Proof. See [Reference Borevič and FaddeevBF60]. For generalizations to other integral domains, see [Reference BassBas63, §7], [Reference Borevič and FaddeevBF65], [Reference LevyLev85], and the survey article [Reference SalceSal02]. ◻

Proof.

1. (a) See [Reference ReinerRei03, Corollary 21.5].

2. (b) This follows from the final statement of [Reference ReinerRei03, Corollary 21.5].

3. (c) This is a classical result due to Eichler [Reference EichlerEic38]; see also [Reference Shioda and LønstedShi79, Theorem 3.5]. ◻

Proof. Taking $L$ -points of (4) yields an exact sequence

$$\begin{eqnarray}0\rightarrow A(L)\rightarrow E(L)^{n}\rightarrow E(L)^{m}.\end{eqnarray}$$

On the other hand, applying $\operatorname{Hom}_{R}(-,E(L))$ to (3) yields an exact sequence

$$\begin{eqnarray}0\rightarrow \operatorname{Hom}_{R}(M,E(L))\rightarrow E(L)^{n}\rightarrow E(L)^{m}.\end{eqnarray}$$

The maps $E(L)^{n}\rightarrow E(L)^{m}$ in both sequences are the same, so the result follows.◻

Proof. For any $n\geqslant 1$ , the presentation $R\stackrel{n}{\rightarrow }R\rightarrow R/nR\rightarrow 0$ shows that $\mathscr{H}\!\mathit{om}_{R}(R/nR,E)\simeq E[n]$ . If $M$ is torsion, then it is a quotient of $(R/nR)^{m}$ for some $m,n\geqslant 1$ ; then $A\subseteq E[n]^{m}$ , so $A$ is finite.

In general, let $r=\operatorname{rk}M$ . There is an exact sequence

$$\begin{eqnarray}0\rightarrow R^{r}\rightarrow M\rightarrow T\rightarrow 0\end{eqnarray}$$

for some torsion module $T$ ; this yields

(6) $$\begin{eqnarray}0\rightarrow \mathscr{H}\!\mathit{om}_{R}(T,E)\rightarrow A\rightarrow E^{r}.\end{eqnarray}$$

By the previous paragraph, $\mathscr{H}\!\mathit{om}_{R}(T,E)$ is finite, so $\dim A\leqslant r\dim E$ . There exists a nonzero $\unicode[STIX]{x1D70C}\in R$ such that $\unicode[STIX]{x1D70C}T=0$ . Since $R$ is f.p. as a $\mathbb{Z}$ -module, it follows that there exists a positive integer $n$ such that $nT=0$ . Then $R^{r}\stackrel{n}{\rightarrow }R^{r}$ factors as $R^{r}{\hookrightarrow}M\rightarrow R^{r}$ , which induces $E^{r}\rightarrow A\rightarrow E^{r}$ whose composition is multiplication by $n$ , which is surjective. Thus $A\rightarrow E^{r}$ is surjective, so $\dim A\geqslant r\dim E$ . Hence $\dim A=r\dim E$ .◻

Proof. (a) Let $r=\operatorname{rk}M=\dim A$ . The proof of Proposition 4.3 shows that $A$ admits a surjection to $E^{r}$ with finite kernel, so if $A$ is an abelian variety, it is isogenous to $E^{r}$ .

The ring $R$ is either $\mathbb{Z}$ , a quadratic order, or a maximal quaternionic order. In the first and third cases, $M$ is projective of rank $r$ over $R$ (the quaternionic case is Theorem 3.3(a)); in other words, $M$ is a direct summand of $R^{n}$ for some $n$ ; thus $A$ is a direct factor of $E^{n}$ , so $A$ is an abelian variety.

So suppose that $R$ is a quadratic order. Let $c$ be the conductor, i.e., the index of $R$ in its integral closure. Let $\ell$ denote a prime. If $\ell \nmid c$ , then the semi-local ring $R\otimes \mathbb{Z}_{(\ell )}$ is a Dedekind domain, but a semi-local Dedekind domain is a principal ideal domain, so $M\otimes \mathbb{Z}_{(\ell )}$ is free of rank  $r$ over $R\otimes \mathbb{Z}_{(\ell )}$ , and $M/\ell M$ is free of rank $r$ over $R/\ell R$ .

