1+I$ , which is contained in  $E$ . This proves $K\subset E^{\times }$ , so $E^{\times }$ has finite index in $F^{\times }$ as well. This proves part (b).◻

Proof. The last assertion is Lemma 3.6. For the first assertion, let $I$ and $D$ be the kernel, respectively the image of the map $E\rightarrow F$ . Then the kernel of $E^{\times }\rightarrow D^{\times }$ is contained in $1+I$ and therefore finite, and by Lemma 3.6 the image is all of $D^{\times }$ . Hence, $E^{\times }\rightarrow D^{\times }$ is a group isogeny. Further, the inclusion map $D^{\times }\rightarrow F^{\times }$ is obviously injective, while by Lemma 3.7(b) the index of $D^{\times }$ in $F^{\times }$ is finite. Hence, $D^{\times }\rightarrow F^{\times }$ is a group isogeny. By Proposition 2.1, the composed map $E^{\times }\rightarrow F^{\times }$ is also a group isogeny.◻

Proof. Let $m\in Z$ be non-zero. First suppose that $L$ is finitely generated. Let $S\subset L$ be a finite subset that generates it as a $Z$ -module, let $T\subset S$ be a maximal subset that is linearly independent over  $Q$ , and let $M\subset L$ be the $Z$ -module generated by  $T$ . Then $M$ is $Z$ -free of rank $\#T$ , so $M/mM$ is finite of order $\#(Z/mZ)^{\#T}\leqslant \#(Z/mZ)^{\dim _{Q}V}$ , with equality if $T$ is a $Q$ -basis of $V$ or, equivalently, if $Q\cdot L=V$ . By maximality of $T$ , we can, for each $s\in S$ , choose a non-zero element $m_{s}\in Z$ such that $m_{s}s\in M$ , and $m^{\prime }=\prod _{s\in S}m_{s}$ is then a non-zero element of $Z$ satisfying $m^{\prime }L\subset M$ . Because $M/m^{\prime }M$ is finite, its subgroup $m^{\prime }L/m^{\prime }M$ is finite as well, and since the latter group is isomorphic to $L/M$ and to $mL/mM$ , we find that $L/M$ and $mL/mM$ are finite of the same order. The group $L/mM$ is finite of order $\#(L/M)\cdot \#(M/mM)$ , so $L/mL$ is also finite, of order

$$\begin{eqnarray}\frac{\#(L/M)\cdot \#(M/mM)}{\#(mL/mM)}=\#(M/mM)=\#(Z/mZ)^{\#T}\leqslant \#(Z/mZ)^{\dim _{Q}V},\end{eqnarray}$$

with equality if $Q\cdot L=V$ .

Passing to the general case, let $U$ be the set of finitely generated sub- $Z$ -modules $L^{\prime }$ of  $L$ , which is a directed partially ordered set by inclusion. Then $L$ is the injective limit of all $L^{\prime }\in U$ , and $L/mL$ is the injective limit of the modules $L^{\prime }/mL^{\prime }$ , all of which have order dividing $\#(Z/mZ)^{\dim _{Q}V}$ . The injective limit has then also order dividing the same number. This completes the proof of Lemma 4.1.◻

Proof. Let $I$ be a left ideal of  $R$ . Then $Q\cdot I$ is a left $A$ -ideal, so by semisimplicity of $A$ it is a direct summand of the left $A$ -module  $A$ . Thus, the endomorphism ring of the latter module contains an idempotent with image $Q\cdot I$ . Since the endomorphisms of the left $A$ -module $A$ are the right multiplications by elements of $A$ , it is equivalent to say that we can choose an idempotent $e\in A$ with $Ae=Q\cdot I$ . We have $e\in Q\cdot I$ , so we can choose a non-zero element $m\in Z$ with $me\in I$ . Multiplying the chain of inclusions $Rme\subset I\subset R$ by $e$ on the right, which when restricted to $I$ is just the identity map, we obtain $Rme\subset I\subset Re$ , where $Rme=mRe$ because $m$ is central. By Lemma 4.1, the group $Re/mRe$ is finite, so $I/Rme$ is finite as well. Hence, $I$ is, as a left $R$ -module, generated by $me$ together with a finite set, and is therefore finitely generated. This proves that $R$ is left-noetherian. Applying this result to $A^{\operatorname{opp}}$ and $R^{\operatorname{opp}}$ , we find that $R$ is right-noetherian as well.◻

