Calculate $\operatorname{Ext}(\mathbb{Z}/n,\mathbb{Z}/m)$ for $n,m \geq 2$.

Question

I can easily find that a free presentation of $\mathbb{Z}/n$ is $\mathbb{Z}$ with $p:\mathbb{Z}\rightarrow\mathbb{Z}/n$ given by the quotient map. And obviously the kernel of this is $\ker(p)=n\mathbb{Z}$. Therefore we have a short exact sequence, $$0\rightarrow n\mathbb{Z}\xrightarrow{i}\mathbb{Z}\xrightarrow{p}\mathbb{Z}/n\rightarrow 0$$ where $i$ is the inclusion. This gives us the exact sequence, $$0\rightarrow\operatorname{Hom}(\mathbb{Z}/n,\mathbb{Z}/m)\xrightarrow{p^*}\operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m)\xrightarrow{i^*}\operatorname{Hom}(n\mathbb{Z},\mathbb{Z}/m)$$ And we can define $\operatorname{Ext}(\mathbb{Z}/n,\mathbb{Z}/m)=\operatorname{Hom}(n\mathbb{Z},\mathbb{Z}/m)/\text{im}(i^*)$.

1. Does the above make sense, is this the right way to approach the problem?
2. If so, what is $\operatorname{Hom}(n\mathbb{Z},\mathbb{Z}/m)$ and what is the image of $i^*$?

Answer 1

It's better to consider $0\to\mathbb{Z}\xrightarrow{\mu_n}\mathbb{Z}\xrightarrow{p}\mathbb{Z}/n\mathbb{Z}\to 0$, where $\mu_n$ is the multiplication by $n$, because $\mu_n^*$ is again multiplication by $n$. Thus you get $$ 0\to \operatorname{Hom}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \xrightarrow{p^*} \operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \xrightarrow{\mu_n} \operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) $$ Now you know that $\operatorname{Hom}(\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\cong\mathbb{Z}/m\mathbb{Z}$ in a “canonical” way so what you need to compute the image of $$ \mu_n\colon\mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z} $$ which is $(n\mathbb{Z}+m\mathbb{Z})/m\mathbb{Z}=d\mathbb{Z}/m\mathbb{Z}$, where $d=\gcd(m,n)$. Finally, $\operatorname{Ext}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z})\cong\mathbb{Z}/d\mathbb{Z}$.

Answer 2

Actually, its more generally true that $${\rm Ext}(\mathbb{Z}/n\mathbb{Z},A)\cong A/nA$$ for any abelian group $A$. Here is the proof (essentially the generalization of @egreg's argument)

Consider the resolution $\mathbb{Z}\overset{\times n}{\to}\mathbb{Z}\to \mathbb{Z}/n$. Apply $\hom(-,A)$ to get the sequence $$0\to \hom(\mathbb{Z},A)\overset{(\times n)^*}{\to} \hom(\mathbb{Z},A)\to 0\to ...$$ But $\hom(\mathbb{Z},A)\cong A$, since a homomorphism $\varphi:\mathbb{Z}\to A$ is uniquely determined by where it sends $1\in \mathbb{Z}$. The induced map $(\times n)^*$ sends an element $a\in A$ to $na\in A$. Hence, $${\rm Ext}(\mathbb{Z}/n\mathbb{Z},A)\cong A/nA\;.$$

Applying that to $A=\mathbb{Z}/m\mathbb{Z}$, we get $\mathbb{Z}/\gcd(n,m)\mathbb{Z}$.