According to Wikipedia, a (non-commutative) ring $R$ is local if and only if there do not exist two proper (principal) (left) ideals $I_1, I_2$ such that $R = I_1 + I_2$. It is easy to see that local rings have this property, but I'm having trouble showing the other way: if there do not exist two proper co-maximal (principal) (left) ideals then the ring is local.
Local rings characterization
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abstract-algebra
1 Answers
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HINT $\ $ If $\rm\ R \ne 0\ $ is nonlocal then there exists an $\rm\: r\:$ such that both $\rm\: r\:$ and $\rm\: 1-r\:$ are nonunits. Choose maximal ideals containing them.
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0Thanks. But this method doesn't seem to work for two-sided ideals. (which is presumably what Wikipedia author meant by the parentheses.) – 2010-12-02
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0@user4341: Wikipedia means that the following are equivalent: $R$ is local; there do not exist two proper ideals $I_1$ and $I_2$ such that $I_1+I_2=R$; there do not exist two proper principal ideals $I_1$ and $I_2$ such that $I_1+I_2=R$; there do not exist two proper left ideals $I_1$ and $I_2$ such that $I_1+I_2=R$; there do not exist two proper principal left ideals $I_1$ and $I_2$ such that $I_1+I_2=R$. What do you mean Bill's hint does not work for ideals? Take $RrR$ and $R(1-r)R$. – 2010-12-02
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0$RrR=R$ and $r$ being a unit are not equivalent statements; a non-unit element can generate the entire ring as a two-sided ideal. – 2010-12-02
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0@user4341: if $R$ is not local, then there are two maximal ideals $\mathfrak{M}_1$, $\mathfrak{M}_2$, distinct. Since $\mathfrak{M}_1+\mathfrak{M}_2 = 1$, there exists $r\in\mathfrak{M}_1$, $s\in\mathfrak{M}_2$ such that $r+s = 1$. Then $RrR\subseteq \mathfrak{M}_1$ and $R(1-r)R = RsR\subseteq\mathfrak{M}_2$ are proper principal ideals, but $RrR + R(1-r)R = R$. This *is* Bill's hint. – 2010-12-03
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0Non-local ring can have a unique two-sided maximal ideal: for example $2\times2$ matrix ring over a field is not local but has a unique two-sided maximal ideal (0). I now realize that this is a counterexample to the Wikipedia's claim. – 2010-12-03