Timeline for A question about "large" indecomposable injectives over commutative rings
Current License: CC BY-SA 3.0
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| Jul 7, 2015 at 20:03 | comment | added | Keith Kearnes | There do exist nonnoetherian rings which do not have a big injective of the type sought. Any infinite power of a finite ring is such an example. If $|R|=n < \omega\leq\kappa$, then any module $M$ over the nonnoetherian ring $S:=R^{\kappa}$ that has a least nonzero submodule must be annihilated by an ultrafilter ideal of $S$. Hence $M$ may be thought of as a module over an ultrapower of $R$. Since $R$ is finite, it is isomorphic to any of its ultrapowers. Thus, $M$ may be considered to be a subdirectly irreducible $R$-module, forcing $|M|$ to divide $|R|=n$. | |
| Jun 23, 2015 at 21:51 | comment | added | Greg Oman | Thanks for taking the time to respond, Dracula (I too appreciated the laugh, as it tipped me off to the fact that I wasn't careful enough in the wording of the question). What I meant to ask was this: does there exist an infinite commutative ring $R$ with the property that there is a large indecomposable injective module $M$ over $R$ with a minimum submodule. Yes, any such ring must be non-Noetherian; this follows from early (and famous) results of Eben Matlis. I don't think the paper you referenced answers the question; my guess is that "most" non-Noetherian rings don't have such a module | |
| Jun 23, 2015 at 15:19 | comment | added | Jason Starr | You have a frightening laugh, Count Dracula. | |
| Jun 23, 2015 at 13:30 | history | answered | Count Dracula | CC BY-SA 3.0 |