Proof. Let $a,b\in Z$ be non-zero. We have an exact sequence of $Z$ -modules

$$\begin{eqnarray}M/bM\stackrel{a}{\rightarrow }M/abM\rightarrow M/aM\rightarrow 0.\end{eqnarray}$$

The left and right terms are finite by assumption, so $M/abM$ is finite. But since $Z$ is infinite, and $M$ is a faithful module, $M$ is also infinite, and so $ab\neq 0$ .◻

Proof. First we prove that assertion (a) implies assertion (b). From assertion (a) it follows, by Lemma 4.4 applied to $M=Z$ , that $Z$ is a domain. Now let $\mathfrak{p}$ be a non-zero prime ideal, and let $m\in \mathfrak{p}$ be non-zero. Then we have $mZ\subset \mathfrak{p}\subset Z$ , and since $Z/mZ$ is finite, the index of $\mathfrak{p}$ in $Z$ is finite and $\mathfrak{p}/mZ$ is finite. Hence, $\mathfrak{p}$ is generated by $m$ together with a finite set, and is therefore finitely generated.

Now we prove that assertion (b) implies assertion (c). By [Reference CohenCoh50, Theorem 2], each commutative ring of which every prime ideal is finitely generated is noetherian. Hence, assertion (b) implies that $Z$ is noetherian. If $\mathfrak{p}$ is a non-zero prime ideal, then $Z/\mathfrak{p}$ is a finite domain, and therefore a field. Hence, each non-zero prime ideal is maximal, so $Z$ has Krull dimension $0$ or  $1$ ; in the former case it must be a field.

Finally, suppose that assertion (c) holds. Then, we will deduce assertion (a) by showing that for any non-zero ideal $I$ of $Z$ , the ring $Z/I$ is finite. Suppose that there exists a non-zero ideal $I$ in $Z$ such that $Z/I$ is infinite. Since $Z$ is noetherian, we may, without loss of generality, assume that $I$ is maximal among ideals with this property. So $Z/I$ is infinite, but its quotient by any non-zero ideal is finite. It follows from Lemma 4.4, applied to $M=Z/I$ , that $Z/I$ is a domain, so $I$ is a prime ideal of $Z$ . It is also non-zero, so assertion (c) implies that $I$ is maximal, and therefore that $Z/I$ is finite, which is a contradiction.◻

Proof. It is clear that each $\unicode[STIX]{x1D6FF}_{j}$ is a group homomorphism, and that for each $x$ all $\unicode[STIX]{x1D6FF}_{j}(x)$ are conjugate to each other, so all $\unicode[STIX]{x1D6FF}_{j}$ induce the same map $D_{\text{ab}}^{\times }\rightarrow \text{M}(n,D)_{\text{ab}}^{\times }$ . It is evidently surjective if $n=1$ .

For $i$ , $j\in \{1,\ldots ,n\}$ , $i\neq j$ , and $x\in D$ , let $B_{ij}(x)\in \text{M}(n,D)$ be the matrix obtained from the unit matrix by replacing the $(i,j)$ -entry by  $x$ ; then one has $B_{ij}(x)\in \text{M}(n,D)^{\times }$ . The subgroup of $\text{M}(n,D)^{\times }$ generated by all $B_{ij}(x)$ is denoted by $\operatorname{SL}_{n}(D)$ .

By [Reference ArtinArt57, ch. IV, Theorem 4.1], we have $\text{M}(n,D)^{\times }=\operatorname{SL}_{n}(D)\cdot \unicode[STIX]{x1D6FF}_{n}(D^{\times })$ . Each $B_{ij}(x)$ is a transvection of the right $D$ -vector space $D^{n}$ in the sense of [Reference ArtinArt57, ch. IV, Definition 4.1]. Assume now that $n>2$ or $\#D\neq 2$ . Then by [Reference ArtinArt57, ch. IV, § 2] each transvection belongs to $[\text{M}(n,D)^{\times },\text{M}(n,D)^{\times }]$ , so $\operatorname{SL}_{n}(D)\subset [\text{M}(n,D)^{\times },\text{M}(n,D)^{\times }]$ , and therefore

$$\begin{eqnarray}\text{M}(n,D)^{\times }=[\text{M}(n,D)^{\times },\text{M}(n,D)^{\times }]\cdot \unicode[STIX]{x1D6FF}_{n}(D^{\times }).\end{eqnarray}$$

