# What is the reason direct product of injective modules are injective, while direct sum not necessarily?

In reviewing my algebra class material, I "discovered" a strange phenomenon, that the direct product of injective modules is injective, while if $R$ is not noetherian then the direct sum is not necessarily injective. This made me felt very surprised. In fact, this is only true if $R$ is noetherian.

Because it is trivial to prove that the direct summand of an injective module is injective, one is tempted to prove the direct sum of injective modules is injective by regarding it as the direct summand of the direct product. But this cannot be done. Similarly there are examples like $\displaystyle\prod^{\infty}_{i=1}\mathbb{Z}$ which is the direct product of projective modules, but nevertheless not projective itself because it is not free.

My question is, is there something deeper behind this seemingly bizarre dichotomy? Homological algebra is not my specialization field and my knowledge is quite superficial, so I cannot answer it satisfactory myself other than saying it is a reverse of arrows, etc.

There is a post which provides most of the background.

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I mean, I can answer the question more thoroughly if you like but you hit the nail on the head. Things that are true categorically are true trivially, and the fact that the arrows are going the wrong way for products is the issue. The only reason you get away with finite products is that they are "biproducts" and so they are both products and coproducts. – Alex Youcis Dec 10 '12 at 1:46
This is fine. But this explanation leaves a problem aside - when it is true for projective modules, the direct product is also projective? Do we have a dual notion of noetherian at here(artician)? – Bombyx mori Dec 10 '12 at 1:49

I think the question is more subtle than first appears. Although I cannot give you a complete characterisation of when a product of projectives is projective, I can answer the question for a fairly interesting class of rings:

1. It obviously works for some rings, like fields. More generally: One sufficient condition for a product of projectives to be projective is that $R$ is quasifrobenius. That is, $R$ is left and right noetherian and injective as a left $R$ module and as a right $R$-module.

Semisimple artinian rings and $R[G]$ where $R$ is quasifrobenius (e.g. field) and $G$ is a finite group are examples, which shows that there are interesting cases beyond fields.

The important property of these rings is that an $R$-module over a quasifrobenius ring $R$ is projective iff it is injective. Hence a direct product of projectives is a direct product of injectives, and hence injective, hence projective.

1. Of course, there are still other questions to ask: if every product of projectives is projective, is the underlying ring quasifrobenius (I'd guess not), and given a specific ring, what conditions on the factors will ensure projectivity?

P.S. Duality in formulating conditions works well for definitions in the category of modules but not with respect to the original ring because the opposite category of the category $R$-Mod is not $S$-Mod for any $S$, and certainly not any $S$ "dually related" to $R$, unless of course $R = 0$.

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