Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let R be a ring. If a direct sum of R-modules is projective, can we imply that the direct sum is also a free module over R ?

share|improve this question

1 Answer 1

I assume you're excluding trivial summands; otherwise the answer would trivially be "no". But it's also "no" with non-trivial summands. Take for instance $\mathbb{Z}$ as a module over $\mathbb{Z}\oplus\mathbb{Z}$, with multiplication defined by $(a,b)\cdot c= a\cdot c$. This is a projective module, but its direct sum with itself is not a free module (which you can see e. g. by noting that many elements of $\mathbb{Z}\oplus\mathbb{Z}$ leave this module invariant whereas only the identity element would leave a free module invariant).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.