Projective resolution of $\mathbb{Z}/(d)$ as a $\mathbb{Z}/(n)$-module

Question

Let $A=\mathbb{Z}/(n)$, $d$ a divisor of $n$ and $M=A/([d])\simeq\mathbb{Z}/(d)$.

I'm trying to find a projective resolution of $M$ as an $A$-module.

The only concrete example I know of a projective resolution is $\cdots\to0\to\mathbb{Z}\xrightarrow{\times d}\mathbb{Z}\to\mathbb{Z}/(d)\to 0$, which regards $\mathbb{Z}/(d)$ as a $\mathbb{Z}$-module.

I tried to do something like $\cdots\to0\to A\xrightarrow{\times [d]}A\to M\to 0$, but that doesn't work, since $\times [d]$ is not injective, now I'm stuck.

Any hints?

Answer 1

There's an obvious exact sequence $$0\to K\to\Bbb Z/(n)\to\Bbb Z/(d)\to0.$$ The kernel $K$ is cyclic of order $n/d$. Repeat $$0\to K_1\to\Bbb Z/(n)\to\Bbb Z/(n/d)\to0.$$ Then $K_1\cong \Bbb Z/(d)$. We get into a cycle.

We get a projective resolution with all terms $\Bbb Z/(n)$ and with differential maps alternately ${}\times d$ and ${}\times n/d$.