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I find the following question: Is the direct sum of two projective and not free $R$-modules projective/free? What if $R$ has got no zero divisors?

The answer to the first question is very simple, because direct sum of projective modules are projective, this is obvious using the definition with direct summand of a free $R$-module.

Is it free: Consider the case $R=\mathbb{Z}_6$ and the free module $\mathbb{Z}_6$-module $\mathbb{Z}_6$ with the composition $\mathbb{Z}_6=\mathbb{Z}_2\oplus\mathbb{Z}_3$, both direct summand are projective $\mathbb{Z}_6$-module, but none of them is a free $\mathbb{Z}_6$-module.

Now I have got no idea, what happens if I assume that $R$ has no zero divisors. Please give me a hint to prove it or a counterexample. Thanks

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You will need projective modules over an integral domain $R$ which are not free. Here is a method to find such rings and modules. Take $R=\mathbb{Z}[\sqrt{-5}]$. Since this is a Dedekind ring, every ideal $I$ in this ring is a projective module. So for example $I=(3,2+\sqrt{-5})$. This would be a free module, if and only if $I$ were a principal ideal. But this is not the case, and $I$ is not free. Over a principal ideal ring $R$ such examples do not exist, because then "projective" and "free" is the same.

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  • $\begingroup$ More generally just choose an algebraic number field $K$ with $\mathcal{O}_K$ not a PID and take a non-principal ideal in there (which is always projective as an $\mathcal{O}_K$ - module.) $\endgroup$
    – user38268
    Aug 9, 2013 at 14:37

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