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Fix a separable closure $k^s$. The Tate module and rational Tate module arewhere the transition maps are multiplication by $\ell$. Since $\ell$ is different from the characteristic, $E[{\ell }^n](k^s)\simeq (\mathbb{Z}/{\ell }^n{)}^2$ and these transition maps are surjective. Choosing compatible bases gives $T_{\ell }(E)\simeq {\mathbb{Z}}_{\ell }^2$, and $T_{\ell }(E)/{\ell }^n{T}_{\ell }(E)\simeq E[{\ell }^n]$. Both spaces carry the natural continuous action of the absolute Galois group of $k$ and of the base-defined endomorphism ring of an elliptic curve.

For (i), choose a nonrational $a\in R$. Such an element exists because the intersection of the given two-dimensional rational subalgebra with the integral endomorphism ring contains a full lattice. The characteristic equation of an elliptic endomorphism isSince $a\in K$ is nonrational, its minimal polynomial over $\mathbb{Q}$ has degree two. Comparing the two monic equations gives $\mathrm{tr}(a)=a+\bar{a}$ and $\mathrm{deg}(a)=a\bar{a}$. The same is the characteristic polynomial on the rational Tate module: the Weil pairing gives determinant $\mathrm{deg}(a)$, and applying it to $1-a$ gives trace $1+\mathrm{deg}(a)-\mathrm{deg}(1-a)$. Thus on $V_{\ell }(E)$ the characteristic polynomial isIf $K_{\ell }$ is a quadratic field over ${\mathbb{Q}}_{\ell }$, a unital $K_{\ell }$-action on this two-dimensional ${\mathbb{Q}}_{\ell }$-space makes it a one-dimensional $K_{\ell }$-space. If $K_{\ell }\simeq {\mathbb{Q}}_{\ell }\times {\mathbb{Q}}_{\ell }$, the two roots above are distinct in ${\mathbb{Q}}_{\ell }$ and their eigenspaces each have dimension one. The two idempotent factors therefore have equal rank one, so the module is again free of rank one. This also proves faithfulness in the split case. Hence

For (ii), first prove an integral divisibility statement. If $a\in R$ maps $T_{\ell }(E)$ into ${\ell }^n{T}_{\ell }(E)$, then its induced action on $E[{\ell }^n]$ is zero. The kernel of an isogeny of $[{\ell }^n]$ is therefore contained in the kernel of $a$. The quotient property of this isogeny of elliptic curves gives a unique endomorphism $b$ withThis factorization is defined over $k$: applying any descent automorphism preserves it, and uniqueness fixes $b$. Inside the rational endomorphism ring, $b=a/{\ell }^n$ belongs to $K$, so $b\in R$. ConsequentlyHere the prime-to-characteristic torsion is étale, so vanishing on its geometric points is vanishing on the entire kernel scheme.

Every element of $R_{\ell }$ preserves the Tate module by continuity. Conversely suppose $\varphi \in K_{\ell }$ preserves it. Choose $n\ge 0$ with ${\ell }^n\varphi \in R_{\ell }$, and approximate it by $a\in R$ modulo ${\ell }^n{R}_{\ell }$. Write ${\ell }^n\varphi =a+{\ell }^{n}c$ with $c\in R_{\ell }$. Then $a{T}_{\ell }(E)\subseteq {\ell }^n{T}_{\ell }(E)$, so the divisibility statement gives $a/{\ell }^n\in R$. It follows that $\varphi =a/{\ell }^n+c\in R_{\ell }$. Thus the CM Tate module stabilizer isThis does not presume freeness of the integral Tate module over a nonmaximal order.

For (iii), every element of $R$ is defined over $k$, so the Galois representation commutes with $R$ and therefore with $K_{\ell }$. By (i), its action on $V_{\ell }(E)$ is multiplication by an element of $K_{\ell }^{\times }$. The action and its inverse preserve $T_{\ell }(E)$; by (ii), this scalar and its inverse belong to $R_{\ell }$. Hence