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The mapis an automorphism of the elliptic curve over $K$, fixes its identity point, and squares to $[-1]$. For the invariant differential on an elliptic curve $\omega =dx/(2y)$, it satisfies $[i{]}^{\ast }\omega =i\omega$. Thus $\mathbb{Z}[i]\subseteq {\mathrm{End}}_K(E)$, with the indicated embedding of $K$.

Under analytic complex uniformization of an elliptic curve $E(\mathbb{C})\simeq \mathbb{C}/L$, this automorphism acts as multiplication by $i$, so $L$ is stable under $\mathbb{Z}[i]$. It is a rank-one torsion-free module over the Gaussian integers. Since that ring is a principal ideal domain, $L=\Omega \mathbb{Z}[i]$. Every analytic endomorphism is multiplication by an $a\in \mathbb{C}$ with $aL\subseteq L$, which here says exactly $a\in \mathbb{Z}[i]$. Therefore the geometric endomorphism ring of an elliptic curve is $\mathbb{Z}[i]$, and all its elements are generated over $\mathbb{Z}$ by the displayed $K$-defined automorphism. Thus ${\mathrm{End}}_K(E)=\mathbb{Z}[i]$.

If an endomorphism is defined over $\mathbb{Q}$, its pullback on the rational differential $\omega$ has scalar in $\mathbb{Q}$. The differential action is injective in characteristic zero, because a nonzero isogeny of elliptic curves is separable and has nonzero differential. Its scalar must therefore lie in $\mathbb{Q}\cap \mathbb{Z}[i]=\mathbb{Z}$. Conversely integer multiplication is defined over $\mathbb{Q}$. Hence

Put $\lambda =1+i$. The valuations of the differences of the three nonidentity fourth roots of unity from $1$ areFor instance $2=-i{\lambda }^2$, and the other two differences are associates of $\lambda$. None is divisible by ${\lambda }^3$. The group $(\mathbb{Z}[i]/{\lambda }^3{)}^{\times }$ has $8-4=4$ elements, so the four units map bijectively to it. Every ideal prime to $\lambda$ consequently has a unique primary Gaussian integer generator.

The Grössencharacter of the square-lattice elliptic curve is explicitlyFor integral ideals prime to $2$, the generator is an odd Gaussian integer; extend multiplicatively to fractional ideals prime to $2$. Products of primary generators are primary, so this is a Hecke character with ideal infinity type $(1,0)$: on a principal ideal with generator $a\equiv 1(\mathrm{mod}{\lambda }^3)$, its value is $a$. Its conductor divides ${\lambda }^3$, and cannot divide ${\lambda }^2$, since $-1\equiv 1(\mathrm{mod}{\lambda }^2)$ would force ${\psi }_E((-1))=-1$, whereas the unit ideal has value $1$. Thus the displayed conductor is exact.

We next verify directly that this character gives the Frobenius values of this particular curve, fixing the unit ambiguity. The pointsatisfies, by the chord-and-tangent addition law,For the first addition, $[i]P=(-i,1+i)$ and the chord has slope $-1$. Since $P$ is killed by ${\lambda }^3$ but not by ${\lambda }^2$, it generates the eight-element $\mathbb{Z}[i]$-module $E[{\lambda }^3]\simeq \mathbb{Z}[i]/{\lambda }^3$. Thus all this torsion is rational over $K$.

The model has discriminant $64$, so it has good reduction at every $\gamma \ne (\lambda )$. First suppose $\gamma$ lies over a split rational prime $p$. Then its residue field is ${\mathbb{F}}_p$, $p\ge 5$, and $\pi ={\psi }_E(\gamma )$ has norm $p$. Reduction of $[\pi ]$ still has degree $p$ and has zero differential, since its differential scalar $\pi$ belongs to $\gamma$. It is therefore purely inseparable of degree $p$. Factoring out the Frobenius isogeny givesfor an automorphism $u$ of the reduction. In characteristic at least five, the automorphisms of this model fixing $O$ are the four Gaussian units: comparing coefficients in $(x,y)\mapsto (v^{2}x,v^{3}y)$ gives $v^4=1$. Prime-to-$p$ torsion reduces injectively. Both $[\pi ]$ and ${\mathrm{Frob}}_p$ fix the reduction of $E[{\lambda }^3]$, the former by the primary congruence and the latter by its rationality over the residue field. Thus $u$ fixes this entire module. The unit congruences proved above force $u=1$.

For an inert rational prime $p\equiv 3(\mathrm{mod}4)$, we have $\gamma =(p)$, $N\gamma =p^2$, and its primary generator is $\pi =-p$, because $4\mid p+1$ implies ${\lambda }^3\mid (-p-1)$. To determine Frobenius without excluding $p=3$, writeReplacing $x$ by $-x$ changes each nonzero quadratic-character value to its negative, because $-1$ is a nonsquare. Hence $a_p=0$. The Frobenius characteristic equation ${\mathrm{Frob}}_p^2-[a_p]{\mathrm{Frob}}_p+[p]=0$ gives ${\mathrm{Frob}}_{p^2}=[-p]$. This proves the asserted Frobenius value at inert primes as well, and identifies the displayed Hecke character with that of $E$.

Finally, the formal parameter $t=-x/y$ is defined over the local ring at $\gamma$. Since the reduced endomorphism is the $q$-power Frobenius isogeny, with $q=N\gamma$, its pullback sends $t$ to $(-x/y{)}^q=t^q$. Passing to the completion at the identity gives the equality of formal series