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Can you tell me why the following is true?

A ring $R$ is local if and only if every principal left $R$-module is indecomposable.


(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)

what is the relation between regular ring and this property:

$ab=1$ then $ba=1$ when $a,b\in R$

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@Arash: Do you have a question, or were you just making a statement? –  Arturo Magidin Dec 12 '10 at 23:32
    
@Willie Wong: That's a good guess, but I would much rather the OP would clarify just what he is doing or expecting. Maybe he was assigned a "prove or disprove"? –  Arturo Magidin Dec 13 '10 at 0:34
    
@Arash: welcome to math.SE! I edited your question to be, well, actually a question. For future reference, it is generally a good idea to state clearly what your question is, and perhaps give a bit of background information. –  Willie Wong Dec 13 '10 at 0:36
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@Arturo: this question was re-directed here from MO. The version on MO, while equally terse, actually asks "Why?". So no, it is not a guess. :-) –  Willie Wong Dec 13 '10 at 0:37
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This question completely changed with the edit 9 mins ago. All the above comments, and the answer below, apply to the previous version, which can be read in the edit history. –  Matt E Dec 14 '10 at 20:35

1 Answer 1

HINT for $\Leftarrow$: Do the contrapositive; if $\mathfrak{m}_1$ and $\mathfrak{m}_2$ are two distinct maximal ideals of $R$, then look at the left $R$-module $R/(\mathfrak{m}_1\cap\mathfrak{m}_2)$; it is principal (generated by $1+(\mathfrak{m}_1\cap\mathfrak{m}_2)$. Is it indecomposable?

HINT for $\Rightarrow$: Suppose $M\neq\{0\}$ is principal, and $M=N_1\times N_2$. Let $m=(n_1,n_2)$ generate $M$. Then there are $a,b\in R$ such that $am = (n_1,0)$ and $bm=(0,n_2)$; so $(a+b)m = m$. Then $a+b-1\in\mathrm{Ann}(m)$. Can $a$ and $b$ both lie in the maximal ideal of $R$? What does that mean?

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