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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. –  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
@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
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

1 Answer 1

up vote 4 down vote accepted

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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Thank you very much –  Miss Independent Jul 21 '12 at 23:16
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. –  Leon Oct 20 '13 at 19:00
@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 math.uga.edu/~pete/Math8030_Exercises.pdf. –  Pete L. Clark Oct 21 '13 at 1:48
@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 2:57
@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

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