If $R$ is a commutative ring, and $I\subset R$ is a non-zero free ideal, then it is principal generated by a non-zerodivisor.

If $R$ is a non-commutative ring having the IBN property and $I\subset R$ is a non-zero free ideal, then $I$ is not necessarily principal. A concrete example of a non-principal free ideal in $R=K\langle x,y\rangle$ is $I=Rx+Ry$. (Thanks to @rschwieb for pointed this out.)

For now I would like to find out an example of a non-principal free ideal in a (non-commutative) ring without the IBN property.

In order to do this I've thought to start with a ring $R$ with the property that $R\simeq R^3$ as (left) $R$-modules. Then the left submodule $I=\{0\}\times R\times R$ of $R^3$ is free (since $I\simeq R^2$), and therefore corresponds to a free left ideal in $R$. It remains to show that $I$ is not principal and here I'm stuck. If $R$ is an integral domain, then this follows immediately, but I don't know such an $R$. (In Lam's, Lectures on Modules and Rings, page 294, one can find an example of integral domain without the IBN property, but this doesn't satisfy the condition $R\simeq R^3$.)

  • $\begingroup$ Hm, I'm not sure what is known about ideals in Leavitt path algebras (which might give you your answer). Hopefully there are some free ideals... $\endgroup$
    – rschwieb
    Jan 10, 2016 at 4:46
  • $\begingroup$ In your proposed example, $I$ will be principal, as it's a quotient of $R^3$, which is isomorphic to $R$. $\endgroup$ Jan 10, 2016 at 11:21
  • $\begingroup$ @JeremyRickard Your remark and my mistake show that there is no domain such that $R\simeq R^3$ (as $R$-modules), and similarly for $R\simeq R^n$ with $n\ge3$. (What about $R\simeq R^2$?) $\endgroup$
    – user26857
    Jan 10, 2016 at 14:39
  • $\begingroup$ If $R\cong R^n$ for any $n>1$ then every free ideal of finite rank will be principal. $\endgroup$ Jan 11, 2016 at 8:50

1 Answer 1


If you don't mind the ideal being of infinite rank, then let $S$ be any ring without the IBN property, and let $R=S\otimes_\mathbb{Z}\mathbb{Z}\langle x_1,x_2,\dots\rangle$. Then $R$ doesn't have the IBN property (if $S^m\cong S^n$ then $R^m\cong R^n$). But the left ideal $\sum_iS\otimes_\mathbb{Z}\mathbb{Z}\langle x_1,x_2,\dots\rangle x_i$ is free of infinite rank and so not principal.

If you want the ideal to have finite rank, then you could probably find an example where $S$ is a ring without the IBN property such that $S\not\cong S^2$ but $S^2\cong S^3$, and $R=S\otimes_\mathbb{Z}\mathbb{Z}\langle x,y\rangle$. The left ideal $S\otimes_\mathbb{Z}\mathbb{Z}\langle x,y\rangle x+ S\otimes_\mathbb{Z}\mathbb{Z}\langle x,y\rangle y$ is free of rank $2$, and I don't see why it would need to be principal.


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