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

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    Then perhaps you might consider editing your question so as to make it a question, not a status update.2011-02-07

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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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    @Jack: thanks for the comment. I'll keep that in mind for whenever I get around to seriously thinking about injective modules.2011-02-07