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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

The construction must first specify the field of definition of the complex multiplication. Let $E/F$ be a CM elliptic curve and let $K$ be its geometric CM field, embedded by its action on an invariant differential. In characteristic zero this action is injective. Consequently every geometric endomorphism is defined over $F^{\prime }=FK$: a Galois automorphism fixing $FK$ fixes its differential scalar and therefore fixes the endomorphism itself. Initially assume $F$ already contains $K$; we return to the other case below.

Choose a complex embedding and an analytic complex uniformization of an elliptic curvewhere $\mathfrak{a}$ is a lattice in $K$, stable under the CM order $R$. It need not be the maximal order or a principal ideal. Its torsion is described by the finite adelic quotients of this lattice.

The input from the CM reciprocity theorem is precise: if $s$ is an idele of $F$, $n=N_{F/K}s$, and $\sigma ={\mathrm{Art}}_F(s)$, then $\sigma$ takes the CM torus to the torus with lattice $n_f^{-1}L$; on torsion the comparison is induced by multiplication by $n_f^{-1}$. This reciprocity statement comes from the ideal action on CM lattices: an ideal $\mathfrak{b}$ gives the isogeny to ${\mathfrak{b}}^{-1}L$, composition multiplies ideals, and adding torsion level structure removes automorphism ambiguity. The resulting action on algebraic CM points is the ray-class Artin action. This is the deep CM input, not an assertion that an arbitrary abelian Galois representation automatically comes from an algebraic Hecke character.

Because $\sigma$ fixes $F$, the algebraic curve $E^{\sigma }$ is $E$. Comparing the reciprocity uniformization with the original one therefore gives multiplication by a scalarThe scalar belongs to $K$: a complex scalar carrying one lattice contained in $K$ to another is the quotient of two nonzero lattice elements. Its torsion action isThe uniqueness of the torsion-compatible comparison is important: two possible scalars would induce the same action on all torsion, hence their difference would annihilate torsion of arbitrarily large order and be zero. Multiplying the torsion-action formulas for $s_f$ and $t_f$ now proves $a(s_f{t}_f)=a(s_f)a(t_f)$.

For $b\in F^{\times }$, Artin reciprocity sends its principal idele to the identity. The archimedean factors are complex and connected, so its finite part also has trivial Artin image. The preceding formula therefore givesIt follows thatis trivial on principal ideles and defines a character of the idèle class group of $F$. This establishes the multiplicativity and principal-idele condition, rather than merely prescribing Frobenius values and hoping that they extend.

We must still prove continuity and a finite conductor. For a finite place $v$ and a prime $\ell$ different from its residue characteristic, an idele supported on the units at $v$ has trivial $\ell$-component of its norm. Thus its action on the Tate module is exactly multiplication by $a(u)$. At a place of good reduction, inertia acts trivially, so $a(u)=1$. At a bad place, a CM curve has potentially good reduction, and inertia on the prime-to-residue-characteristic Tate module has finite image. Hence $a$ is trivial on an open subgroup of the local units there. There are only finitely many bad places. Together these facts give an open finite-congruence kernel on the unit ideles and prove continuity. The archimedean norm factor is already continuous. The potentially-good-reduction and inertia facts are the reduction-theoretic inputs in Serre and Tate's treatment of CM abelian varieties.

Consequently ${\psi }_E$ is a Hecke character. In ideal notation, for $b\equiv 1(\mathrm{mod}\mathfrak{f})$,so its infinity type of a Hecke character is the type determined by the chosen CM embedding, with the inverse archimedean norm in the idele convention. Nonmaximal orders merely contribute additional finite-level conditions; the proof retains their lattice and does not replace it by the maximal-order lattice without justification.

At a good prime $v$, choose a uniformizer idele and a prime $\ell$ different from the residue characteristic. Its norm has trivial $\ell$-component, so arithmetic Frobenius acts on the Tate module by ${\psi }_E(v)$. Specialization of prime-to-residue-characteristic torsion identifies this with the reduced Frobenius isogeny. ThusThe corresponding Hasse-Weil L-function factors as $L(E/F,s)=L({\psi }_E,s)L({\bar{\psi }}_E,s)$, including the local factors determined by inertia. This explains why the Hecke character encodes both the torsion representations and the Euler factors.

If $F$ does not contain $K$, apply the construction over $FK$. The two conjugate CM characters are exchanged by the nontrivial automorphism of $FK/F$, and the two-dimensional Galois representation over $F$ is their induced representation. In that case the natural description is $L(E/F,s)=L({\psi }_E/FK,s)$ by induction, rather than a pair of one-dimensional characters over $F$. Question 3 supplies a concrete example: $\mathbb{Q}$ does not define $[i]$, whereas $\mathbb{Q}(i)$ does. This completes the existence argument for an arbitrary CM curve over a number field, with its field of definition correctly accounted for.