If is Noetherian and is finitely generated, then has a finitely generated free resolution.
Proof
Let be a set of generators of (this may not be finite, for example always works).
Then there is a surjective -module homomorphism .
Let be its kernel.
Repeating the above argument, we get a surjective -module homomorphism which fits into the diagram
This means by the way we constructed.
Now you can repeat this process for , and onward.
In this way the horizontal subdiagram of
is a free resolution of .
Proof for part 2:
The Neotherian condition on implies that any submodule of a finitely generated R-module is again finitely generated.
This means that if we apply the process above then we know that we can find a finite generating set i.e. .
Then, is a submodule of a finitely generated free module so that means there is a finite generating set .
Keep doing that…