There are all sorts of notions of dimension that can be applied to rings. Whatever notion you use though, the ones with dimension zero are usually fairly simple compared with the rings of higher dimension. Here we’ll look at three types of dimension and state what the rings of zero dimension look like with respect to each type. Of course, several examples are included.

All rings are associative with identity but not necessarily commutative. Some basic homological algebra is necessary to understand all the definitions.

Global Dimension

The left global dimension of a ring $R$ is the supremum over the projective dimensions of all left $R$-modules. The right global dimension is the same with “left” replaced by “right”. And yes, there are rings where the left and right global dimensions differ.

However, $R$ has left global dimension zero if and only if it has right global dimension zero. So, it makes sense to say that such rings have global dimension zero. Here is their characterisation:

A ring $R$ has global dimension zero if and only if it is semisimple; that is, if and only if it is a finite direct product of full matrix rings over division rings.

Examples of such rings are easy to generate by this characterisation:

  1. Fields and finite products of fields
  2. $M_2(k)$, the ring of $2\times 2$ matrices over a division ring $k$
  3. etc.

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Let $R$ be a ring. The projective dimension $\mathrm{pd}_R(M)$ of an $R$-module $M$ is the infimum over the lengths of projective resolutions of $M$. The left global dimension of $R$ is the supremum over the projective dimensions of all left $R$-modules. There is a notion of right global dimension where left modules are replaced with right modules. Since we’ll be talking about commutative rings only, we’ll just use global dimension to refer to both kinds, and write $\mathrm{g\ell.dim}(R)$ for the global dimension of $R$.

As an example, if $R$ is a field then $\mathrm{g\ell.dim}(R) = 0$ because every module is free. If $R$ is a principal ideal domain (PID), then $\mathrm{g\ell.dim}(R) = 1$. This is because any module $M$ admits a surjection $F\to M$ where $F$ is a free module. But the kernel of this map is also free since over a PID, a submodule of a free module is free. One of the first results in the theory of global dimension is that $\mathrm{g\ell.dim}(R[x]) = 1 + \mathrm{g\ell.dim}(R)$. So far then we have examples of rings with any finite global dimension.
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