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sniffnoySome days ago Heidi asked me how to show that every module is a direct sum of indecomposable modules. Sounds like a triviality, I said. Well, OK, not a total triviality; I assume in general you'd have to use choice, probably in the form of Zorn's Lemma.
But I'd never actually seen that particular statement proved, so I tried several ways of applying Zorn's Lemma, but I couldn't quite get any of them to work.
Well, today I tried looking things up, and I found that a module is called "completely decomposable" if it can be written as a direct sum of indecomposable modules. We wouldn't give a name to the concept if every module were like that, were we? Google didn't easily turn up any concrete counterexample, and I was about to ask math.SE, but I decided I could probably solve this myself.
So, yeah -- let F be a field with more than 2 elements -- having only 2 elements is probably not a problem, but best just to avoid the problem rather than solve it -- let R be the product of infinitely many copies of F, and consider R as a module over itself. Yay, this is not completely decomposable. (Hint: Which ideals of R are direct summands in the first place?)
Now, if it turns out that Heidi asked me if every module can be written as a direct sum of indecomposable modules, rather than how to prove that they are, I am going to be a little embarrassed...
-Harry
But I'd never actually seen that particular statement proved, so I tried several ways of applying Zorn's Lemma, but I couldn't quite get any of them to work.
Well, today I tried looking things up, and I found that a module is called "completely decomposable" if it can be written as a direct sum of indecomposable modules. We wouldn't give a name to the concept if every module were like that, were we? Google didn't easily turn up any concrete counterexample, and I was about to ask math.SE, but I decided I could probably solve this myself.
So, yeah -- let F be a field with more than 2 elements -- having only 2 elements is probably not a problem, but best just to avoid the problem rather than solve it -- let R be the product of infinitely many copies of F, and consider R as a module over itself. Yay, this is not completely decomposable. (Hint: Which ideals of R are direct summands in the first place?)
Now, if it turns out that Heidi asked me if every module can be written as a direct sum of indecomposable modules, rather than how to prove that they are, I am going to be a little embarrassed...
-Harry
