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I need two counterexamples.

First, a direct sum of $R$-modules is projective iff each one is projective. But I need an example to show that, “an arbitrary direct product of projective modules need not be a projective module.”

If I let $R= \mathbb Z$ then $\mathbb Z$ is a projective $R$-module, but the direct product $\mathbb Z \times \mathbb Z \times \cdots$ is not free, hence it is not a projective module. We have a theorem which says that every free module over a ring $R$ is projective. Am I correct?

Second, a direct product of $R$-modules is injective iff each one is injective but I need an example to show that the direct sum of injective modules need not be injective.

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    "its not free ,hence it is not projective" All free modules are projective, but not all projective modules are free.2012-07-21
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    I rewrote most of this. Let me know if I made any mistakes. Please try to take more care in your writing and formatting, for my sake! I second Alex's observation.2012-07-21
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    @AlexBecker Correct, but over a PID a module is projective iff it is free, and $\mathbb{Z}$ is a PID, so maybe this was a hidden step in OP's argument.2012-07-21
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    So the example is right????? i need an example on injective you didn't answer my question2012-07-21
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    The direct product of infinitely many copies of $\mathbb{Z}$ is indeed not projective, but the reason you give is incorrect. You know that we *always* have that free implies projective, and that the module here is not free. But from $P\to Q$ and $\neg P$ you **cannot** conclude $\neg Q$: if it rains, then you get wet; that does not mean that if it doesn't rain, then you don't get wet (maybe you fall into a pool?) (cont)2012-07-21
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    (cont) Instead, what you need to use is the fact that, for the special case of the ring $\mathbb{Z}$ (or more generally, a PID), you have the **other** implication: projective implies free; hence, not free implies not projective. For the injective example, you're going to need a ring that is not noetherian.2012-07-21
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    i uses ALGEBRA by Thomas W. Hungerford there is a Theorem says :every free module is projective this mean free implies projective2012-07-21
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    @MissIndependent Right, but do you understand Arturo's comment? A statment is equivalent to its contrapositive, so your theorem implies "not projective implies not free." It does *not* imply "not free implies not projective." Although, in this case, this is true because you are working over $\mathbb{Z}$, which is a PID.2012-07-21
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    Thanks for the clarification ^_~2012-07-21

1 Answers 1

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As for the first question: yes, $P = \prod_{i=1}^{\infty} \mathbb{Z}$ is a direct product of free $\mathbb{Z}$-modules which is not free. Since $\mathbb{Z}$ is a PID, $P$ is also not projective. The proof that $P$ is not free is nontrivial, but I believe it has already been given either here or on MathOverflow.

As for the second question: the Bass-Papp Theorem asserts that a commutative ring $R$ is Noetherian iff every direct sum of injective $R$-modules is injective. Thus every non-Noetherian ring carries a counterexample. The proof of the result -- given for instance in $\S 8.9$ of these notes -- is reasonably constructive: if

$I_1 \subsetneq I_2 \subsetneq \ldots \subsetneq I_n \subsetneq \ldots$

is an infinite properly ascending chain of ideals of $R$, then for all $n$ let $E_n = E(R/I_n)$ be the injective envelope (see $\S 3.6.5$ of loc. cit.) of $R/I_n$, and let $E = \bigoplus_{n=1}^{\infty} E_n$. Then $E$ is a direct sum of injective modules and (an argument given in the notes shows) that $E$ is not itself injective.

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    Is the dual statement also true? Namely $R$ is left/right Noetherian (or maybe Artinian) iff every direct product of projective left/right $R$-modules is projective.2013-10-20
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    @Leon: No (for Noetherian rings). The countable product of copies of $\mathbb{Z}$ is not a projective (equivalently, free) $\mathbb{Z}$-module. See e.g. Theorem 2.4 in http://math.uga.edu/~pete/Math8030_Exercises.pdf.2013-10-21
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    @Leon: But, **yes**, a commutative ring is Artinian iff every product of projective modules is projective. This is a 1960 theorem of Chase. So far as I can see this result is hard to find in standard references, but e.g. it appears as an exercise on p. 161 of T.Y. Lam's *Lectures on Modules and Rings*.2013-10-21
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    Thank you! You're right, the theorem is hard to find (e.g. Rotman, *An Introduction to Homological Algebra* 2nd ed., p.186, but no proof). I did not find it in Chase's original 1960 article. I assume the version for noncommutative rings is also true?2013-10-21
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    @Leon: The theorem takes place in the non-commutative case, but the result is a bit more complicated there. I recommend you take a look at Lam's text.2013-10-21
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    @PeteL.Clark Could it be that the exercise following Theorem 2.4 has a typo, and $\bigoplus R$ is meant to be a $\prod R$? Regards,2014-06-12