Let $R$ be a ring and $M$ and $R$-module. If every finitely generated submodule of $M$ is flat, then so is $M$, because direct limits commute with the $\mathrm{Tor}$-functor. What about the converse? If $M$ is flat, are all its finitely generated submodules flat too?

Not necessarily! In fact, here’s a roundabout argument without an actual counterexample: we’ve already seen that the weak dimension of a ring is less than or equal to one iff every ideal is flat. And, for Noetherian rings, the weak dimension is the same as the global dimension. For a field, the global dimension of $k[X]:=k[x_1,\dots,x_n]$ is $n$ and so if $n\geq 2$ then $k[X]$ must have ideals that are not flat, and yet each ideal is finitely generated. Hence $k[X]$ as a $k[X]$-module is flat (as it’s free) but has finitely generated $k[X]$-submodules that cannot be flat.

Amusingly, this counterexample is also a counterexample to the statement that to any conjecture one should give either a proof or an explicit counterexample!

Hint: for an actual counterexample, $(x,y)$ in $k[x,y]$ works!

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.
More »