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Write $\tau =(1+\sqrt{-3})/2$. The parity condition gives $R=\mathbb{Z}[\tau ]$, since $(a+b\sqrt{-3})/2=(a-b)/2+b\tau$. For any such element,If $a,b$ are even the numerator is divisible by four; if they are odd it is congruent to $1+3=0$ modulo four. Therefore $N(z)$ is a nonnegative integer. It is zero only at zero, and the Eisenstein-integer norm is multiplicative because complex conjugation respects multiplication.

If $z$ is a unit, $N(z)N(z^{-1})=1$, so $N(z)=1$. Conversely $N(z)=1$ makes $\overline{z}$ its inverse in $R$. The equation $a^2+3b^2=4$ permits exactly $(a,b)=(\pm 2,0)$ or $(\pm 1,\pm 1)$. HenceIndeed ${\tau }^2=\tau -1$ and ${\tau }^3=-1$, so this unit group is cyclic of order six.

For the prime-ideal assertion, ${\tau }^2-\tau +1=0$ and the polynomial $t^2-t+1$ has distinct roots $3,5$ modulo seven. Consequently evaluation at either root defines a surjective ring map $R\to {\mathbb{F}}_7$. Its kernelsare distinct maximal ideals, hence prime ideals. If $A+B\tau$ belongs to both, then $A+3B\equiv A+5B\equiv 0(\mathrm{mod}7)$. Subtraction gives $2B\equiv 0$, so both $A$ and $B$ are divisible by seven. The converse inclusion is immediate, provingEquivalently, the two distinct factors of the reduced polynomial give the Chinese remainder theorem decomposition $R/7R\cong {\mathbb{F}}_7\times {\mathbb{F}}_7$. This proof does not require the permitted unique-factorization assumption.

In a principal ideal domain, an ascending chain of ideals stabilizes: its union is an ideal $(d)$, and its generator belongs to one member, which then contains the union. This proves the needed ascending chain condition. If a nonzero nonunit had no finite irreducible factorization, split it into two nonunits and select a factor that still has no such factorization. Continuing would give a strictly ascending chain of principal ideals, because $a=bc$ with $c$ a nonunit implies $(a)⊊(b)$. This contradicts the chain condition, so factorization exists.

An irreducible element $p$ generates a maximal ideal: any intermediate ideal is $(d)$ with $p=dc$, and irreducibility forces $d$ or $c$ to be a unit. Therefore $p$ is a prime element. In any two irreducible factorizations, the first prime factor divides a factor of the other, hence is associate to it. Cancel and continue. This proves every principal ideal domain is a unique factorization domain.

In $\mathbb{Z}[\sqrt{-3}]$, the multiplicative field norm is $N(a+b\sqrt{-3})=a^2+3b^2$. The only units are $\pm 1$. There is no element of norm two, so every element of norm four is irreducible: a factorization into nonunits would require norms two and two. Consequentlyis a pair of genuinely different irreducible factorizations. Neither $1+\sqrt{-3}$ nor $1-\sqrt{-3}$ is associate to $2$.

Take the Eisenstein integers $R=\mathbb{Z}[\omega ]$, where $\omega =(-1+\sqrt{-3})/2$ and ${\omega }^2+\omega +1=0$. Since $\sqrt{-3}=1+2\omega$, the embedded subring is $\mathbb{Z}+2\mathbb{Z}\omega$, whose additive index in $R$ is two. Its larger ring has Eisenstein-integer norm $N(a+b\omega )=a^2-ab+b^2$ and is Euclidean. Indeed, write a complex quotient in the real basis $1,\omega$ and round both coordinates to integers. The coordinate errors $u,v$ satisfy $|u|,|v|\le 1/2$, giving $|u+v\omega {|}^2=u^2-uv+v^2\le 3/4<1$. Thus division leaves a remainder of smaller norm. A Euclidean domain is a principal ideal domain, because division by a nonzero element of least norm in an ideal leaves zero remainder. This gives the requested index-two embedding into a principal ideal domain.

The Eisenstein integers are $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$. Their multiplicative algebraic norm isGiven $\alpha ,\beta \ne 0$, choose an Eisenstein integer $q$ nearest to the complex number $\alpha /\beta$. The hexagonal lattice has covering radius $1/\sqrt{3}$, soWith $r=\alpha -q\beta$, multiplicativity givesThus division with remainder decreases the nonnegative integer norm, proving that $\boxed{\mathbb{Z}[\omega ]\text{ is a Euclidean domain}}$.

For $E:y^2=x^3+1$, let $\alpha (x,y)=(\zeta x,y)$. The three points $P,\alpha P,{\alpha }^{2}P$ lie on the horizontal line through $P$, so their sum is zero. Thusin the endomorphism ring of an elliptic curve. Since $\mathrm{deg}\alpha =1$ and complex conjugation sends $\alpha$ to ${\alpha }^2$, degree on $\mathbb{Z}[\alpha ]$ is the Eisenstein-integer norm:

A separable isogeny is determined by its kernel up to unique isomorphism of its target, and every finite Galois-stable subgroup of an elliptic curve is the kernel of the corresponding quotient isogeny. Put $T=(-1,0)$. The three nonzero points of $E[2]$ are $T,\alpha T,{\alpha }^{2}T$, andHence $\phi (\alpha -{\alpha }^2)$ kills $\ker \phi =\{O,T\}$ and factors uniquely through $\phi$:for an isogeny $\psi :E^{\prime }\to E^{\prime }$. Degrees giveMoreover,Composing the factorization twice and using the surjectivity of $\phi$ gives ${\psi }^2=[-3]$.