We claim that $A$ is smooth. This is automatic if $\operatorname{char}k=0$ . So suppose that $\operatorname{char}k=p>0$ . By [Reference WaterhouseWat69, Theorem 4.2], we have $p\nmid c$ , so by the above, $M/pM$ is free of rank $r$ over $R/pR$ . By Proposition 4.2, applying $\operatorname{Hom}_{R}(M,-)$ to

$$\begin{eqnarray}0\longrightarrow \operatorname{Lie}E\longrightarrow E(k[\unicode[STIX]{x1D716}]/(\unicode[STIX]{x1D716}^{2}))\longrightarrow E(k)\longrightarrow 0\end{eqnarray}$$

yields

$$\begin{eqnarray}0\longrightarrow \operatorname{Hom}_{R}(M,\operatorname{Lie}E)\longrightarrow A(k[\unicode[STIX]{x1D716}]/(\unicode[STIX]{x1D716}^{2}))\longrightarrow A(k)\longrightarrow 0.\end{eqnarray}$$

Thus

$$\begin{eqnarray}\operatorname{Lie}A\simeq \operatorname{Hom}_{R}(M,\operatorname{Lie}E)\simeq \operatorname{Hom}_{R/pR}(M/pM,\operatorname{Lie}E)\simeq (\operatorname{Lie}E)^{r}.\end{eqnarray}$$

In particular, $\dim \operatorname{Lie}A=r$ , so $A$ is smooth.

Since $A$ is also proper, it is an extension of a finite étale commutative group scheme $\unicode[STIX]{x1D6F7}$ by an abelian variety $B$ . The constructed surjection $A\rightarrow E^{r}$ with finite kernel restricts to a homomorphism $B\rightarrow E^{r}$ with finite kernel, and it must still be surjective since $E^{r}$ does not have algebraic subgroups of finite index; thus $B$ is isogenous to $E^{r}$ . Since $B(\overline{k})$ is divisible, the extension splits over $\overline{k}$ . In particular, for each prime $\ell$ ,

(7) $$\begin{eqnarray}\#A(\overline{k})[\ell ]=\#E(\overline{k})[\ell ]^{r}\#\unicode[STIX]{x1D6F7}[\ell ].\end{eqnarray}$$

On the other hand, Proposition 4.2 implies

(8) $$\begin{eqnarray}A(\overline{k})[\ell ]=\operatorname{Hom}_{R}(M,E(\overline{k})[\ell ])=\operatorname{Hom}_{R/\ell R}(M/\ell M,E(\overline{k})[\ell ]).\end{eqnarray}$$

We claim that

(9) $$\begin{eqnarray}\#A(\overline{k})[\ell ]=\#E(\overline{k})[\ell ]^{r}.\end{eqnarray}$$

If $\ell \nmid c$ , then $M/\ell M$ is free of rank  $r$ over $R/\ell R$ , so (9) holds. Now suppose that $\ell |c$ ; in particular, $\ell \neq p$ . Then $R/\ell R\simeq \mathbb{F}_{\ell }[e]/(e^{2})$ . Every $(R/\ell R)$ -module is a direct sum of copies of $\mathbb{F}_{\ell }$ and $\mathbb{F}_{\ell }[e]/(e^{2})$ . Since $R$ is saturated in $\operatorname{End}E$ and $\ell \neq p$ , the homomorphisms

$$\begin{eqnarray}\frac{R}{\ell R}\longrightarrow \frac{\operatorname{End}E}{\ell (\operatorname{End}E)}\longrightarrow \operatorname{End}E(\overline{k})[\ell ]\end{eqnarray}$$

are injective. On the other hand, $\#E(\overline{k})[\ell ]=\ell ^{2}=\#(R/\ell R)$ . The previous three sentences imply that $E(\overline{k})[\ell ]$ is free of rank  $1$ over $R/\ell R$ . The equality $\#\operatorname{Hom}_{R/\ell R}(N,R/\ell R)=\#N$ holds for $N=\mathbb{F}_{\ell }$ and $N=\mathbb{F}_{\ell }[e]/(e^{2})$ , so it holds for every finite $(R/\ell R)$ -module $N$ , and in particular for $M/\ell M$ . Thus (8) implies

$$\begin{eqnarray}\#A(\overline{k})[\ell ]=\#(M/\ell M)=\#(R/\ell R)^{r}=\#E(\overline{k})[\ell ]^{r};\end{eqnarray}$$

the middle equality holds since $M$ and $R^{r}$ are torsion-free $\mathbb{Z}$ -modules of the same rank. Hence (9) holds for all $\ell$ .

Comparing (7) and (9) shows that $\#\unicode[STIX]{x1D6F7}[\ell ]=1$ for all $\ell$ , so $\unicode[STIX]{x1D6F7}$ is trivial. Thus $A=B$ , an abelian variety.

