Let and let be considered as an -submodule of . Show that is not a basis of . Show that has rank 1, but is not free with free rank 1.

Note that , so that and are not -linearly independent. Thus has no hope of being a basis of .

Now suppose are nonzero. Similarly, , so that is not -linearly independent. In particular, the rank of as an -module is at most 1. On the other hand, since is an integral domain, every singleton set is linearly independent. So the rank of over is exactly 1.

Suppose now that is free with free rank 1. That is, for some . Now for some , so that has degree 0. That is, , and we have . If , then . This is a contradiction since (for example) . So, without loss of generality, . But then , and so , a contradiction. Thus is not free of free rank 1 as an -module.