Arithmetics in the ring \( \mathbb Z[\omega] \)

FTA holds in the ring \( \mathbb{Z}[\omega] \).

By the theorem 5, it suffices to show that a division with remainder is possible, i.e. for all \( a,b\in\mathbb{Z}[\omega] \), \( b\neq0 \) there exist \( p\in\mathbb{Z}[\omega] \) such that \( N(a-pb)< N(b) \).

Like in the proof for the Gaussian integers, let \( \sigma,\tau\in\mathbb{R} \) be such that \( a/b=\sigma+\tau i \), and let \( s,t\in\mathbb{Z} \) be such that \( |\sigma-s|\leq 1/2 \) and \( |\tau-t|\leq1/2 \). Setting \( p=s+ti \) gives us \( N(a-pb)\leq 3N(b)/4< N(b), \) implying the theorem.

If \( p\equiv1 \) (mod 6) is a prime number, prove that there exist \( a,b\in\mathbb{Z} \) such that \( p=a^2-ab+b^2 \).

It suffices to show that \( p \) is composite in \( \mathbb{Z}[\omega] \). Indeed, if there is a prime element \( z=a+b\omega\in\mathbb{Z}[\omega] \) (\( a,b\in\mathbb{Z} \)) that divides \( p \), then also \( \overline{z}\mid\overline{p}=p \). Note that \( z \) and \( \overline{z} \) are coprime; otherwise \( z\mid\overline{z} \), so there exists a unit element \( \epsilon \) with \( \overline{z}=\epsilon z \), and hence \( z\sim (1-\omega)\mid 3 \), which is false. Therefore \( a^2-ab+b^2=z\overline{z}\mid p \), which implies \( a^2-ab+b^2=p \).

Thus we need to prove that \( p \) is composite in \( \mathbb{Z}[\omega] \). It follows from the condition on \( p \) that -3 is a quadratic residue modulo \( p \), so there exist \( m,n\in\mathbb{Z} \) which are not divisible by \( p \) such that \( p\mid(2m-n)^2+3n^2=4(m^2-mn+n^2) \), i.e. \( p\mid(m-n\omega) \overline{m-n\omega} \). However, \( p \) does not divide any of the elements \( (m-n\omega),\overline{m-n\omega} \), so it must be composite.

Element \( x\in\mathbb{Z}[\omega] \) is prime if and only if \( N(x) \) is prime or \( |x| \) is a prime integer of the form \( 3k-1 \), \( k\in\mathbb{N} \).

Number \( x=3 \) is composite, as \( N(1-\omega)= (1-\omega)(2+\omega)=3 \). Moreover, by problem 4, every prime integer \( p\equiv1 \) (mod 6) is composite in \( \mathbb{Z}[\omega] \).

The rest of the proof is similar to the proof of Theorem 7 and is left as an exercise.

The equation \[ x^3+y^3=z^3\quad\quad\quad\quad\quad(\ast)\] has no nontrivial solutions in \( \mathbb{Z}[\omega] \), and consequently has none in \( \mathbb{Z} \) either.

Suppose that \( x,y,z \) are nonzero elements of \( \mathbb{Z}[\omega] \) that satisfy \( (\ast) \). We can assume w.l.o.g. that \( x,y,z \) are pairwise coprime.

Consider the number \( \rho=1-\omega \). It is prime, as its norm equals \( (1-\omega)(1-\omega^2)=3 \). We observe that \( \overline{\rho}=1-\omega^2=(1-\omega)(1+\omega)\sim\rho \); hence \( \alpha\in\mathbb{Z}[\omega] \) is divisible by \( \rho \) if and only if so is \( \overline{alpha} \). Each element in \( \mathbb{Z}[\omega] \) is congruent to \( -1,0 \) or 1 (mod \( \rho \)): indeed, \( a+b\omega\equiv a+b=3q+r\equiv r \) (mod \( \rho \)) for some \( q\in\mathbb{Z} \) and \( r\in\{-1,0,1\} \).

