Algebras and modules
Model category presentations
Geometry on formal duals of algebras
A ring is a principal ideal domain if
it is an integral domain;
every ideal in is a principal ideal.
Often pid is used as an abbreviation of “principal ideal domain”.
Structure theory of modules
Free and projective modules
If is a pid, then any submodule of a free module over is also free. (For the converse statement, see here.)
By freeness of , there exists an isomorphism , a coproduct of copies of (as a module over the ring ) indexed over a set , which we assume well-ordered using the axiom of choice. Define submodules of :
Any element of can be written uniquely as where and . Define a homomorphism
by . The kernel of is , and we have an exact sequence
where is a submodule (i.e., an ideal) of , hence generated by a single element . Let consist of those such that , and for , choose such that . We claim that forms a basis for .
First we prove linear independence of . Suppose , with . Applying , we get . Since , we have (since we are working over a domain). The assertion now follows by induction.
Now we prove that the generate . Assume otherwise, and let be the least such that some cannot be written as a linear combination of the , for . If , then , so that for some , but this contradicts minimality of . Therefore . Now, we have for some ; put . Clearly
and so . Thus for some . At the same time, cannot be written as a linear combination of the ; again, this contradicts minimality of . Thus the generate , as claimed.
Since the integers for a pid, and abelian groups are the same as -modules, we have
Also, since projective modules are retracts of free modules, we have
Projective modules over a pid are free. In particular, submodules of projective modules are projective.
A finitely generated torsionfree module over a pid is free.
Let be a finite set of generators of , and let be a maximal subset of linearly independent elements. (Unless , then has at least one element, because is torsionfree.) We claim that can be embedded as a submodule of the free module generated by (which in turn is the span of as a submodule ). By Theorem 1, it follows that is free.
Let be the elements of . It follows from maximality of that for any , there is a linear relation
with . For each in the complement , pick such an , and form . Then the image of the scalar multiplication factors through , and is monic because is torsionfree. This completes the proof.
Let be a pid. Then an -module is torsionfree if and only if it is flat.
Suppose is flat. Let be the field of fractions of ; since is a domain, we have a monic -module map . By flatness, we have an induced monomorphism . For any nonzero , the naturality square
commutes, and since the map is multiplication by a non-zero scalar on a vector space, it follows that this map and therefore also is monic, i.e., is torsionfree.
In the other direction, suppose is torsionfree. Any module is the filtered colimit over the system of finitely generated submodules and inclusions between them; in this case all the finitely generated submodules of are torsion-free and hence free, by Proposition 1. Thus is a filtered colimit of free modules; it is therefore flat by a standard result proved here.
Structure theory of finitely generated modules