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I can't find an example of a countable injective module over a non-Noetherian ring.

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And, presumably, you would like to know one? (I suspect you weren't just posting a status update... have you considered actually asking the question, then?) – Arturo Magidin Feb 6 '11 at 22:11
In other words: This is not twitter :) – Mariano Suárez-Alvarez Feb 6 '11 at 22:45
Yeah, I would like to know one – Wesley Farrel Feb 6 '11 at 22:46
Then perhaps you might consider editing your question so as to make it a question, not a status update. – Arturo Magidin Feb 7 '11 at 0:45
up vote 2 down vote accepted

Another type of example: Let A be any non-Noetherian ring, let B be the field of 2 elements, and let C be the field of rational numbers. Then R = A×B×C is a non-Noetherian ring, its module 0×B×0 is injective and finite, and its module 0×0×C is injective and countably infinite. Injective modules are at least somewhat local ideas, so even if the ring is bad somewhere, it doesn't mean all the injectives are bad.

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Let $R$ be a countable non-Noetherian integral domain, e.g. a polynomial ring in countably infinitely many indeterminates over $\mathbb{Z}$ (or, more appealingly to a number theorist, the ring $\overline{\mathbb{Z}}$ of all algebraic integers).

Then the fraction field $K$ of $R$ is a countable, injective $R$-module. In fact it is the injective hull of the $R$-module $R$: see this wikipedia article on injective hulls.

In general, I confess that I have not thought as much as I should on injective hulls -- I haven't even gone through the proof that they always exist! -- but I would have to think that if $M$ is any infinite $R$-module, then its injective hull has the same cardinality as $M$. This would give many examples.

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I'm not much into the infinite, but apparently Eklof showed that even if you assume R is commutative (associative, unital) and card(M) ≥ card(R), then card(E(M)) need not equal card(M). In general though, I think you cannot expect to do better than a bound depending both on card(M) and the number of generators of the ideals of R. If you need to invert a continuum of independent generators, even a countable module is going to get big. If R is commutative and noetherian, then things are much better. – Jack Schmidt Feb 7 '11 at 0:25
@Jack: thanks for the comment. I'll keep that in mind for whenever I get around to seriously thinking about injective modules. – Pete L. Clark Feb 7 '11 at 1:32

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