(b) By Lemma 4.5 below, it suffices to show that if $M\rightarrow P$ is an injection of modules with $P$ projective, then $\mathscr{H}\!\mathit{om}_{R}(P,E)\rightarrow \mathscr{H}\!\mathit{om}_{R}(M,E)$ is surjective. We have an exact sequence

$$\begin{eqnarray}0\longrightarrow \mathscr{H}\!\mathit{om}_{R}(P/M,E)\longrightarrow \mathscr{H}\!\mathit{om}_{R}(P,E)\longrightarrow \mathscr{H}\!\mathit{om}_{R}(M,E).\end{eqnarray}$$

By (a), $\mathscr{H}\!\mathit{om}_{R}(P,E)$ and $\mathscr{H}\!\mathit{om}_{R}(M,E)$ are abelian varieties, so the image $I$ of $\mathscr{H}\!\mathit{om}_{R}(P,E)\rightarrow \mathscr{H}\!\mathit{om}_{R}(M,E)$ is an abelian subvariety of $\mathscr{H}\!\mathit{om}_{R}(M,E)$ . By Proposition 4.3,

$$\begin{eqnarray}\dim \mathscr{H}\!\mathit{om}_{R}(P,E)=\dim \mathscr{H}\!\mathit{om}_{R}(P/M,E)+\dim \mathscr{H}\!\mathit{om}_{R}(M,E),\end{eqnarray}$$

so $\dim I=\dim \mathscr{H}\!\mathit{om}_{R}(M,E)$ . Thus $I=\mathscr{H}\!\mathit{om}_{R}(M,E)$ ; i.e., $\mathscr{H}\!\mathit{om}_{R}(P,E)\rightarrow \mathscr{H}\!\mathit{om}_{R}(M,E)$ is surjective.

(c) Since $\mathscr{H}\!\mathit{om}_{R}(-,E)$ is exact, it transforms the co-image of $g$ into the image of $f$ . (Co-image equals image in any abelian category, though the proof above does not need this.)

(d) The proof of [Reference WaterhouseWat69, Proposition A.2] shows that $\mathscr{H}\!\mathit{om}_{R}(R/I,E)\simeq E[I]$ (there $R$ is equal to $\operatorname{End}E$ , but this is not used). The proof of [Reference WaterhouseWat69, Proposition A.3] shows that $E/E[I]$ is the connected component of $\mathscr{H}\!\mathit{om}_{R}(I,E)$ , but $\mathscr{H}\!\mathit{om}_{R}(I,E)$ is already connected, by (a).

(e) The function $(\#T)^{2/\text{rk}\,R}$ of $T$ is multiplicative in short exact sequences. So is the quantity $\#\mathscr{H}\!\mathit{om}_{R}(T,E)$ , since $\mathscr{H}\!\mathit{om}_{R}(-,E)$ is exact. Thus we may reduce to the case in which $T$ is simple, i.e., $T\simeq R/I$ for some maximal ideal $I$ . Then $\mathscr{H}\!\mathit{om}_{R}(T,E)=E[I]$ by (d). We have $I\supseteq \ell R$ for some prime $\ell$ . If $I=\ell R$ , then $E[I]=E[\ell ]$ , which has order $\ell ^{2}=\#(R/I)^{2/\text{rk}\,R}$ .

Now suppose that $I\neq \ell R$ . If $R$ has rank  $2$ , then $\#(R/I)=\ell$ ; if $R$ has rank  $4$ , then $\#(R/I)=\ell ^{2}$ . Choose $f\in I\setminus \ell R$ ; then $f$ does not kill $E[\ell ]$ , so $E[I]\subsetneq E[\ell ]$ . Thus $\#E[I]\leqslant \ell =\#(R/I)^{2/\text{rk}\,R}$ . Thus $\#\mathscr{H}\!\mathit{om}_{R}(T,E)\leqslant (\#T)^{2/\text{rk}\,R}$ holds for each Jordan–Hölder factor of $R/\ell R$ , but for $T=R/\ell R$ equality holds, so all the inequalities must have been equalities.

(f) Start with the exact sequence

$$\begin{eqnarray}0\rightarrow E[n]\rightarrow E\stackrel{n}{\rightarrow }E.\end{eqnarray}$$

Given $S\in {\mathcal{C}}$ , apply the left exact functors $\operatorname{Hom}_{{\mathcal{C}}}(S,-)$ and then $\operatorname{Hom}_{R}(M,-)$ ; taken for all $S$ , this produces an exact sequence of representable functors

$$\begin{eqnarray}0\longrightarrow \mathscr{H}\!\mathit{om}_{R}(M,E[n])\longrightarrow \mathscr{H}\!\mathit{om}_{R}(M,E)\stackrel{n}{\longrightarrow }\mathscr{H}\!\mathit{om}_{R}(M,E).\end{eqnarray}$$

Hence $\mathscr{H}\!\mathit{om}_{R}(M,E[n])\simeq A[n]$ .

