Calculating Ext and Tor for $\mathbb Z/m\mathbb Z$ and $\mathbb Z/n\mathbb Z$

Question

Looking at $\mathbb Z/m\mathbb Z, \mathbb Z/n\mathbb Z$ as $\mathbb Z$ modules, how do we determine what $Ext^i(\mathbb Z/m\mathbb Z, \mathbb Z/n\mathbb Z)$ and $Tor_i(\mathbb Z/m\mathbb Z,\mathbb Z/n\mathbb Z)$ are for all $i$?

I know that $\mathbb Z$ is a projective module over itself and therefore, we get the following projective resolution: $$\cdots \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \xrightarrow{\times m} \mathbb{Z} \rightarrow Z/mZ \rightarrow 0$$

I am not sure what to do from here.

Answer 1

I'm not sure what you're trying to say with that exact sequence. The following is a projective resolution of $\Bbb Z/m\Bbb Z$:

$$0\to\Bbb Z\overset{\times m}{\to}\Bbb Z\to\Bbb Z/m\Bbb Z\to0$$

Then taking $\hom(-,\Bbb Z/n\Bbb Z)$, you get

$$0\to\hom(\Bbb Z/m\Bbb Z,\Bbb Z/n\Bbb Z)\to\hom(\Bbb Z,\Bbb Z/n\Bbb Z)\overset{\times m}{\to}\hom(\Bbb Z,\Bbb Z/n\Bbb Z)\to0$$

where the latter is not exact in general. Simplifying, you need to compute the cohomology groups of the chain

$$0\to\Bbb Z/n\Bbb Z\overset{\times m}{\to}\Bbb Z/n\Bbb Z\to0$$

which should be managable. Recall that by left exactness, the kernel of the left map is equal $\hom(\Bbb Z/m\Bbb Z,\Bbb Z/n\Bbb Z)$.

Answer 2

$\mathbf Z$ has global dimension $1$. Therefore $\DeclareMathOperator{\Ext}{Ext}\DeclareMathOperator{\Tor}{Tor}\DeclareMathOperator{\Hom}{Hom}\;\Ext^i_{\mathbf Z}(\cdot,\cdot)=\Tor^{\mathbf Z}_i(\cdot,\cdot)=0$ for all $i\ge 2$.

Now consider the long exact sequences deduced from the short exact sequence $$0\longrightarrow \mathbf Z \xrightarrow{{}\times m\;}\mathbf Z\longrightarrow \mathbf Z/m \mathbf Z \longrightarrow 0$$

• $\Hom(\mathbf Z,\mathbf Z/n \mathbf Z)\xrightarrow{{}\times m\;} \Hom(\mathbf Z,\mathbf Z/n \mathbf Z)\longrightarrow \Ext^1_{\mathbf Z}(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)\longrightarrow \Ext^1_{\mathbf Z}(\mathbf Z,\mathbf Z/n \mathbf Z)=0$
• $0=\Tor^{\mathbf Z}_1(\mathbf Z,\mathbf Z/n\mathbf Z) \longrightarrow \Tor^{\mathbf Z}_1(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)\longrightarrow \mathbf Z\otimes_{\mathbf Z}\mathbf Z/n \mathbf Z\xrightarrow{{}\times m\;} \mathbf Z\otimes_{\mathbf Z} \mathbf Z/n \mathbf Z$

With the canonical identifications, the first one shows $ \Ext^1_{\mathbf Z}(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)$ is the cokernel of the morphism of multiplication by $m$ in $\mathbf Z/n \mathbf Z$, i.e. $$\Ext^1_{\mathbf Z}(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)\simeq(\mathbf Z/n \mathbf Z)/(m\cdot\mathbf Z/n \mathbf Z)=(\mathbf Z/n \mathbf Z)/(d\mathbf Z/n \mathbf Z)\simeq \mathbf Z/d\mathbf Z.$$

The second exact sequence shows $\Tor^{\mathbf Z}_1(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)$ is the kernel of multiplication by $m$ in $\mathbf Z/n \mathbf Z$, i.e., setting $n'=\dfrac nd$: $$\Tor^{\mathbf Z}_1(\mathbf Z/m \mathbf Z,\mathbf Z/n \mathbf Z)\simeq n'\mathbf Z/n \mathbf Z= n'\mathbf Z/n'd \mathbf Z\simeq\mathbf Z/d \mathbf Z.$$