This implies that $\bar{\unicode[STIX]{x1D6FF}}$ is surjective.◻

Proof. It is clear that $\unicode[STIX]{x1D704}$ is a group homomorphism. If we have $n=\#D=2$ , then $\text{M}(n,D)^{\times }$ is a non-abelian group of order six, in which case $\text{M}(n,D)_{\text{ab}}^{\times }$ has order two and the conclusion of the lemma is valid. Assume now that $n\neq 2$ or $\#D\neq 2$ , so that the map $\bar{\unicode[STIX]{x1D6FF}}$ from Lemma 5.1 is surjective.

Denote by $\bar{\unicode[STIX]{x1D704}}:D_{\text{ab}}^{\times }\rightarrow \text{M}(n,D)_{\text{ab}}^{\times }$ the map induced by $\unicode[STIX]{x1D704}$ . For each $x\in D^{\times }$ one has $\unicode[STIX]{x1D704}(x)=\prod _{j=1}^{n}\unicode[STIX]{x1D6FF}_{j}(x)$ and therefore $\bar{\unicode[STIX]{x1D704}}(x)=\bar{\unicode[STIX]{x1D6FF}}(x)^{n}$ , so the surjectivity of $\bar{\unicode[STIX]{x1D6FF}}$ yields

$$\begin{eqnarray}\bar{\unicode[STIX]{x1D704}}(D_{\text{ab}}^{\times })=\bar{\unicode[STIX]{x1D6FF}}(D_{\text{ab}}^{\times })^{n}=(\text{M}(n,D)_{\text{ab}}^{\times })^{n},\end{eqnarray}$$

and the lemma is proved. ◻

Proof. See [Reference LamLam01, Theorem 16.9]. ◻

Proof. Since $D^{\times }/(\text{Z}(D)^{\times }\cdot [D^{\times },D^{\times }])$ is a quotient of $D_{\text{ab}}^{\times }$ , it is an abelian group. Let $a\in D^{\times }$ . It will suffice to show that the image $\bar{a}$ of $a$ in the quotient $D^{\times }/(\text{Z}(D)^{\times }\cdot [D^{\times },D^{\times }])$ has order dividing  $m$ . The subfield $\text{Z}(D)(a)$ of $D$ is contained in a maximal subfield of  $D$ , and each maximal subfield of $D$ is an extension field of $\text{Z}(D)$ of degree  $m$ , by [Reference Curtis and ReinerCR81, (7.22)]. Hence, we have $[\text{Z}(D)(a):\text{Z}(D)]=l$ for some divisor $l$ of  $m$ , and $a$ is a zero of an irreducible polynomial $f\in \text{Z}(D)[X]$ of degree  $l$ with leading coefficient  $1$ . Using Theorem 5.3 we find $b_{1}$ , …, $b_{l}\in D^{\times }$ such that $b_{1}ab_{1}^{-1}\cdot \cdots \cdot b_{l}ab_{l}^{-1}=(-1)^{l}f(0)\in \text{Z}(D)^{\times }$ . Mapping this identity to the abelian group $D^{\times }/(\text{Z}(D)^{\times }\cdot [D^{\times },D^{\times }])$ we obtain $\bar{a}^{l}=1$ , so $\bar{a}^{m}=1$ , as required. This proves Lemma 5.4.◻

Proof. By [Reference BourbakiBou12, § 14, Theorem 1], there are a positive integer $n$ and a division ring $D$ with $\text{Z}(D)=k$ such that $B$ is, as an algebra over $k$ , isomorphic to $\text{M}(n,D)$ . Then $D$ has finite degree $m^{2}$ over $k$ , and $nm=d$ . By Lemma 5.4, the cokernel of the natural group homomorphism $k^{\times }\rightarrow D_{\text{ab}}^{\times }$ has exponent dividing $m$ , and by Lemma 5.2 the cokernel of the natural group homomorphism $D_{\text{ab}}^{\times }\rightarrow \text{M}(n,D)^{\times }$ has exponent dividing  $n$ . It follows that the cokernel of the natural group homomorphism $k^{\times }\rightarrow \text{M}(n,D)_{\text{ab}}^{\times }$ has exponent dividing $nm=d$ .◻