(g) We have

$$\begin{eqnarray}\displaystyle T_{\ell }A & := & \displaystyle \underset{e}{\varprojlim }\,A[\ell ^{e}](k_{s})\nonumber\\ \displaystyle & \simeq & \displaystyle \underset{e}{\varprojlim }\,\mathscr{H}\!\mathit{om}_{R}(M,E[\ell ^{e}])(k_{s})\quad \text{(by (f))}\nonumber\\ \displaystyle & \simeq & \displaystyle \underset{e}{\varprojlim }\,\operatorname{Hom}_{R}(M,E[\ell ^{e}](k_{s}))\quad (\text{by Proposition 4.2 with }E\text{ replaced by }E[\ell ^{e}])\nonumber\\ \displaystyle & \simeq & \displaystyle \operatorname{Hom}_{R}\Bigl(M,\mathop{\varprojlim }\nolimits_{e}E[\ell ^{e}](k_{s})\Bigr)\nonumber\\ \displaystyle & =: & \displaystyle \operatorname{Hom}_{R}(M,T_{\ell }E).\square\nonumber\end{eqnarray}$$

Proof. Given $A\in {\mathcal{C}}$ , choose an epimorphism $P\rightarrow A$ with $P$ projective, and let $K$ be the kernel. The sequence $0\rightarrow K\rightarrow P\rightarrow A\rightarrow 0$ yields

$$\begin{eqnarray}0\rightarrow FA\rightarrow FP\rightarrow FK\rightarrow (R^{1}F)A\rightarrow (R^{1}F)P=0,\end{eqnarray}$$

and the hypothesis implies that $FP\rightarrow FK$ is surjective, so $(R^{1}F)A=0$ . This holds for all $A$ , so $F$ is exact.◻

Proof. Let $M$ be a f.p. torsion-free left $R$ -module. Choose a presentation as in (3),

(10) $$\begin{eqnarray}R^{m}\stackrel{X}{\longrightarrow }R^{n}\longrightarrow M\longrightarrow 0,\end{eqnarray}$$

in which the first homomorphism is right-multiplication by some $X\in \operatorname{M}_{m\times n}(R)$ on row vectors. Apply $\mathscr{H}\!\mathit{om}_{R}(-,E)$ to obtain

$$\begin{eqnarray}0\longrightarrow A\longrightarrow E^{n}\stackrel{X}{\longrightarrow }E^{m},\end{eqnarray}$$

in which $X$ now acts on the left. Taking dual abelian varieties yields

(11) $$\begin{eqnarray}E^{m}\stackrel{X^{\dagger }}{\longrightarrow }E^{n}\longrightarrow A^{\vee }\longrightarrow 0,\end{eqnarray}$$

where $X^{\dagger }\in \operatorname{M}_{n\times m}(R)$ is obtained from $X$ by taking the transpose and applying the Rosati involution entrywise.

On the other hand, applying $\operatorname{Hom}_{R}(-,R)$ to (10) yields an exact sequence of right $R$ -modules

$$\begin{eqnarray}0\longrightarrow M^{\ast }\longrightarrow R^{n}\stackrel{X}{\longrightarrow }R^{m}\end{eqnarray}$$

involving left-multiplication by $X$ on column vectors. This may be reinterpreted via the Rosati involution as an exact sequence of left $R$ -modules

$$\begin{eqnarray}0\longrightarrow M^{\ast }\longrightarrow R^{n}\stackrel{X^{\dagger }}{\longrightarrow }R^{m}\end{eqnarray}$$

involving right-multiplication by $X^{\dagger }$ on row vectors. Applying the exact functor $\mathscr{H}\!\mathit{om}_{R}(-,E)$ yields

$$\begin{eqnarray}E^{m}\stackrel{X^{\dagger }}{\longrightarrow }E^{n}\longrightarrow \mathscr{H}\!\mathit{om}_{R}(M^{\ast },E)\longrightarrow 0.\end{eqnarray}$$

Comparing with (11) yields an isomorphism

(12) $$\begin{eqnarray}\mathscr{H}\!\mathit{om}_{R}(M^{\ast },E)\simeq A^{\vee }=\mathscr{H}\!\mathit{om}_{R}(M,E)^{\vee }.\end{eqnarray}$$

Given a homomorphism of f.p. torsion-free left $R$ -modules $M\stackrel{f}{\rightarrow }N$ , we can build a commutative diagram

and apply the constructions above to show that (12) is functorial in $M$ .◻

Proof. (a) The ring $R$ is $\mathbb{Z}$ , a quadratic order, or a maximal quaternionic order. By Theorem 3.1, Theorem 3.2(i), or Theorem 3.3(b), respectively, every f.p. torsion-free left $R$ -module is a finite direct sum of nonzero left $R$ -ideals. Thus, (a) follows if, for any two nonzero $R$ -ideals $I$ and $J$ , the natural map

$$\begin{eqnarray}\operatorname{Hom}_{R}(J,I)\longrightarrow \operatorname{Hom}(\mathscr{H}\!\mathit{om}_{R}(I,E),\mathscr{H}\!\mathit{om}_{R}(J,E))\end{eqnarray}$$

is an isomorphism. If $R=\mathbb{Z}$ , this is trivial. If $R$ is a quadratic order, this is the elliptic curve case of the isomorphism given in [Reference KaniKan11, Proposition 17, (48)]. If $R$ is a maximal quaternionic order, then by Theorem 3.3(a) all f.p. torsion-free left $R$ -modules are projective, i.e., direct summands of f.p. free left $R$ -modules; since $\mathscr{H}\!\mathit{om}_{R}(-,E)$ is fully faithful when restricted to free modules, it is also fully faithful on projective modules.

