Compute Ext functor $\mathrm{Ext}^i_{\mathbb Z}(\mathbb Q, \mathbb Z/2\mathbb Z)$

Question

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.

Answer 1

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 $i\geq 0$.

Answer 2

If you must do it with a free resolution, consider the free group $\oplus_{i=1}^\infty \mathbb Z$ and morphism:

$$\phi:(a_i)_{i=1}^\infty \mapsto \sum_{i=1}^\infty \frac{a_i}{i!}$$

Then the kernel has as a basis $e_n=(e_{ni})_i$ with $$e_{ni}=\begin{cases}1&i=n\\-(n+1)&i=n+1\\0&\text{otherwise}\end{cases}$$ Then: $$\phi(e_n)=\frac1{n!}-\frac{n+1}{(n+1)!}=0.$$

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Showing this is a free basis is not hard.

The $e_n$ are independent because if $b=(b_i)=\sum c_ne_n$, then for the smallest $k$ such that $c_k\neq 0,$ then $b_k=c_k\neq0,$ so $b\neq0.$

Showing they generate the kernel is only a little harder.

If $a=(a_i)$ is a non-zero element in the kernel, let $$m(a)=\min\{i\mid a_i\neq0\}\\n(a)=\max\{i\mid a_i\neq 0\}.$$

If $m(a)=n(a),$ then $a$ can’t be non-zero, because $\phi(a)=\frac{a_m}{m!},$ and $a_m\neq0.$

So $m(a)<n(a).$

We can define $$a’=a-a_{m(a)}e_{m(a)}.$$

Then $a’$ is in the kernel, and if non-zero, then $m(a’)>m(a)$ and $n(a’)=n(a).$

So by induction on $n(a)-m(a),$ you eventually get to zero, and the original element was a combination of the $e_m.$