Proof. In the case the semisimple ring $B$ is simple, our hypothesis that it be finite over its centre implies that it is a central simple algebra over $\text{Z}(B)$ , and the assertion follows from Theorem 5.5. Generally, by [Reference LamLam01, ch. 1, Theorem 3.5] the ring $B$ is a product of finitely many semisimple rings that are simple, and the result follows from the case we just did.◻

Notation 6.1. The following assumptions will be in force throughout the present section: $Z$ is an infinite commutative ring that satisfies the equivalent conditions of Theorem 4.5, with field of fractions $Q$ ; further, $A$ is a $Q$ -algebra of finite vector space dimension over $Q$ , and $R$ is a left-noetherian sub- $Z$ -algebra of $A$ with the property that $Q\cdot R=A$ . By an $R$ -module we shall always mean a left $R$ -module. We call a module finite if its cardinality is finite. If $L$ is a finitely generated $R$ -module, let $L_{\text{tors}}$ denote the set of all elements of $L$ that have a non-zero annihilator in $Z$ . Since the image of $Z$ in $R$ is central, $L_{\text{tors}}$ is a sub- $R$ -module.

Proof. First, we show that $L_{\text{tors}}$ is finite. Since $R$ is left-noetherian, $L_{\text{tors}}$ is finitely generated as an $R$ -module. So there exists a non-zero $m\in Z$ that annihilates $L_{\text{tors}}$ , and $L_{\text{tors}}$ is then a finitely generated module over the ring $R/mR$ , which is finite by Lemma 4.1. This proves one implication.

For the converse, let $U\subset L$ be a finite sub- $R$ -module. Then for each $x\in U$ , the set $\{zx:z\in Z\}$ is finite, so the annihilator of $x$ in $Z$ has finite index in $Z$ ; in particular it is non-zero, since $Z$ is assumed to be infinite, so $x\in L_{\text{tors}}$ .◻

Proof. First, suppose that $f:L\rightarrow M$ is an isogeny. Then $c_{f}=(L,\operatorname{id},f):L\rightleftharpoons M$ is a commensurability.

Next, suppose that we have a commensurability $(X,f,g):L\rightleftharpoons M$ . Then the kernels and cokernels of $f$ , $g$ are finite $R$ -modules, and so are $Z$ -torsion modules by Lemma 6.2. They are therefore annihilated by the functor $Q\otimes _{Z}-$ , so the maps $Q\otimes _{Z}f$ and $Q\otimes _{Z}g$ are isomorphisms.

Finally, suppose that we have an isomorphism $\unicode[STIX]{x1D719}:Q\otimes _{Z}L\rightarrow Q\otimes _{Z}M$ of $A$ -modules. It follows from Lemma 6.2 that the quotient map $L\rightarrow \bar{L}=L/L_{\text{tors}}$ is an isogeny. Since $\bar{L}$ is $Z$ -torsion free, it embeds into $Q\otimes _{Z}L$ , and similarly for $\bar{M}$ . By ‘clearing denominators’, we can find non-zero elements $m_{1}$ , $m_{2}\in Z$ such that $m_{1}\unicode[STIX]{x1D719}(\bar{L})$ is contained in $\bar{M}\subset Q\otimes _{Z}M$ , and $\unicode[STIX]{x1D719}(\bar{L})$ contains $m_{2}\bar{M}$ . Since $\bar{M}/m_{1}m_{2}\bar{M}$ is finite by Lemma 4.1, it follows that $m_{1}\unicode[STIX]{x1D719}:\bar{L}\rightarrow \bar{M}$ is an isogeny. Let  $m_{3}\in Z$ be a non-zero element that annihilates $M_{\text{tors}}$ . Then $m_{3}M$ is canonically isomorphic to $\bar{M}$ , and since $M/m_{3}M$ is finitely generated and torsion, Lemma 6.2 implies that the embedding $\bar{M}\cong m_{3}M\subset M$ is an isogeny. The composition of the three isogenies $L\rightarrow \bar{L}\rightarrow \bar{M}\rightarrow M$ is an isogeny by Proposition 2.1, as claimed.◻