(b) Let $M$ be a f.p. torsion-free left $R$ -module. Then

$$\begin{eqnarray}\operatorname{Hom}(\mathscr{H}\!\mathit{om}_{R}(M,E),E)=\operatorname{Hom}(\mathscr{H}\!\mathit{om}_{R}(M,E),\mathscr{H}\!\mathit{om}_{R}(R,E))=\operatorname{Hom}_{R}(R,M)=M\end{eqnarray}$$

since $E=\mathscr{H}\!\mathit{om}_{R}(R,E)$ and $\mathscr{H}\!\mathit{om}_{R}(-,E)$ is fully faithful by (a).

(c) As remarked in the proof of (a), every f.p. torsion-free left $R$ -module is a finite direct sum of nonzero left $R$ -ideals $I$ .◻

Proof. (a) We have $\bigcap A[I_{i}]=A[\sum I_{i}]$ for any left ideals $I_{i}$ .

(b) Let $A=\prod A_{i}$ and $G=\prod G_{i}$ . Suppose that $G=A[I]$ . For each $f\in I$ , the composition $A_{i}{\hookrightarrow}A\stackrel{f}{\rightarrow }A{\twoheadrightarrow}A_{i}$ defines $\bar{f}\in \operatorname{End}A_{i}$ , and $G_{i}$ is the intersection of the kernels of all such  $\bar{f}$ .

Conversely, suppose that $G_{i}=A_{i}[I_{i}]$ for each $i$ . Let $I:=\prod I_{i}$ denote the set of ‘diagonal’ endomorphisms $(f_{1},\ldots ,f_{n}):A\rightarrow A$ with $f_{i}\in I_{i}$ . Then $G=A[I]$ .

(c) By induction, we may assume $n=2$ . Since $I_{1}$ and $I_{2}$ are coprime $2$ -sided ideals, we have $A[I_{1}I_{2}]=A[I_{1}]\oplus A[I_{2}]$ , and every subgroup scheme $H\subseteq A[I_{1}I_{2}]$ decomposes as $H_{1}\oplus H_{2}$ , where $H_{i}\subseteq A[I_{i}]$ ; namely, $H_{i}=H\cap A[I_{i}]$ .

If $G_{1}+G_{2}$ is a kernel subgroup, then so is $G_{i}=(G_{1}+G_{2})\cap A[I_{i}]$ , by (a).

Conversely, suppose that $G_{i}=A[J_{i}]$ for some left ideal $J_{i}$ . Replace $J_{i}$ by $J_{i}+I_{i}$ to assume that $J_{i}\supseteq I_{i}$ . Let $K:=I_{2}J_{1}+I_{1}J_{2}$ . We claim that $A[K]=G_{1}+G_{2}$ . First, $I_{2}J_{1}\subseteq J_{1}$ , which kills $G_{1}$ ; also, $I_{1}J_{2}\subseteq I_{1}$ , which kills $G_{1}$ . Thus $K$ kills $G_{1}$ . Similarly, $K$ kills $G_{2}$ . Thus $G_{1}+G_{2}\subseteq A[K]$ . On the other hand, if we write $A[K]=H_{1}\oplus H_{2}$ with $H_{i}\subseteq A[I_{i}]$ , we will show that $H_{i}\subseteq G_{i}$ , so that $A[K]\subseteq G_{1}+G_{2}$ . Write $1=e_{1}+e_{2}$ with $e_{i}\in I_{i}$ . Then the subsets $e_{1}J_{1}\subseteq I_{2}J_{1}\subseteq K$ and $e_{2}J_{1}\subseteq I_{2}$ kill $H_{1}$ , so $J_{1}$ kills $H_{1}$ ; i.e., $H_{1}\subseteq A[J_{1}]=G_{1}$ . Similarly $H_{2}\subseteq G_{2}$ . So $A[K]\subseteq G_{1}+G_{2}$ . Hence $G_{1}+G_{2}=A[K]$ , a kernel subgroup.◻

Proof. (i)  $\Rightarrow$  (ii) Suppose that $G$ is a kernel subgroup, say $A[I]$ . Let $f_{1},\ldots ,f_{n}$ be generators for $I$ . Then $G$ is the kernel of

(ii)  $\Rightarrow$  (iii) This is a special case of Theorem 4.4(c).

(iii)  $\Rightarrow$  (iv) If $E^{r}/G\simeq \mathscr{H}\!\mathit{om}_{R}(M,E)$ for some f.p. torsion-free $M$ , then by Theorem 4.8(a), the natural surjection $E^{r}{\twoheadrightarrow}E^{r}/G$ comes from some injection $M{\hookrightarrow}R^{r}$ . Applying $\mathscr{H}\!\mathit{om}_{R}(-,E)$ to

$$\begin{eqnarray}0\rightarrow M\rightarrow R^{r}\rightarrow R^{r}/M\rightarrow 0\end{eqnarray}$$

yields

$$\begin{eqnarray}0\rightarrow H\rightarrow E^{r}{\twoheadrightarrow}E^{r}/G\end{eqnarray}$$

for some $H$ , which must be isomorphic to $G$ .

