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.

Perhaps the simplest example of a ring with $\mathrm{g\ell.dim}(R) = \infty$ is $\mathbb{Z}/4$. Indeed, consider the surjection $\mathbb{Z}/4\to\mathbb{Z}/2$ given by reducing modulo $2$, and consider $\mathbb{Z}/2$ as a $\mathbb{Z}/4$-module. There’s the following projective resolution:
\cdots\xrightarrow{2}\mathbb{Z}/4\xrightarrow{2}\mathbb{Z}/4\xrightarrow{2}\mathbb{Z}/4\to \mathbb{Z}/2\to 0.
Here, the $2$ means multiplication by $2$. Applying the functor $\mathrm{Hom}(-,\mathbb{Z}/2)$ shows that $\mathrm{Ext}^i(\mathbb{Z}/2,\mathbb{Z}/2)\cong \mathbb{Z}/2$ for all $i$, so that $\mathrm{pd}_{\mathbb{Z}/4}(\mathbb{Z}/2) = \infty$, and hence $\mathrm{g\ell.dim}(\mathbb{Z}/4) = \infty$. In doing this we’ve shown that $\mathbb{Z}/2$ is not a projective $\mathbb{Z}/4$-module, so $\mathbb{Z}/4$ is not semisimple and hence not a product of fields. There we have it: examples of rings of every possible global dimension.

For commutative rings, there is the notion of Krull dimension. Actually, there is a notion for noncommutative rings too, but talking about that would lead us too far into the noncommutative realm, better known as the dark side. So, the question is, how do the Krull dimension and global dimension compare?

Let’s recall the notion of Krull dimension of $R$ first, which we will simply denote by $\dim(R)$: it is the supremum over the lengths of all the chains of prime ideals in $R$. This notion is a little more geometric in nature, since for rings of the form $k[x_1,\dots,x_n]/I$ where $k$ is a field, the Krull dimension seems to accurately capture what we intuitively perceive as dimension of the corresponding algebraic variety. As an example, the Krull dimension of $k[x_1,\dots,x_n]$ where $k$ is a field is $n$. However, we’ve already noted that $\mathrm{g\ell.dim}(k[x_1,\dots,x_n]) = n$ as well, because, as I mentioned earlier, $\mathrm{g\ell.dim}(R[x]) = \mathrm{g\ell.dim}(R) + 1$.

We’ve also seen $\mathbb{Z}/4$, which has infinite global dimension. However, since $(2)$ is the unique prime ideal of this ring, $\mathbb{Z}/4$ also has Krull dimension zero. So, the Krull dimension can be strictly less than the global dimension. In fact, for Noetherian rings this is always true:

Theorem. If $R$ is a commutative Noetherian ring then $\dim(R) \leq \mathrm{g\ell.dim}(R)$.

Actually, the theorem is not so difficult, once we accept a modest but important fact: the global dimension of a commutative Noetherian ring can be computed via the formula $\mathrm{g\ell.dim}(R) = \sup_m\mathrm{g\ell.dim}(R_m)$ where $m$ runs over the maximal ideals of $R$. One can check the best book in the world on rings and modules, Lam’s Lectures on Rings and Modules, Theorem 5.92, for this fact. Anyways, this is good because it follows from the definition of Krull dimension that the Krull dimension can also be computed this way.

So, we localize! If $R$ is a commutative Noetherian ring, then $R_m$ is a commutative Noetherian local ring, and there are only two possibilities for $R_m$. The first is that $R_m$ has infinite global dimension, in which case there is nothing to prove. The second case is that $R_m$ has finite global dimension; but then $R_m$ is a regular local ring and the global dimension and Krull dimension of $R_m$ coincide (this is a very nontrivial fact, however). So, this means that either $\dim(R) = \mathrm{g\ell.dim}(R)$ or $\dim(R) \leq \mathrm{g\ell.dim}(R) = \infty$.

This of course also shows that $\mathbb{Z}/4$ is a Noetherian local ring that is not regular. Of course, one could always observe this directly too, by just calculating that $\dim(\mathbb{Z}/4) = 0$ whereas $\dim_{(\mathbb{Z}/4)/(2)}( (2)/(2)^2) = 1$. However, I actually prefer the infinite global dimension proof, since it involves the homological properties of $\mathbb{Z}/4$.

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