How to show $\operatorname{Ext}^1(I, \mathbb{Z})\cong\mathbb{Z}/2$ for $\mathbb{Z}[C_2]$?

Question

I am trying to become better acquainted with $Ext^1$ and thought the best place to start is with actual computations. This was an exercise in some brief notes I found and wanted to see if my thinking is correct. Here, $\mathbb{Z}$ is the trivial $\mathbb{Z}[C_2]$-module, and $I$ the argumentation ideal. As I understand it, I need to show there are only two extensions up to isomorphism/equivalence. The one that splits and the one that doesn't. Now if I can show $\operatorname{Ext}^1(\mathbb{Z},I)\cong\mathbb{Z}/2$, then the result should follow as the dual of any such SES is another SES, and both $\mathbb{Z}$ and $I$ are self-dual. So I focus on extensions of the form

$0\to I\to ?\to\mathbb{Z}\to 0.$

Clearly we have $?=\mathbb{Z}[C_2], \mathbb{Z}\oplus I$ which are the two classes. But how do we show there cannot be any other?

As a further question, does anyone know any good computation-focused resources for this material?

Answer 1

A projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[C_2]$ module is

$$ \dots \xrightarrow{g+1} \mathbb{Z}[C_2] \xrightarrow{g-1} \mathbb{Z}[C_2] \xrightarrow{g+1} \mathbb{Z}[C_2] \xrightarrow{g-1} \mathbb{Z}[C_2] \xrightarrow{\varepsilon} \mathbb{Z}. $$ where $\varepsilon(k+lg) = k+l$. This is a concrete instance of a more general fact: for any ring $R$ we have a $R[C_n]$-free resolution of $R$ (with trivial action) via

$$ \dots \xrightarrow{N} R[C_n] \xrightarrow{g-1} R[C_2] \xrightarrow{N} R[C_2] \xrightarrow{g-1} R[C_2] \xrightarrow{\varepsilon} R. $$

where $\varepsilon(a_0+a_1g + \cdots+ a_{n-1}g^{n-1}) = a_0+\cdots+a_{n-1}$ and $N = 1 + g + \cdots + g^{n-1}$.

In particular we obtain that

$$ \dots \xrightarrow{g+1} \mathbb{Z}[C_2] \xrightarrow{g-1} \mathbb{Z}[C_2] \xrightarrow{g+1} \mathbb{Z}[C_2] \xrightarrow{g-1} I \to 0 $$

is a projective resolution of $I$.

Now, applying the hom functor we get a complex

$$ 0 \to \hom_{\mathbb{Z}[C_2]}(\mathbb{Z}[C_2],\mathbb{Z}) \xrightarrow{(g+1)_\ast} \hom_{\mathbb{Z}[C_2]}(\mathbb{Z}[C_2],\mathbb{Z})\xrightarrow{(g-1)_\ast} \hom_{\mathbb{Z}[C_2]}(\mathbb{Z}[C_2],\mathbb{Z}) \to \cdots $$

and this complex is isomorphic to the following complex $C$,

$$ 0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \cdots $$

Hence

$$ \mathrm{Ext}^1_{\mathbb{Z}[C_2]}(I,\mathbb{Z}) \simeq H^1(C) = \mathbb{Z}/2\mathbb{Z} $$

as desired.