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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The tensor product is one of the most important constructions in mathematics, and here we shall see my favourite examples of the tensor product in action, hopefully to illuminate its properties for beginners. Proofs or references are provided, but since the emphasis is on examples, the proofs that are given are terse and details are left to the interested reader.

Let $ R$ be a ring.

Proposition. If $ M$ is a left $ R$-module and we consider $ R$ as a right $ R$-module then $ R\otimes_R M \cong M$.
Proof. Multiplication $ R\times M\to M$ is bilinear, so it extends to a map $ R\otimes_R M\to M$. The map $ M\to R\otimes_R M$ given by $ m\mapsto 1\otimes m$ is its inverse.
Proposition. If $ M$ is a left (resp. right) $ R$-module then the functor $ -\otimes_R M$ (resp. $ M\otimes_R-$) is right-exact.
Proof. The tensor functor is a left-adjoint so it is right-exact.

Here is an application of the above result.

Proposition. Let $ m,n\geq 1$ be integers. In the category of abelian groups $ \mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m\cong \mathbb{Z}/\mathrm{gcd}(m,n)$.

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