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We knew that the direct sum of a family of projective modules is a projective module, and the direct product of a family of injective modules is also injective.

My question is, is the infinite direct product of an infinite family of projective modules also projective? If it is, please give an elegant proof, or else please give some counter-examples.

Thank for reading my question.

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Are you asking for an example that a direct product of an infinite family of projective modules is not projective? See Baer–Specker group en.wikipedia.org/wiki/Baer-Specker_group –  wxu Dec 11 '11 at 5:19
    
Yes, it is also my purpose. Thanks wxu –  Knumber10 Dec 11 '11 at 5:36
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chase's ams.org/mathscinet-getitem?mr=120260 shows that projective is closed under direct products if and only if the ring is perfect (if and only if every module has a projective cover, etc.). –  Jack Schmidt Dec 11 '11 at 18:03

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up vote 3 down vote accepted

Note: The question originally asked about infinite direct sums.

If you meant "infinite direct product", on the other hand, then the answer is "no." Consider $\mathbb{Z}$-modules; then a module is projective if and only if it is free. So $\mathbb{Z}$ itself is projective, but $$\prod_{i=1}^{\infty}\mathbb{Z}$$ is not free, hence not a projective $\mathbb{Z}$-module.


Here's the answer to the original question:

The direct sum of free modules is free, even if the family is infinite.

Assume that $\{P_i\}_{i\in I}$ is a family of projective modules; for each $i$, there exists $M_i$ such that $P_i\oplus M_i$ is free. Then $$\left(\bigoplus_{i\in I}P_i\right) \oplus \left(\bigoplus_{i\in I}M_i\right) \cong \bigoplus_{i\in I}(P_i\oplus M_i)$$ is a direct sum of free modules, hence free. Since there is a module $M$ (namely, $M=\oplus M_i$) such that $(\oplus P_i)\oplus M$ is free, it follows that $\oplus P_i$ is projective.

That is, any direct sum of projective modules is projective.

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Yes, I mean infinite direct product. Thank for answering my question. –  Knumber10 Dec 11 '11 at 5:30
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@NguyễnDuyKhánh: Perhaps you can add the correction to your question, and fix the title? –  Arturo Magidin Dec 11 '11 at 5:31
    
Thank you very much. The example is more simple than I had thought. I had confusion in English term. My question can be understood as ''an arbitrary direct product of projective modules need not be a projective module''. One of my friend has just told me it's in fact a similar to a problem in Dummit and Foote's book, which concerns free module. I have taken a glimpse at it and it is more vague than a counter-example. –  Knumber10 Dec 11 '11 at 7:28
    
I'm sorry, can you tell me why is the product of infinite Z is not free? I can't see it direct. –  lee Mar 3 '13 at 13:58

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