## Επεκτάσεις σωμάτων και εφαρμογές - Arithmetics in the ring $$\mathbb Z[\omega]$$

### Article Index

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 $$k$$.