Proof. Let $(W,p,q):X\rightleftharpoons Y$ be an equivalence between $(X,f,g)$ and $(Y,h,j)$ . Since $p$ and $q$ are isogenies, Lemma 6.2 implies that $Q\otimes _{Z}p$ and $Q\otimes _{Z}q$ are both invertible. Moreover, we have

$$\begin{eqnarray}\displaystyle Q\otimes _{Z}f & = & \displaystyle (Q\otimes _{Z}h)\circ (Q\otimes _{Z}q)\circ (Q\otimes _{Z}p)^{-1},\nonumber\\ \displaystyle Q\otimes _{Z}g & = & \displaystyle (Q\otimes _{Z}j)\circ (Q\otimes _{Z}q)\circ (Q\otimes _{Z}p)^{-1},\nonumber\end{eqnarray}$$

so $(Q\otimes g)\circ (Q\otimes f)^{-1}=(Q\otimes j)\circ (Q\otimes h)^{-1}$ .◻

Proof. Let $V$ be an $A$ -module with finite generating set $S$ . Let $L$ be the sub- $R$ -module of $V$ generated by $S$ over $R$ . Then the $A$ -module ${\mathcal{F}}(L)$ is isomorphic to $V$ .◻

Proof. Let $\bar{L}$ be the image of $L$ in $Q\otimes _{Z}L$ and let $\bar{M}$ be the image of $M$ in $Q\otimes _{Z}M$ . By Lemma 6.2, the natural map $f:L\rightarrow Q\otimes _{Z}L$ gives rise to a commensurability $c_{L}=(L,\operatorname{id},f):L\rightleftharpoons \bar{L}$ , and similarly we have a commensurability $c_{M}:M\rightleftharpoons \bar{M}$ . Since $\bar{L}$ and $\bar{M}$ are finitely generated as $R$ -modules, and since $\bar{M}$ generates $Q\otimes _{Z}M$ over $Q$ , we may choose a non-zero $m\in Z$ such that $m\unicode[STIX]{x1D719}(\bar{L})$ is contained in $\bar{M}$ . Let $g$ be the inclusion $m\unicode[STIX]{x1D719}(\bar{L})\subset \bar{M}$ , and define the correspondence $c_{\unicode[STIX]{x1D719}}=(\bar{L},m,gm\unicode[STIX]{x1D719}):\bar{L}\rightleftharpoons \bar{M}$ . It follows from Lemma 4.1 that $c_{\unicode[STIX]{x1D719}}$ is a skew correspondence. By Proposition 2.6, the composition $c=c_{M}^{-1}\circ c_{\unicode[STIX]{x1D719}}\circ c_{L}:L\rightleftharpoons M$ is also a skew correspondence, and it is easy to see that ${\mathcal{F}}(c)=\unicode[STIX]{x1D719}$ .◻

Proof. Let $c=(X,f,g)$ and $d=(Y,h,j)$ . We will show that $c$ and $d$ are equivalent by showing that the fibre product $(X\times _{L\oplus M}Y,p_{0},p_{1}):X\rightleftharpoons Y$ is a commensurability.

First, assume that the images of $f$ , $g$ , $h$ and $j$ are $Z$ -torsion free. Then $f$ and $g$ factor through $X/X_{\text{tors}}$ , and similarly for $h$ and $j$ . By Lemma 6.2, the quotient maps $X\rightarrow X/X_{\text{tors}}$ and $Y\rightarrow Y/Y_{\text{tors}}$ are isogenies, so after replacing $c$ and $d$ by equivalent commensurabilities, we may assume that $X$ and $Y$ are $Z$ -torsion free. It then follows from Lemma 6.2 that $f$ , $g$ , $h$ and $j$ are injective. Since ${\mathcal{F}}(c)={\mathcal{F}}(d)$ , we have

$$\begin{eqnarray}(Q\otimes _{Z}g)\circ (Q\otimes _{Z}f)^{-1}=(Q\otimes _{Z}j)\circ (Q\otimes _{Z}h)^{-1},\end{eqnarray}$$

and it follows that the canonical injection $X\times _{L\oplus M}Y\rightarrow X\times _{L}Y$ is an isomorphism. By Proposition 2.6, the fibre product $(X\times _{L}Y,p_{0},p_{1}):X\rightleftharpoons Y$ of the diagram $X\rightarrow M\leftarrow Y$ is a commensurability, which proves this special case of the lemma.

