symmetric monoidal (∞,1)-category of spectra
A ring is a principal ideal domain if
Often pid is used as an abbreviation of “principal ideal domain”.
the ring of integers
a discrete valuation ring (for example, a ring of formal power series over a field)
That both the integers and the polynomial rings over finite fields are principal integral domains with finite group of units is one aspect of the close similarity between the two that is the topic of the function field analogy. That also the holomorphic functions on the complex plane form a Bézout domain may then be viewed as part of the further similarity that relates the previous two to topics such as geometric Langlands duality. See at function field analogy -- table for more on this.
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
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.
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.
In constructive mathematics, many important rings may fail to be principal ideal domains in the naïve sense; the notion of Bézout domain, in which only the finitely generated ideals are required to be principal, is better behaved.
For instance, the ring of integers is a principal ideal domain if and only if the law of excluded middle holds: In one direction, the usual proofs rely on being able to decide whether any particular integer belongs to the ideal or not. For the converse, let be an arbitrary proposition. Consider the ideal . By assumption, it is generated by some number . Since the integers are discrete, it holds that or . In the first case holds, in the second .
However, this ideal cannot be proved to be finitely generated either. If an ideal is generated by , then we may form their gcd? one step at a time, which we can do algorithmically. Therefore, remains a Bézout domain.
On the other hand, we could try to modify the concept of principal ideal domain to recover a concept that is identical to the usual one in classical mathematics but also includes . For instance, we could demand that every decidable ideal is principal, or (potentially more strongly) that any ideal generated by a decidable subset is principal. While these seem to work at first, they are too weak to prove that every PID is a Bézout domain, so we should try to think of something better.
O. Helmer, Divisibility properties of integral functions, Duke Math. J. 6 (1940), 345-356.
Wikipedia, Principal ideal domain
Eric Wofsey, Principal Ideal Domains, (written for Mathcamp 2009) pdf