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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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    I've started a meta discussion on these perplexing edits. http://meta.math.stackexchange.com/questions/1343/what-to-do-with-a-user-who-is-editing-existing-questions-and-replacing-with-entir2010-12-15

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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?