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I was reading Foundations of Module and Ring Theory and i found this equivalence of maximal left ideal as exercise in the the first chapter:

A left ideal $I$ of a ring $R$ is a maximal if and only if for each $r ∈ R-I$ there is some $s∈R$ such that $1-sr∈I$.

Which im trying to figure out the proof :/

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Suppose $I$ is a maximal left ideal in the ring $R$ and let $r\in R-I$; then $Rr+I$ is a left ideal of $R$ properly containing $I$, so, by the maximality assumption, $Rr+I=R$ and therefore $1\in Rr+I$.

Conversely, suppose $I$ satisfies the condition and let $J$ be a left ideal of $R$ properly containing $I$. Choose $r\in J-I$; by the condition, there is $s\in R$ such that $t=1-sr\in I$. Thus $1=t+sr\in J$ and $J=R$.

Fill in the details.

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From aspects of modules,

Let $I$ maximal left ideal then $R/I$ is a simple $R$ module.

Let $r\in R- I$ then $\bar r=r+I$ is nonzero elements of $R/I$ by simplicity of $R/I$,

$<\bar r>= R/I$ which means that there exist an $s\in R$ such that

$$\bar s \bar r =\bar 1$$ $$sr+I=1+I\implies 1-sr\in I$$

I left converse diraction as an exercise for you.

Note: Notice that $R/I$ need not be ideal but it is abelian group and $<x>$ represent the $R$ module generated by $x$.

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