I would like to compute $\mathrm{Ext}^i_{\mathbb Z}(\mathbb Q, \mathbb Z/2\mathbb Z)$.
So from the definition I learned in my class, I need to find a free resolution of $\mathbb Q$ over $\mathbb Z$, the first step is to find a surjection of $P^0\to\mathbb Q$ where $P^0$ is a free $\mathbb Z$-module. Then apply $\text{Hom}_\mathbb Z(-,\mathbb Z/2\mathbb Z)$ and compute the cohomology. However, I find it hard to come up with an easy surjection onto $\mathbb Q$. Since $\mathbb Q$ is countable, we can find a bijection $\phi:\mathbb Z\to\mathbb Q$ and consider the infinite direct sum $\oplus_{i=1}^{\infty} \mathbb Z$ and a map $$\psi:\oplus_{i=1}^{\infty}\mathbb Z\to \mathbb Q$$ given by $\psi((a_1, a_2,\ldots))=\sum_{i=1}^\infty a_i\phi(i).$ This is clearly a surjection and a morphism. However, it is hard to compute the cohomology coming from this map since I do not know how to characterize its kernel.
So I am just wondering is there a "better" resolution which can helps me compute Ext functor easily? Any hint is appreciated.
Best Answer
Taking an explicit free resolution of $\mathbb{Q}$ is not impossible but difficult. Instead we can use the functoriality of $\operatorname{Ext}^i$ to determine $\operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)$. Consider the map $$ \varphi\colon \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)\to \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2);\quad x\mapsto 2x. $$ This map is induced by the isomorphism $\mathbb{Q}\to \mathbb{Q};\;x\mapsto 2x$, so $\varphi$ is an isomorphism. However, it is also induced by the zero map $\mathbb{Z}/2\to \mathbb{Z}/2;\;x\mapsto 2x$, so $\varphi$ is a zero map. Putting these together we get $$ \operatorname{Ext}^i(\mathbb{Q},\mathbb{Z}/2)=0 $$ for all $i\geq 0$.