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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.

  • 0
    You know those $n$'s are characterized in number theory, do you?2012-12-05
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    I know that for $p$ congruent to $1$ modulo $4$ we have that $-1$ is a square in $\mathbb{Z}/p\mathbb{Z}$.2012-12-05
  • 0
    But, is the above correct?2012-12-05
  • 0
    I see no problem, but I see that you have found sufficient conditions for $n$ to give you a $\Bbb Z[i]$-module, not necessary ones. Have you done the work? Maybe I am just tired and am not seeing that you have actually shown everything.2012-12-05
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    Well I think by the argument above I showed that it is necessary that $-1$ is a square because we have $-1=(-1)\cdot 1=(i\cdot i)(1)=i\cdot x =i\cdot (1+...+1)=x+..+x=x^2$.2012-12-05
  • 0
    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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