An ideal of an integral domain is a free -module if and only if it is generated by one element. Indeed, given any two non-zero elements , . So any two elements are linear dependent. Hence, if were free module, it must generated by one element, or is a principle ideal. Our strategy is to find an ideal which is not principle and show that is projective.

Let . Obviously is an integral domain. Let be the ideal generated by 5 and . .

is not a principal ideal.

Define by , then it is straightforward to verify , for all . If is a unit, then , which means and . So the only units in are .

Now suppose is a principle ideal, then such that . Then such that , which means . Since has no solution in , or can not be 5. So we must have and or and . The second case is impossible, because means . Then . We can obtain , this is impossible.

The only case we left is , which means . Then . We have , corresponds to the ideal . If , then . This fact leads to , for some , . If we set , then in , this is impossible.

In conclusion, can not be principle.

is a projective module.

Define by . If we can find such that , then is a direct summand of , which is a free -module. Consequently, is a projective module.

Let , . There exist unique such that

A straightforward calculation shows that, , . Define . It is easy to verify that is an -module homomorphism. Moreover,

This means . Thus is a projective module.

As discussed in the beginning of this section, since is not principle, we get a non-free projective module.