We now prove the general case. By applying Lemma 6.2 with $U=f(X)_{\text{tors}}$ , and similarly for $g$ , $h$ and $j$ , we may choose a non-zero $m\in Z$ such that the images of $mf$ , $mg$ , $mh$ and $mj$ are $Z$ -torsion free. It is easy to see that $c$ is equivalent to $(X,mf,mg)$ and $d$ is equivalent to $(Y,mh,mj)$ . So the general case follows from the special case above.◻

Proof of Theorem 6.5.

The result follows by combining Lemmas 6.6–6.8. ◻

Proof. By Proposition 2.15, the category $_{R}\mathbf{Mod}_{\text{com}}$ is the maximal subgroupoid of $_{R}\mathbf{Mod}_{\operatorname{skew}}$ . So Theorem 6.5 implies that the functor ${\mathcal{F}}$ induces an equivalence of categories from $_{R}\mathbf{Mod}_{\text{com}}$ to the category whose objects are the finitely generated $A$ -modules, and whose morphisms are the $A$ -module isomorphisms. The corollary follows by restricting ${\mathcal{F}}$ to the full subgroupoid $G_{L}$ of $_{R}\mathbf{Mod}_{\operatorname{skew}}$ .◻

Proof. We first show that $p_{1}$ has finite kernel. We have

$$\begin{eqnarray}\ker p_{1}=\{(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706},0)\in \operatorname{End}L\times \operatorname{End}L\times \operatorname{End}M:f\unicode[STIX]{x1D706}=0\}\cong \operatorname{Hom}(L,\ker f),\end{eqnarray}$$

which is finite since $L$ is finitely generated and $\ker f$ is finite by assumption.

Next, we show that the image of $p_{1}$ has finite additive index in $\operatorname{End}M$ . The modules $L_{\text{tors}}$ and $M/f(L)$ are finite, so by Lemma 6.2 there exist non-zero $m_{1},m_{2}\in Z$ such that $m_{1}$ annihilates $L_{\text{tors}}$ and $m_{2}$ annihilates $M/f(L)$ . Thus, $f:m_{1}L\rightarrow m_{1}M$ is injective, and the image contains $m_{1}m_{2}M$ , so $f^{-1}$ defines a homomorphism $m_{1}m_{2}M\rightarrow m_{1}L$ . Given $\unicode[STIX]{x1D707}\in \operatorname{End}M$ , we may therefore define $\unicode[STIX]{x1D706}:L\rightarrow L,x\mapsto f^{-1}(m_{1}m_{2}\unicode[STIX]{x1D707}(f(x)))$ , which has the property that $(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706},m_{1}m_{2}\unicode[STIX]{x1D707})\in \operatorname{End}c_{f}$ . So the image of $p_{1}$ contains $m_{1}m_{2}\operatorname{End}M$ , which has finite additive index in $\operatorname{End}M$ by Lemma 4.1. This proves that $p_{1}$ is an isogeny.

We now show that the image of $p_{0}$ has finite additive index in $\operatorname{End}L$ . Given any $\unicode[STIX]{x1D706}\in \operatorname{End}L$ , we may define $\unicode[STIX]{x1D707}:M\rightarrow M$ , $y\mapsto f(\unicode[STIX]{x1D706}(f^{-1}(m_{1}m_{2}y)))$ , where $m_{1},m_{2}$ are as before. We then have $(m_{1}m_{2}\unicode[STIX]{x1D706},m_{1}m_{2}\unicode[STIX]{x1D706},\unicode[STIX]{x1D707})\in \operatorname{End}c_{f}$ . So the image of $p_{0}$ contains $m_{1}m_{2}\operatorname{End}L$ , which has finite additive index in $\operatorname{End}L$ by Lemma 4.1.

