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We want to find all $n>0$ such that $\mathbb{Z}/n\mathbb{Z}$ has a $\mathbb{Z}[i]$-module structure.

This is what I have. First of all we have that for $k\in \mathbb{Z}$ we have that for $m\in \mathbb{Z}/n\mathbb{Z}$, $k\cdot m=[km]\in \mathbb{Z}/n\mathbb{Z}$.

Note that if we manage to answer what $i\cdot 1$ is equal to then we would be done since $(a+bi)m=am+b(im)$.

Say $i\cdot 1=x$. Then we have that $i\cdot x=i\cdot(1+...+1)=i\cdot 1+...+i\cdot 1=x+...+x=x^2$. That is, $i\cdot (i\cdot 1)=x^2$On the other hand, $(i\cdot i)\cdot 1=(-1)\cdot 1=-1$So we want $x^2=-1$. So any $n$ that satisfies $\mathbb{Z}/n\mathbb{Z}$ such that $-1$ is a square would work because we can define $i\cdot 1=x$ where $x$ is such that $x^2=-1$. Is the above correct?

Thanks.

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    Okay, like I said, I was tired ; I actually meant that you have found necessary conditions, but are they sufficient. In other words, you've found some conditions on $n$ (i.e. that $x^2 \equiv - 1$ must have a solution) but what does that tell you about $n$? Are you sure that having a $\Bbb Z[i]$-module structure and having such an $x$ in $\Bbb Z / n \Bbb Z$ is equivalent?2012-12-05

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$\mathbb Z/n\mathbb Z$ is a $\mathbb Z[i]$-module iff there exists a ring homomorphism $\mathbb Z[i]\to\operatorname{End}(\mathbb Z/n\mathbb Z)$.

Obviously $\operatorname{End}(\mathbb Z/n\mathbb Z)\simeq \mathbb Z/n\mathbb Z$, so we reduced the problem to the existence of ring homomorphisms $\mathbb Z[i]\to\mathbb Z/n\mathbb Z$. Of course, a necessary condition is that the equation $x^2=-1$ has solutions in $\mathbb Z/n\mathbb Z$, but this is also sufficient: take $x_0\in\mathbb Z/n\mathbb Z$ such that $x_0^2=-1$ and define $f:\mathbb Z[i]\to\mathbb Z/n\mathbb Z$ by $f(a+bi)=a+bx_0$ (here $a,b$ are taken modulo $n$). It's easily seen that $f$ is a ring homomorphism.

Remark. For which $n$ the equation $x^2\equiv-1\pmod n$ has solutions is a well known fact.