(iv)  $\Rightarrow$  (ii) Choose a surjection $h:R^{s}{\twoheadrightarrow}M$ . Applying $\mathscr{H}\!\mathit{om}_{R}(-,E)$ to the composition $R^{s}\stackrel{h}{{\twoheadrightarrow}}M{\hookrightarrow}R^{r}$ produces a homomorphism $E^{r}{\twoheadrightarrow}E^{r}/G{\hookrightarrow}E^{s}$ with kernel $G$ .

(ii)  $\Rightarrow$  (i) We may increase $s$ to assume that $r|s$ . Then $G$ is an intersection of $s/r$ endomorphisms of $E^{r}$ , so it is a kernel subgroup by Proposition 6.2(a).◻

Proof. (i)  $\Rightarrow$  (ii) Trivial.

(ii)  $\Rightarrow$  (iii) Suppose that $A$ is an abelian variety isogenous to $E^{r}$ . Then $A\simeq E^{r}/G$ for some finite subgroup scheme $G$ . By assumption, $G$ is a kernel subgroup. Proposition 6.3(i) $\Rightarrow$ (iii) implies that $A$ is in the image of $\mathscr{H}\!\mathit{om}_{R}(-,E)$ . The result now follows from Theorem 4.8.

(iii)  $\Rightarrow$  (i) Let $G$ be a subgroup scheme of $E^{r}$ . Then $E^{r}/G$ is isogenous to $E^{s}$ for some $s\leqslant r$ . By assumption, $\mathscr{H}\!\mathit{om}_{R}(-,E)$ is an equivalence of categories, so $E^{r}/G$ is of the form $\mathscr{H}\!\mathit{om}_{R}(M,E)$ . By Proposition 6.3(iii) $\Rightarrow$ (i), $G$ is a kernel subgroup.◻

Proof. If $C=\mathbb{Z}_{\ell }$ , this is trivial. For any ring $A$ and positive integer $n$ , the category of $A$ -modules is equivalent to the category of $\operatorname{M}_{n}(A)$ -modules [Reference LamLam99, Theorem 17.20], and the equivalence preserves injections, finite generation, and projectivity [Reference LamLam99, Remark 17.23(A)]; applying this to $A=\mathbb{Z}_{\ell }/\ell ^{e}\mathbb{Z}_{\ell }$ and $n=2$ shows that the case $C=\mathbb{Z}_{\ell }$ implies the case $C=\operatorname{M}_{2}(\mathbb{Z}_{\ell })$ .

Finally, suppose that $C$ is of rank  $2$ . Then $C/\ell ^{e}C$ is free of rank  $2$ over $\mathbb{Z}/\ell ^{e}\mathbb{Z}$ ; say, with basis $1$ , $\unicode[STIX]{x1D6FC}$ . For $c\in C/\ell ^{e}C$ , let $\unicode[STIX]{x1D706}(c)$ be the coefficient of $\unicode[STIX]{x1D6FC}$ in $c$ . Multiplying any nonzero element of $\ker \unicode[STIX]{x1D706}$ by $\unicode[STIX]{x1D6FC}$ gives an element outside $\ker \unicode[STIX]{x1D706}$ . Therefore the pairing

$$\begin{eqnarray}\displaystyle C/\ell ^{e}C\times C/\ell ^{e}C & \longrightarrow & \displaystyle \mathbb{Z}/\ell ^{e}\mathbb{Z},\nonumber\\ \displaystyle x,y & \longmapsto & \displaystyle \unicode[STIX]{x1D706}(xy)\nonumber\end{eqnarray}$$

is a perfect pairing. In other words, the Pontryagin dual $(C/\ell ^{e}C)^{D}$ is isomorphic to $C/\ell ^{e}C$ as a $C/\ell ^{e}C$ -module. If $M$ is a finitely generated $C/\ell ^{e}C$ -module, there exists a surjection $(C/\ell ^{e}C)^{r}{\twoheadrightarrow}M^{D}$ for some $r\in \mathbb{Z}_{{\geqslant}0}$ ; taking Pontryagin duals yields an injection

$$\begin{eqnarray}M{\hookrightarrow}((C/\ell ^{e}C)^{r})^{D}\simeq (C/\ell ^{e}C)^{r}.\end{eqnarray}$$