Finally, we show that $p_{0}$ has finite kernel. We have

$$\begin{eqnarray}\displaystyle \ker p_{0} & = & \displaystyle \{(0,0,\unicode[STIX]{x1D707})\in \operatorname{End}L\times \operatorname{End}L\times \operatorname{End}M:\unicode[STIX]{x1D707}f=0\}\nonumber\\ \displaystyle & \cong & \displaystyle \operatorname{Hom}(M/f(L),M)\cong \operatorname{Hom}(M/f(L),M_{\text{tors}}),\nonumber\end{eqnarray}$$

where the last isomorphism follows from Lemma 6.2 and the assumption that $M/f(L)$ is finite. Invoking Lemma 6.2 again, it follows that $\ker p_{0}$ is finite, so $p_{0}$ is an isogeny.◻

Proof. The correspondence $\text{e}(c)$ is canonically isomorphic to the composition of $\text{e}(c_{f})^{-1}:\operatorname{End}L\rightleftharpoons \operatorname{End}X$ with $\text{e}(c_{g}):\operatorname{End}X\rightleftharpoons \operatorname{End}M$ . The correspondences $\text{e}(c_{f})$ and $\text{e}(c_{g})$ are commensurabilities by Lemma 7.1, so $\text{e}(c)$ is a commensurability by Proposition 2.6. The assertion on $\text{a}(c)$ follows from Theorem 3.8 by passing to the unit groups.◻

Proof. (a) Write $c=(X,f,g):L\rightleftharpoons M$ , $d=(Y,h,j):$ $M\rightleftharpoons N$ . We claim that there is an isogeny

$$\begin{eqnarray}i:\operatorname{End}c\times _{\operatorname{End}M}\operatorname{End}d\rightarrow \operatorname{End}(d\circ c)\end{eqnarray}$$

that makes the following diagram of endomorphism rings commute:

where all unlabelled morphisms are those defined in the introduction.

An element of $\operatorname{End}c\times _{\operatorname{End}M}\operatorname{End}d$ is a pair of triples

$$\begin{eqnarray}\displaystyle & & \displaystyle ((\unicode[STIX]{x1D706},\unicode[STIX]{x1D709},\unicode[STIX]{x1D707}),(\unicode[STIX]{x1D707}^{\prime },\unicode[STIX]{x1D710},\unicode[STIX]{x1D708})),\nonumber\\ \displaystyle & & \displaystyle \unicode[STIX]{x1D706}\in \operatorname{End}L,\quad \unicode[STIX]{x1D709}\in \operatorname{End}X,\quad \unicode[STIX]{x1D707},\unicode[STIX]{x1D707}^{\prime }\in \operatorname{End}M,\quad \unicode[STIX]{x1D710}\in \operatorname{End}Y,\quad \unicode[STIX]{x1D708}\in \operatorname{End}N,\nonumber\end{eqnarray}$$

satisfying $\unicode[STIX]{x1D706}f=f\unicode[STIX]{x1D709}$ , $\unicode[STIX]{x1D707}g=g\unicode[STIX]{x1D709}$ , $\unicode[STIX]{x1D707}^{\prime }h=h\unicode[STIX]{x1D710}$ , $\unicode[STIX]{x1D708}j=j\unicode[STIX]{x1D710}$ , and the fibre product condition in fact demands that $\unicode[STIX]{x1D707}=\unicode[STIX]{x1D707}^{\prime }$ .

An element of $\operatorname{End}(d\circ c)$ is a triple $(\unicode[STIX]{x1D706}^{\prime },\unicode[STIX]{x1D701}^{\prime },\unicode[STIX]{x1D708}^{\prime })\in \operatorname{End}L\times \operatorname{End}(X\times _{M}Y)\times \operatorname{End}N$ satisfying $\unicode[STIX]{x1D706}^{\prime }fp_{0}=\unicode[STIX]{x1D701}^{\prime }fp_{0}$ , $\unicode[STIX]{x1D708}^{\prime }jp_{1}=jp_{1}\unicode[STIX]{x1D701}^{\prime }$ , where $p_{0}$ , $p_{1}$ are the canonical projection maps from $X\times _{M}Y$ to $X$ , respectively $Y$ . Define

