Free resolution of $\mathbb{Z}/2\mathbb{Z}$ as $\mathbb{Z}/4\mathbb{Z}$-module

Question

Let $R = \mathbb{Z}/4\mathbb{Z}$ and consider $M = \mathbb{Z}/2\mathbb{Z}$ as an $R$-module. In my course notes, we have constructed a free resolution of $M$ that has infinite length:

\begin{equation} ... \stackrel{\cdot 2}{\to} \mathbb{Z}/4\mathbb{Z} \stackrel{\cdot 2}{\to} \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \end{equation} and now they state that any free resolution of $M$ as an $R$-module has infinite length. However, I have no idea how I could prove this.

Answer 1

As ThePuix explained, the simplest idea is to use some derived functor constructed from this resolution, as it will not depend on the choice of resolution.

Explicitly, if $F$ is a right exact functor on the category of $R$-modules, and $$\dots \to P_2\to P_1\to P_0\to \mathbb{Z}/2\mathbb{Z}\to 0$$ is a projective resolution of $R$-module (in particular, it can be a free resolution), then the homology of the complex $$ \dots\to F(P_2)\to F(P_1)\to F(P_0)\to 0$$ does not depend on the choice of resolution (up to a canonical isomorphism), and the homology group at $F(P_n)$ is by definition the $n$th derived functor $L_nF(\mathbb{Z}/2\mathbb{Z})$.

If there was a finite length resolution (say of length $d$) then obviously we would have $L_nF(\mathbb{Z}/2\mathbb{Z}) = 0$ for any $F$ and for every $n>d$.

But we can give a counter-example: let us choose for instance $F(X)=X\otimes_R A$ for some $R$-module $A$. The derived functors are then by definition $Tor_n^R(\mathbb{Z}/2\mathbb{Z},A)$. Using your free resolution, we easily see that $Tor_n^R(\mathbb{Z}/2\mathbb{Z},A)$ is given by the homology of the complex $$ \dots\to A\xrightarrow{\cdot 2} A\xrightarrow{\cdot 2} A\to 0$$ so $Tor_n^R(\mathbb{Z}/2\mathbb{Z},A)= A[2]/2A$ where $A[2]$ is the $2$-torsion subgroup of $A$.

If we take $A=\mathbb{Z}/2\mathbb{Z}$, then $Tor_n^R(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}\neq 0$, which concludes.

Answer 2

Let $R=\mathbb{Z}/4\mathbb{Z}$ and $M=\mathbb{Z}/2\mathbb{Z}$ for simplicity. There is a non-split exact sequence $0\to M\to R\to M\to 0$, so by dimension shifting we have, for $n\ge1$, $$ \operatorname{Ext}_R^{n+1}(M,M)\cong\operatorname{Ext}_R^n(M,M) $$

On the other hand, a finite free resolution would imply that $\operatorname{Ext}_R^k(M,M)$ vanishes for sufficiently high $k$. This would imply that $\operatorname{Ext}_R^1(M,M)$ vanishes, which it doesn't because the given exact sequence is not split.