# Example of a non-free module over some Laurent polynomial ring

This is probably a naive question. What is an example of a non-free finitely generated module $M$ over some Laurent polynomial ring $$L_n=K[X_1,X_1^{-1},\ldots,X_n,X_n^{-1}]$$ where $K$ is a field. It would be nice to have an example with $n=1$.

By a hard theorem of Swan every finitely generated projective module over $L_n$ is free and therefore it suffices to find a non-projective module.

Since $L_n$ is a commutative domain, you can use any quotient of $L_n$ by a nontrivial ideal, say $I$.

$L_n/I$ can't be projective since that would imply that this short exact sequence splits: $$0\to I\to L_n\to L_n/I\to 0$$

But nontrivial ideals of domains cannot be summands.

For any ring $R$ with a maximal right ideal $M$, $R/M$ is going to be free iff $R/M=R$ is a simple right $R$ module, in which case it is a division ring. Thus $R/M$ is never free for any non-division ring $R$. (And of course, it is cyclic.)

• Even without knowing that it is enough to find a non-projective, it is still easy to see that such a quotient cannot be free, as it has torsion. – Tobias Kildetoft Mar 19 '13 at 13:15

Concerning the question there is nothing special about the ring of Laurent polynomials.

For a commutative ring $R \neq 0$, the following are equivalent:

• $R$ is a field.
• Every $R$-module is free.
• Every finitely generated $R$-module is free.
• For every ideal $I \subseteq R$ the $R$-module $R/I$ is free.
• $R$ has exactly two ideals.
• $R$ has exactly two principal ideals.

If these statements are $1,\dotsc,6$, then $1 \Rightarrow 2$ follows from Zorn's Lemma, $2 \Rightarrow 3 \Rightarrow 4$ are trivial, $4 \Rightarrow 5$ follows from the observation that free modules have annihilator $0$ (in case of rank $>0$) or $R$ (rank $=0$), and $5 \Rightarrow 6 \Rightarrow 1$ are trivial.

Of course $k[X,X^{-1}]$ is not a field, see here for various reasons. Unwinding the proofs, one finds for example that the $k[X,X^{-1}]$-module $k[X,X^{-1}]/(X-1) \cong k$ is not free.

• @ron It's pretty easy to do the same thing (with the first three - I didn't bother to check the last three) in noncommutative rings. $R$ is a division ring iff all right $R$ modules are free iff $R/M$ is free for a maximal right ideal $M$. If $R/M$ is free, it is just $R$ by simplicity of $R/M$, thus $R$ is a simple $R$ module, aka a division ring. So you can see here that simple modules make great candidates for nonfree modules. – rschwieb Mar 19 '13 at 17:19