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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.

Arturo Magidin
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Miss Independent
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    "its not free ,hence it is not projective" All free modules are projective, but not all projective modules are free. – Alex Becker Jul 21 '12 at 21:25
  • 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. – Dylan Moreland Jul 21 '12 at 21:29
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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. – Derek Allums Jul 21 '12 at 21:34
  • So the example is right?????

    i need an example on injective

    you didn't answer my question

     – Miss Independent Jul 21 '12 at 21:36
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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) – Arturo Magidin Jul 21 '12 at 22:02
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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. – Arturo Magidin Jul 21 '12 at 22:03
  • i uses ALGEBRA by Thomas W. Hungerford there is a Theorem says :every free module is projective this mean free implies projective – Miss Independent Jul 21 '12 at 22:22
  • @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. – Derek Allums Jul 21 '12 at 22:25
  • Thanks for the clarification ^_~ – Miss Independent Jul 21 '12 at 22:31

1 Answers1

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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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Pete L. Clark
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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. – Leo Oct 20 '13 at 19:00
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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. – Pete L. Clark Oct 21 '13 at 01:48
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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. – Pete L. Clark Oct 21 '13 at 02:57
  • 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? – Leo Oct 21 '13 at 15:41
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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. – Pete L. Clark Oct 21 '13 at 15:54
  • @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, – Pedro Jun 12 '14 at 14:59