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

Here \( \omega \) denotes a primitive cubic root of unity. Then the norm
of an element \( a+b\omega\in\mathbb{Z}[\omega] \) (\( a,b\in\mathbb{Z} \))
is \( N(a+b\omega)=a^2-ab+b^2 \) and the units are \( \pm1 \), \( \pm \omega \)
and \( \pm(1+\omega)=\mp\omega^2 \).

** Theorem 8 **

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.

** Problem 5 **

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.

** Theorem 9 **

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.

Maybe the most famous application of the elementary arithmetic of
the ring \( \mathbb{Z}[\omega] \) is the Last Fermat Theorem for the
exponent \( n=3 \). This is not unexpected, having in mind that
\( x^3+y^3 \) factorizes over \( \mathbb{Z}[\omega] \) into linear factors:
\[ x^3+y^3=(x+y)(x+\omega y)(x+\omega^2y)=(x+y)(\omega x+\omega^2y)
(\omega^2x+\omega y).\quad\quad\quad\quad\quad(1)\] The proof we present was first given
by Gauss.

** Theorem 10 **

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 \( k \).