How many Eisenstein integers modulo 3 are there?

Question

I'm trying to find a system of representatives for the Eisenstein integers modulo 3. What would the set S be and how can I determine the number of solutions in S of the equation x^2=0 mod 3?

Answer 1

The ring of Eisenstein integers is isomorphic to $\Bbb{Z}[X]/(X^2+X+1)$, so the ring of Eisenstein integers mod 3 is isomorphic to \begin{eqnarray*} (\Bbb{Z}[X]/(X^2+X+1))/(3)&\cong&(\Bbb{Z}[X]/(3))/(X^2+X+1)\\ &\cong&\Bbb{F}_3[X]/(X^2+X+1)\\ &\cong&\Bbb{F}_3[X]/((X-1)^2). \end{eqnarray*} This is a ring of $9$ elements, so counting the solutions to $x^2=0$ is quickly done. In fact we have $$(aX+b)^2=a^2X^2+2abX+b^2=2a(a+b)X+(b^2-a^2)=(a+b)(2aX+a-b),$$ which quickly shows that the solutions are precisely the elements of the form $a(X-1)$.

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Alternatively, without any reference to ring theory:

Two Eisenstein integers $x=a\omega+b$ and $y=c\omega+d$ are congruent modulo $3$ if and only if $a\equiv c\pmod{3}$ and $b\equiv d\pmod{3}$. From this it follows that a system of representatives is, for example, the set $$S=\{a\omega+b:\ 0\leq a,b<3\}.$$ For the second part of your question, we are looking for Eisenstein integers $x$ satisfying $x^2\equiv0\pmod{3}$. Write $x=a\omega+b$ with $a$ and $b$ integers. Then because $\omega^2=-\omega-1$ we find that \begin{eqnarray*} x^2=(a\omega+b)^2&=&a^2\omega^2+2ab\omega+b^2\\ &=&(2ab-a^2)\omega+(b^2-a^2) \end{eqnarray*} So to get $x^2\equiv0\pmod{3}$ we need that both $$2ab-a^2\equiv0\pmod{3}\qquad\text{ and }\qquad b^2-a^2\equiv0\pmod{3}.$$ The latter implies that $b\equiv\pm a\pmod{3}$. If $b\equiv a\pmod{3}$ then the first congruence becomes $a^2\equiv0\pmod{3}$ and hence $a\equiv b\equiv0\pmod{3}$. If $b\equiv-a\pmod{3}$ then the first congruence becomes $-3a^2\equiv0\pmod{3}$, which holds for all $a$. So we get the solutions $$0,\ \omega-1,\ 2\omega-2.$$ That is, all integer multiples of $\omega-1$.