Dimension zero rings for three types of dimension

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.

Weak Dimension

The weak dimension is like the global dimension: it is the supremum over flat dimensions of all left $R$-modules. This time, I just called it “weak dimension” instead of “left weak dimension”. That’s because unlike global dimension in general, weak dimension doesn’t depend on whether you use left or right $R$-modules.

Weak dimension zero rings also have a very easy characterisation:

A ring $R$ has weak dimension zero if and only if for every $a\in R$ there is a $b\in R$ such that $aba = a$.

Rings with this property are called von Neumann Regular. This condition is pretty neat because it is actually expressible in first-order logic. Here are the examples:

1. Fields
2. Semisimple rings (i.e. rings of global dimension zero)
3. Infinite product of fields (these are not semisimple)
4. Endomorphism rings of vector spaces of arbitrary dimension

Krull Dimension

Now we come to the tricky one! Krull dimension. For a commutative ring $R$, the Krull dimension is the supremum over the lengths of all chains of prime ideals in $R$. Actually it is possible to define Krull dimension for some noncommutative rings as well, but I won’t go there since I don’t know much about it.

Anyways, Krull dimension is the most tricky of the dimensions so far, in the sense that the complexity of the ring grows more quickly with the increase in Krull dimension, compared to the increase in the othe types of dimension. To at least get a sense of why Krull dimension is tricky, it corresponds geometrically to the intuitive notion of dimension for algebraic varieties like curves and hypersurfaces, and these are already quite difficult to get a handle on.

There are a few characterisations of commutative rings of Krull dimension zero.

For a commutative ring $R$, the following are equivalent:

1. $R$ has Krull dimension zero
2. Every element of the Jacobson radical ${\rm Jac}(R)$ of $R$ is nilpotent and $R/{\rm Jac}(R)$ is von Neumann regular
3. For any $a\in R$, the chain of ideals $(a)\supseteq (a^2)\supseteq\cdots$. That is, $R$ satisfies the descending chain condition on principal ideals.
4. (P. Maroscia, 1974) The topological space ${\rm Spec}(R)$ is Hausdorff

The last condition is cool because I’ve heard people say many times that the spectrum is almost never Hausdorff but they never elaborate!! Well, now you know exactly when it is Hausdorff!

Note that $R$ having the descending chain condition on principal ideals is strictly weaker than $R$ having the descending chain condition on all ideals (i.e. being Artinian), since $R$ is Artinian if and only if $R$ is Noetherian and Krull zero-dimensional. And of course, there are examples of rings that have Krull dimension zero and yet are not Noetherian. There are also examples of Noetherian rings that are infinite-dimensional. Go figure!