Is there a notion of a dimension associated to free resolutions like projective and injective dimensions associated to projective and injective resolutions? My guess is that it coincides with projective dimension but couldn't find any reference talking about this. Any help or tips will be appreciated!
3 Answers
I'm sort of stealing the idea from t3suji, but here goes:
If the module is projective, i.e. $PD = 0$, then $FD \leq 1$.
If the module is not projective, i.e. $PD > 0$, then $PD = FD$.
The first statement is t3suji's comment. The second statement follows by first observing that if $F$ is the free module on infinitely many generators, there is an exact sequence of length $n+1$,
$0\rightarrow F\rightarrow F\rightarrow\ldots\rightarrow F\rightarrow 0$,
so long as $n\geq 1$. Then if we have a projective resolution
$0\rightarrow P_n\rightarrow\ldots\rightarrow P_0\rightarrow M\rightarrow 0$,
we can make the free resolution
$0\rightarrow P_n\oplus F\rightarrow\ldots\rightarrow P_0\oplus F\rightarrow M\rightarrow 0$.

$\begingroup$ Nice. It may be better to replace the condition that F has countably many generators by the condition that it has infinitely many generators (otherwise I do not see how you can make sure that P_i+F is free  it seems that you may need more than countably many generators if P_i's are not countably generated?) $\endgroup$– t3sujiJun 8, 2010 at 2:06

When requiring finitelygeneratedness of the resolution, then the free dimension of a projective module can be infinite.
As a simple example, take the ring $R=k\oplus k$, and let $e_1=(1,0)$ and $e_2=(0,1)$ be the obvious idempotents. Then the module $ke_1$ is projective. However, any surjection $$ R^{\oplus n}\rightarrow ke_1 $$ will have a kernel $S_1$ which is isomorphic to $R^{n1}\oplus ke_2$. Therefore, any surjection $$ R^{\oplus m}\rightarrow S_1 $$ will have a kernel $S_2$ which is isomorphic to $R^{mn}\oplus ke_1$. Thus, any f.g. free resolution of $ke_1$ must infinite.
Of course, as noted above, $ke_1$ has an infinitelygenerated free resolution of length 1, as it is projective.

$\begingroup$ Greg: nice example +1! Just a quick note, one can, in a very similar way, show that R=Z/6 and M=Z/2 work. I think, for commutative R, a necessary condition is connectedness of Spec(R), see my question linked above. $\endgroup$ Jun 8, 2010 at 19:33
It does not coincide with projective dimension because a projective module is not necessarily isomorphic to a free module (while a projective $P$ has a trivial projective resolution $0 \to P \to P \to 0$). In fact, projective modules that admit a free resolution are precisely the ones that are stably free (i.e. become free when a free module is added). It is known that this is always the case for finitely generated modules over a polynomial ring (this is a theorem of Serre). Cf. the last chapter of Lang's Algebra.
It is, incidentally, still true that the free dimension is finite in many interesting cases, such as finitely generated modules over a regular local ring or a polynomial ring over a field. This is because, in the former case, the residue class field $k$ of a regular local ring $R$ has a finite free resolution (namely, the Koszul complex), and any f.g. module $M$ with $\mathrm{Tor}_R(M,k) = 0$ can be shown to be projective, hence free (because $R$ is local, by a Nakayama argument). (Added admittedly in the local case this follows from the fact that projective dimension is finite. In the graded (polynomial) case a projective module is still free, but this is at least harder to prove.)
Edit: As pointed out in the comments, "free dimension" should require finitely generatedness of the resolution.

10$\begingroup$ On the other hand, if you remove the finitegeneratedness assumption, then every projective module has a free resolution of length two (if P is a direct summand of a free module F, then P+(infinitely many copies of F) is free). Therefore, in this case, the difference between projective dimension and `free dimension' is at most one. $\endgroup$– t3sujiJun 8, 2010 at 0:31