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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    $\begingroup$ "its not free ,hence it is not projective" All free modules are projective, but not all projective modules are free. $\endgroup$ Jul 21, 2012 at 21:25
  • $\begingroup$ 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. $\endgroup$ Jul 21, 2012 at 21:29
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    $\begingroup$ @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. $\endgroup$ Jul 21, 2012 at 21:34
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    $\begingroup$ 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) $\endgroup$ Jul 21, 2012 at 22:02
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    $\begingroup$ (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. $\endgroup$ Jul 21, 2012 at 22:03

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Leo
    Oct 20, 2013 at 19:00
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    $\begingroup$ @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. $\endgroup$ Oct 21, 2013 at 1:48
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    $\begingroup$ @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. $\endgroup$ Oct 21, 2013 at 2:57
  • $\begingroup$ 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? $\endgroup$
    – Leo
    Oct 21, 2013 at 15:41
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    $\begingroup$ @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. $\endgroup$ Oct 21, 2013 at 15:54

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