◻

Proof. Since $E[\ell ^{e}](k_{s})=T_{\ell }E/\ell ^{e}T_{\ell }E$ , by Nakayama’s lemma it is enough to check that $E[\ell ](k_{s})^{2}$ is free as an $A$ -module, for $A=R/\ell R$ and for $A=C/\ell C$ . Identify $E[\ell ](k_{s})^{2}$ with $\mathbb{F}_{\ell }^{2}$ , so that $A\subseteq \operatorname{M}_{2}(\mathbb{F}_{\ell })$ . The case $A=\mathbb{F}_{\ell }$ is trivial. If $A$ is $\mathbb{F}_{\ell }\oplus \mathbb{F}_{\ell }\unicode[STIX]{x1D6FC}$ for some $\unicode[STIX]{x1D6FC}\in \operatorname{M}_{2}(\mathbb{F}_{\ell })$ , then every faithful $A$ -module of dimension  $2$ over $\mathbb{F}_{\ell }$ is free. If $A=\operatorname{M}_{2}(\mathbb{F}_{\ell })$ , then the free $A$ -module $A$ is a direct sum of two copies of $\mathbb{F}_{\ell }^{2}$ (the two column spaces).◻

Proof. The first map is an isomorphism since $C=\operatorname{End}_{R_{\ell }}T_{\ell }E$ and $C$ and $R_{\ell }$ are saturated in $\operatorname{End}T_{\ell }E\simeq \operatorname{M}_{2}(\mathbb{Z}_{\ell })$ . Lemma 6.6 and [Reference LamLam99, Theorem 18.8(3) $\Rightarrow$ (1)] imply that $E[\ell ^{e}](k_{s})$ is a generator of the category of finitely generated $R/\ell ^{e}R$ -modules, so [Reference LamLam99, Proposition 18.17(2)(d)] yields the second isomorphism.◻

Proof. Suppose that $G(k_{s})$ is a $C/\ell ^{e}C$ -submodule of $E[\ell ^{e}]^{r}(k_{s})$ . Let $H:=E[\ell ^{e}]^{r}/G$ . Then $H(k_{s})$ is a finitely generated $C/\ell ^{e}C$ -module. By Lemma 6.5, $H(k_{s})$ injects into a free $C/\ell ^{e}C$ -module, which in turn injects into $E[\ell ^{e}]^{s}(k_{s})$ for some $s$ . Because of the homomorphisms ${\mathcal{G}}_{k}\rightarrow C^{\times }\rightarrow (C/\ell ^{e}C)^{\times }$ , the $C/\ell ^{e}C$ -module homomorphism $H(k_{s})\rightarrow E[\ell ^{e}]^{s}(k_{s})$ is a ${\mathcal{G}}_{k}$ -module homomorphism, so it comes from a homomorphism $H\rightarrow E[\ell ^{e}]^{s}$ of étale group schemes. The composition $E[\ell ^{e}]^{r}{\twoheadrightarrow}H{\hookrightarrow}E[\ell ^{e}]^{s}$ is given by an $s\times r$ matrix $N_{e}$ with entries in $\operatorname{End}_{C/\ell ^{e}C}E[\ell ^{e}](k_{s})=R/\ell ^{e}R$ (the equality is Lemma 6.7). Lift $N_{e}$ to $N\in \operatorname{M}_{s\times r}(R)$ . Then $G$ is the intersection of the kernel subgroups $E[\ell ^{e}]^{r}$ and $\ker (N:E^{r}\rightarrow E^{s})$ . By Propositions 6.2(a) and 6.3,  $G$ is a kernel subgroup.

Conversely, if $G$ is a kernel subgroup, say the kernel of $E^{r}\rightarrow E^{s}$ , then it is also the kernel of $E[\ell ^{e}]^{r}\rightarrow E[\ell ^{e}]^{s}$ , which is a homomorphism of $C/\ell ^{e}C$ -modules, so $G$ is a $C/\ell ^{e}C$ -module.◻

Proof. (i)  $\Rightarrow$  (ii) Nakayama’s lemma.

(ii)  $\Rightarrow$  (iii) Let $G$ be an $\ell$ -power subgroup scheme of $E^{r}$ , say $G\subseteq E[\ell ^{e}]^{r}$ . Then $G(k_{s})$ is a $\mathbb{Z}_{\ell }[{\mathcal{G}}_{k}]$ -module. Since $\mathbb{Z}_{\ell }[{\mathcal{G}}_{k}]\rightarrow C$ is surjective, $G(k_{s})$ is also a $C$ -module, and hence a $C/\ell ^{e}C$ -module. By Proposition 6.8, $G$ is a kernel subgroup.

