Calculating $ \operatorname{Ext}(\mathbb{Z}/2, \mathbb{Z}/2)$ over $\mathbb{Z}$:
I just need someone to confirm that I have calculated $ \operatorname{Ext}(\mathbb{Z}/2, \mathbb{Z}/2)$ correctly. I know this is an easy, common calculation, however I cannot find any actual answer to confirm my results against anywhere. Which is very frustrating.
I have calculated: $ \operatorname{Ext}^0(\mathbb{Z}/2, \mathbb{Z}/2) = \mathbb{Z}/2$, $ \operatorname{Ext}^1(\mathbb{Z}/2, \mathbb{Z}/2) = \mathbb{Z}/2$, and $ \operatorname{Ext}^n(\mathbb{Z}/2, \mathbb{Z}/2) = 0$ for $n \geq 2$. Is this correct?
I used the projective resolution:
$...\rightarrow 0 \rightarrow 0 \rightarrow \mathbb{Z} = P_1 \xrightarrow {\cdot 2} \mathbb{Z} = P_0 \xrightarrow {\cdot 1} \mathbb{Z}/2$
which yielded Hom groups:
$...\leftarrow 0 \leftarrow 0 \leftarrow \operatorname{Hom}(\mathbb{Z},\mathbb{Z}/2) \xleftarrow {\cdot 2} \operatorname{Hom}(\mathbb{Z},\mathbb{Z}/2) \leftarrow 0$
which is equivalent to:
$...\leftarrow 0 \leftarrow 0 \leftarrow \mathbb{Z}/2 \xleftarrow {\cdot 0} \mathbb{Z}/2 \leftarrow 0$
The homology at the 0th and 1st points is then $\mathbb{Z}/2$, and $0$ elsewhere.
Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $\mathbb Z/2$ by itself are the trivial one and $\mathbb Z/4$, up to isomorphism, and this is isomorphic to $\mathbb Z/2$, of course. Note, however, that if you try to do the same with $\mathbb Z/p$ (the very same computation works), then you get Ext is $\mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $\mathbb Z/p^2$ which are not-isomorphic. Can you find them?