The importance of number \( \rho \) lies in the following property: \[ \alpha\equiv\pm1\; (\mbox{mod }\rho)\; (\alpha\in\mathbb{Z}[\omega])\;\;\mbox{implies}\;\; \alpha^3\equiv\pm1\;(\mbox{mod }\rho^4). \quad\quad\quad\quad\quad(2)\] Indeed, if \( \alpha=\pm1+\beta\rho \), we have \( a^3\mp1=(a\mp1)(a\mp\omega) (a\mp\omega^2)=\rho^3\beta(\beta\pm1)(\beta\pm(1+\omega)) \), where the elements \( b \), \( b\pm1 \), \( b\pm(1+\omega) \) leave three distinct remainders modulo \( \rho \), implying that one of them is also divisible by \( \rho \), thus justifying our claim.

Among the numbers \( x,y,z \), (exactly) one must be divisible by \( \rho \): otherwise, by (2), \( x^3,y^3,z^3 \) would be congruent to \( \pm1 \) (mod \( \rho^4 \)), which would imply one of the false congruences \( 0\equiv\pm1 \), \( \pm1\equiv\pm2 \) (mod \( \rho^4 \)). We assume w.l.o.g. that \( \rho\mid z \). Moreover, (2) also gives us that \( \rho^2\mid z \).

Let \( k\geq2 \) be the smallest natural number for which there exists a solution to \( (\ast) \) with \( (x,y,z)=1 \) and \( \rho^k\mid z \), \( \rho^{k+1}\nmid z \). Consider this solution \( (x,y,z) \).

The factors \( x+y \), \( \omega x+\omega^2y \), \( \omega^2x+\omega y \) from (1) are congruent modulo \( \rho \) and have the sum 0. It follows from \( \rho\mid z \) that each of them is divisible by \( \rho \) and that \( \rho \) is their greatest common divisor. Let \[ x+y=A\rho,\;\;\;\omega x+\omega^2y=B\rho,\;\;\;\omega^2x+\omega y=C\rho,\] where \( A,B,C\in\mathbb{Z}[\omega] \) are pairwise coprime and \( A+B+C=0 \). The product \( ABC \) is a perfect cube (equal to \( (z/\rho)^3 \)), and hence each of \( A,B,C \) is adjoint to a cube: \[ A=\alpha\zeta^3,\;\;\;B=\beta\eta^3,\;\;\;C=\gamma\xi^3\] for some pairwise coprime \( \zeta,\eta,\xi\in\mathbb{Z}[\omega] \) and units \( \alpha,\beta,\gamma \). Therefore, \[ \alpha\zeta^3+\beta\eta^3 +\gamma\xi^3=0.\quad\quad\quad\quad\quad(3)\] Since \( \alpha\beta\gamma \) is a unit and a perfect cube, we have \( \alpha\beta\gamma=\pm1 \). Furthermore, \( ABC=(z/\rho)^3 \) is divisible by \( \rho \) (since \( \rho^2\mid z \)), so (exactly) one of the numbers \( \zeta,\eta,\xi \), say \( \xi \), is divisible by \( \rho \). In fact, \( \xi^3 \) divides \( ABC \) which is divisible by \( \rho^{3k-3} \) and not by \( \rho^{3k-2} \), so \( \rho^{k-1} \) is the greatest power of \( \rho \) that divides \( \xi \). The numbers \( \zeta \) and \( \eta \) are not divisible by \( \rho \); consequently, \( \zeta^3 \) and \( \eta^3 \) are congruent to \( \pm1 \) modulo \( \rho^4 \). Thus the equality \( A+B+C=0 \) gives us \( \alpha\pm\beta\equiv0 \) (mod \( \rho^4 \)), therefore \( \beta=\pm \alpha \); now \( \alpha\beta\gamma=\pm1 \) implies \( \gamma=\pm\alpha \).

Canceling \( \alpha \) in equation (3) yields \( \zeta^3\pm\eta^3\pm\xi^3=0 \), which gives us another nontrivial solution to \( (\ast) \) with \( (\zeta,\eta,\xi)=1 \). However, in this solution we have \( \rho^{k-1}\mid\xi \) and \( \rho^k\nmid\xi \), which contradicts the choice of \( k \).