(iii)  $\Rightarrow$  (i) Suppose that $\mathbb{F}_{\ell }[{\mathcal{G}}_{k}]\rightarrow C/\ell C$ is not surjective; let $D$ be the image. The algebra $C/\ell C$ is one of $\mathbb{F}_{\ell }$ , $\left\{\!\left(\begin{smallmatrix}a & b\\ 0 & a\end{smallmatrix}\right)\!\right\}\simeq \mathbb{F}_{\ell }[\unicode[STIX]{x1D716}]/(\unicode[STIX]{x1D716}^{2})$ , $\left\{\!\left(\begin{smallmatrix}a & 0\\ 0 & b\end{smallmatrix}\right)\!\right\}\simeq \mathbb{F}_{\ell }\times \mathbb{F}_{\ell }$ , $\mathbb{F}_{\ell ^{2}}$ , or $\operatorname{M}_{2}(\mathbb{F}_{\ell })$ . The first is excluded since it has no nontrivial subalgebras. In the second, third, and fourth cases, $D$ can only be $\mathbb{F}_{\ell }$ , and it is easy to find a subspace of $\mathbb{F}_{\ell }^{2}\simeq E[\ell ](k_{s})$ that is not a $C/\ell C$ -module. In the fifth case, $D$ is contained in a copy of either $\left\{\!\left(\begin{smallmatrix}a & b\\ 0 & c\end{smallmatrix}\right)\!\right\}$ or $\mathbb{F}_{\ell ^{2}}$ . Now $\left\{\!\left(\begin{smallmatrix}a & b\\ 0 & c\end{smallmatrix}\right)\!\right\}$ fixes a line in $\mathbb{F}_{\ell }^{2}$ not fixed by $\operatorname{M}_{2}(\mathbb{F}_{\ell })$ . And $\mathbb{F}_{\ell ^{2}}$ fixes an $\mathbb{F}_{\ell ^{2}}$ -line in $\mathbb{F}_{\ell ^{2}}^{2}\simeq E^{2}[\ell ](k_{s})$ that is not fixed by $\operatorname{M}_{2}(\mathbb{F}_{\ell })$ . Thus in each case, there is a subgroup scheme of $E[\ell ]$ or $E^{2}[\ell ]$ that is not a $C/\ell C$ -module, and hence by Proposition 6.8 not a kernel subgroup.◻

Proof. The ring $R:=\operatorname{End}E\simeq \operatorname{End}E_{\overline{k}}$ is a quadratic order. Although $R$ is not necessarily a Dedekind domain, its conductor is prime to $p$ , so it makes sense to speak of the splitting behavior of $(p)$ in $R$ . In fact, since $E$ is ordinary, $(p)$ splits, say as $\mathfrak{p}\mathfrak{q}$ . So $E[p]$ is the direct sum of group schemes $E[\mathfrak{p}]$ and $E[\mathfrak{q}]$ , each of order  $p$ by Theorem 4.4(e). Since $E$ is ordinary, one of them, say $E[\mathfrak{p}]$ , is étale, and the other is connected. For any $e\in \mathbb{Z}_{{\geqslant}0}$ , we have $(p^{e})=\mathfrak{p}^{e}\mathfrak{q}^{e}$ so $E[p^{e}]\simeq E[\mathfrak{p}^{e}]\oplus E[\mathfrak{q}^{e}]$ . The Jordan–Hölder factors of $E[\mathfrak{p}^{e}]$ are isomorphic to $E[\mathfrak{p}]$ , so $E[\mathfrak{p}^{e}]$ is étale; similarly $E[\mathfrak{q}^{e}]$ is connected. We have $G\subseteq E[p^{e}]^{r}$ for some $e$ . By Proposition 6.2(c), we may assume that $G\subseteq E[\mathfrak{p}^{e}]^{r}$ or $G\subseteq E[\mathfrak{q}^{e}]^{r}$ .

In the first case, $E[\mathfrak{p}^{e}](k^{s})\simeq \mathbb{Z}/p^{e}\mathbb{Z}$ , so $G$ is the kernel of a homomorphism $E[\mathfrak{p}^{e}]^{r}\rightarrow E[\mathfrak{p}^{e}]^{s}$ given by a matrix in $\operatorname{M}_{s\times r}(\mathbb{Z})$ . Since $E[\mathfrak{p}^{e}]$ is a kernel subgroup, so is $E[\mathfrak{p}^{e}]^{r}$ , and so is $G$ , by Propositions 6.2(a) and 6.3.

In the second case, we take Cartier duals: $E[\mathfrak{p}^{e}]^{r}{\twoheadrightarrow}G^{\vee }$ . Then $G^{\vee }$ is the cokernel of some homomorphism $E[\mathfrak{p}^{e}]^{s}\rightarrow E[\mathfrak{p}^{e}]^{r}$ given by a matrix $N\in \operatorname{M}_{r\times s}(\mathbb{Z})$ . So $G$ is the kernel of the homomorphism $E[\mathfrak{q}^{e}]^{r}\rightarrow E[\mathfrak{q}^{e}]^{s}$ given by the transpose $N^{T}\in \operatorname{M}_{s\times r}(\mathbb{Z})$ . Since $E[\mathfrak{q}^{e}]$ is a kernel subgroup, so is $E[\mathfrak{q}^{e}]^{r}$ , and so is $G$ , by Propositions 6.2(a) and 6.3.◻