Example of module/ring with long projective resolution

Question

During the seminary I had some homological algebra, but only with groups (not modules) and on the tutorials from algebraic topology course I got such exercise:

Give an example of a ring $R$ and a module $M$ over that ring that does not have a two-step projective resolution: $$0\longrightarrow P_1 \longrightarrow P_0 \longrightarrow M \longrightarrow 0,$$ i.e., the resolution must be longer. Remember, the ring $R$ cannot be PID (why?).

Any hint?

Answer 1

A ring is (right) hereditary if every submodule of a projective (right) module is projective. An equivalent condition is that every (right) ideal is projective. Over such a ring, every module obviously has a two step projective resolution.

In particular, you see that a PID $R$ is hereditary, because ideals are principal, so a nonzero ideal is isomorphic to $R$ as a module, hence projective.

The simplest example of a non hereditary ring is $R=\mathbb{Z}/4\mathbb{Z}$. The maximal ideal $M$ is not projective, because it is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ and the obvious surjective homomorphism $R\to M$ does not split. Now, what module doesn't have a two step projective resolution?

Answer 2

As requested, here's a hint: try $R= k[x]/x^2$ for your favourite field $k$ and $M=k$ with $x$ acting on $M$ as zero.

Answer 3

Two examples:

1. Let $k$ be a field and $R: =k[x,y]$ the polynomial ring in two variables. Set $M:=R/(x,y)\cong k$. Consider the following projective resolution of $M$ $$0\rightarrow R\xrightarrow{d_2} R\oplus R\xrightarrow{d_1} R\rightarrow M\rightarrow 0,$$ where $d_2(f)=(yf,-xf)$ and $d_1(f,g)=xf+yf$. (This is up to sign the Koszul complex of the regular sequence $(x,y)$ in $R$.) By applying the functor $-\otimes_R M$ one finds $$\operatorname{Tor}_2^{R}(M,M)\cong k.$$ In particular, $M$ does not have a projective resolution of length less than two.
   NB: This example can be generalized to polynomial rings in more variables to show that any projective resolution of $k\cong k[x_1,\ldots x_n]/(x_1,\ldots, x_n)$ as a $k[x_1,\ldots,x_n]$-module has length at least $n$. This implies that the bound on the global dimension of $k[x_1,\ldots,x_n]$ in Hilbert's syzygy theorem is sharp.

2. An example coming from group (co)homology. Let $C_n$ be the cyclic group of order $n$. Consider its integral group ring $R:=\mathbb{Z}[C_n]\cong \mathbb{Z}[t]/(t^n-1)$. The the integers $M:=\mathbb{Z}$, with trivial $R$-module structure, have the following projective resolution $$\ldots \rightarrow R\xrightarrow{1-t}R\xrightarrow{N}R\xrightarrow{1-t} R\xrightarrow{\epsilon}M\rightarrow 0.$$ Here we let $\epsilon(t):=1$ and $N:=1+t+\ldots+t^{n-1}\in R.$ By tensoring with $M$ over $R$, we find $$\operatorname{Tor}_i^R(M,M)=\begin{cases}\mathbb{Z} & i=0, \\ C_n & i \text{ odd},\\ 0 & \text{ else.}\end{cases}$$ In particular, $M$ has infinite projective dimension.

Answer 4

Let $k$ be a field and consider the following quiver algebra over $k$ in the picture：

Consider the ideal I = <$\beta\alpha$> and the quotient algebra A = $kQ/I$. After a simple calculation, we can see that the projective dimension of the simple module $S_1$ (i.e. the simple module at position 1 in the picture) is 2.

As for quiver algebra, we can see in https://en.wikipedia.org/wiki/Quiver_(mathematics) or the book Schiffler, Ralf, Quiver representations., CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC. Cham: Springer (ISBN 978-3-319-09203-4/hbk; 978-3-319-09204-1/ebook). xi, 230 p. (2014). ZBL1310.16015.