Posted by Jason Polak on 02. January 2018 · Write a comment · Categories: commutative-algebra · Tags: ,

A finitely-generated module over a principal ideal domain is always isomorphic to $R^n\oplus R/a_1\oplus\cdots\oplus R/a_n$ where $n$ is a nonnegative integer and $a_i\in R$ for $i=1,\dots,n$. This is called the structure theorem for modules over a principal ideal domain. Examples of principal ideal domains include fields, $\Z$, $\Z[\sqrt{2}]$, and the polynomial ring $k[x]$ when $k$ is a field.

If $a\in R$ is not a unit, then $R/a$ is not projective, since $a$ annihilates any element of $R/a$ and therefore $R/a$ cannot be the direct summand of any free module. Therefore, we can conclude from the structure theorem that any finitely-generated projective module over a principal ideal domain is a free module. Don’t get your hopes up though: there are many examples of non-free projective modules.

But let’s stick with principal ideal domains. It is actually true that every projective module over a principal ideal domain is free. Kaplansky in [1] proved the following even stronger theorem:

Theorem. If $R$ is an integral domain in which every finitely generated ideal is principal, then every projective $R$-module is free.

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A commutative ring $R$ can be non-Noetherian and have all of its localisations at prime ideals Noetherian, such as the infamous $\prod_{i=1}^\infty \mathbb{Z}/2$. So being Noetherian is not a local property. However, there is an interesting variant of ‘local’ that does work, which I learnt from Yves Lequain’s paper [1]. It goes like this:

Theorem. Let $R$ be a ring and fix a left maximal ideal $M$ of $R$. Then $R$ is left Noetherian if and only if every left ideal contained in $M$ is finitely generated.

The nice thing about this statement is that it avoids localisation so it’s easy to state for noncommutative rings.
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