$$\begin{eqnarray}\begin{array}{@{}llll@{}}i:\quad & \text{End}\,c\times _{\operatorname{End}M}\operatorname{End}d\ & \rightarrow \ & \operatorname{End}(d\circ c)\\ \quad & ((\unicode[STIX]{x1D706},\unicode[STIX]{x1D709},\unicode[STIX]{x1D707}),(\unicode[STIX]{x1D707},\unicode[STIX]{x1D710},\unicode[STIX]{x1D708}))\ & \mapsto \ & (\unicode[STIX]{x1D706},(\unicode[STIX]{x1D709},\unicode[STIX]{x1D710}),\unicode[STIX]{x1D708}).\end{array}\end{eqnarray}$$

A routine verification, which we leave to the reader, shows that the image of $i$ is indeed contained in $\operatorname{End}(d\circ c)$ .

To see that this definition of $i$ makes the above diagram of endomorphism rings commute is also routine, and will also be omitted. It remains to check that $i$ is an isogeny. The correspondence $\text{e}(d)\circ \text{e}(c):\operatorname{End}L\rightleftharpoons \operatorname{End}N$ consists of $\operatorname{End}c\times _{\operatorname{End}M}\operatorname{End}d$ , together with the maps to $\operatorname{End}L$ and $\operatorname{End}N$ . By Theorem 7.2, the correspondences $\text{e}(c)$ and $\text{e}(d)$ are commensurabilities, so by Proposition 2.6 the correspondence $\text{e}(d)\circ \text{e}(c)$ is a commensurability. In particular, the morphism $\operatorname{End}c\,\times _{\operatorname{End}M}\operatorname{End}d\rightarrow \operatorname{End}L$ is an isogeny. Also, $\operatorname{End}(d\circ c)\rightarrow \operatorname{End}L$ is an isogeny by Theorem 7.2. The fact that $i$ is an isogeny therefore follows from Proposition 2.1. This proves our claim.

The isogeny $i$ defines an equivalence between $\text{e}(d\circ c)$ and $\text{e}(d)\circ \text{e}(c)$ . This proves part (a) for endomorphism rings. By passing to the unit groups and applying Theorem 3.8 to the isogeny $i$ , we also obtain part (a) for automorphism groups.

Part (b) immediately follows from part (a) by Propositions 2.12 and 2.6. ◻

Proof. Let $c=(X,f,g)$ and $d=(Y,h,j)$ . First, assume that an equivalence between $c$ and $d$ is given by an isogeny $p:Y\rightarrow X$ , so that we have the following commutative diagram:

As before, write $c_{f}=(X,\operatorname{id},f):X\rightleftharpoons L$ , and define $c_{g}$ , $c_{p}$ similarly. Then $d$ is canonically isomorphic to $(c_{g}\circ c_{p})\circ (c_{p}^{-1}\circ c_{f}^{-1})$ . By Theorem 7.3, the commensurability $\text{e}(d)$ is equivalent to $\text{e}(c_{g})\circ \text{e}(c_{p})\circ \text{e}(c_{p})^{-1}\circ \text{e}(c_{f}^{-1})$ . By Proposition 2.13, the composition $\text{e}(c_{p})\circ \text{e}(c_{p})^{-1}$ is equivalent to $(\operatorname{End}X,\operatorname{id},\operatorname{id}):\operatorname{End}X\rightleftharpoons \operatorname{End}X$ . So by Proposition 2.11 the commensurability $\text{e}(c_{g})\circ \text{e}(c_{p})\circ \text{e}(c_{p})^{-1}\circ \text{e}(c_{f}^{-1})$ is equivalent to $\text{e}(c_{g})\circ (\operatorname{End}X,\operatorname{id},\operatorname{id})\circ \text{e}(c_{f}^{-1})$ , which is canonically isomorphic to $\text{e}(c_{g})\circ \text{e}(c_{f}^{-1})$ . Applying Theorem 7.3 again, we find that $\text{e}(c_{g})\circ \text{e}(c_{f}^{-1})$ is equivalent to $\text{e}(c_{g}\circ c_{f}^{-1})$ . Finally, $c_{g}\circ c_{f}^{-1}$ is canonically isomorphic to $c$ , and the special case of the proposition follows.