5

Let $m,n$ be positive integers with $m\mid n$. I want to compute $$ \mathrm{Ext}_{\mathbb{Z}/n\mathbb{Z}}^i (\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \qquad \mathrm{Tor}^{\mathbb{Z}/n\mathbb{Z}}_i (\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) $$ for $i \ge 0$. At first, I thought that the way to do this was to find a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module, and I didn't know how to do that, so asked about it in an earlier question (What is a projective resolution of $\mathbb{Z}/m\mathbb{Z}$ as a $\mathbb{Z}/n\mathbb{Z}$-module?).

However, the answer seems to depend on whether $m$ has repeated prime factors of $n$, so this doesn't seem like the way to compute these Ext and Tor groups. Is there a better way to do this?

JJR
  • 1,269
Joshua Ruiter
  • 2,982
  • Think of all of this as being analogous to number theory arguments involving multiplicative functions, where you work out how everything behaves for prime powers, and factor everything else into prime powers. –  May 01 '17 at 16:06

1 Answers1

3

I'll write $Z_n$ for $\Bbb Z/n\Bbb Z$ to reduce typing. Then a projective resolution of $Z_m$ as a $Z_n$-module goes like this $$\cdots\to Z_n\xrightarrow{\times n/m} Z_n\xrightarrow{\times m} Z_n\xrightarrow{\times n/m} Z_n\to Z_m\to0.$$ Dropping the final $Z_m$ and tensoring with $Z_m$, the Tor groups are the homology groups of $$\cdots\to Z_m\xrightarrow{\times n/m} Z_m\xrightarrow{\times m} Z_m\xrightarrow{\times n/m} Z_m\to0.$$ Then $\times m$ maps are 0, so the even Tor groups are all equal the cokernel of the multiplication of $n/m$ on $Z_m$ and the odd Tor groups equal the kernel of this map. I don't have the energy to compute these further or look at the Ext groups...

Angina Seng
  • 158,341