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If a non-trivial ring $R$ has a unique maximal left ideal $J$ , then $J$ is two-sided and is also the unique maximal right ideal in $R$.

I can prove that it is two sided, but I can't prove that it is unique.

My proof:

Let $r \in R$. Then $Jr$ is a left ideal. If $Jr = R$, then $jr = 1$ for some $j \in J$. Note that $rj \in J$ since J is a left ideal, so $rj \neq 1$. Using a lemma, $1 - rj$ is not left invertible, which is to say that $R(1-rj) \neq R$. But then R(1-rj) is contained in a maximal left ideal, i.e. $1-rj \in R(1-rj) \subseteq J$, so $1= rj +(1-rj) \in J$, which gives a contradiction. Hence $Jr \neq R$. This implies that $Jr$ is contained in $J$. Since the choice of $r$ is arbitrary, $jr \in J$ for all $j \in J$ and $r \in R$. Hence, $J$ is a two sided ideal.

Can someone tell me how I can prove uniqueness?

massy255
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user10024395
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    Wouldn't the argument be symmetric? Meaning: if there's a maximal right ideal $;M;$ containing $;J;$, then $;M;$ is also two sided and also maximal left ideal, contradicting that $;J;$ is maximal left? – Timbuc Apr 20 '15 at 10:07
  • @Timbuc it is symmetric but my proof is not complete because i have not prove uniqueness, I don't think you can't just use this to arrive at a contradiction unless the uniqueness part is proven. – user10024395 Apr 20 '15 at 10:16
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    But again: if $;J;$ is not a unique maximal right ideal and $;L\lneq M;,;;M;$ a maximal right ideal, then using the proof's symmetry show $;M;$ is also two-sided and a left maximal ideal, contradicting both uniqueness of $;J;$ a maximal left ideal and also its maximality... – Timbuc Apr 20 '15 at 10:19
  • @Timbuc why must J be strictly < M, what happens when J is not unique but maximal – user10024395 Apr 20 '15 at 10:46
  • That too: if $;M;$ is another max. right ideal different from $;J;$ , then you can prove by symmetry that $;M;$ is two sided nad maximal left...again contradiction! I can't see what's the problem here...perhaps I'm missing something. – Timbuc Apr 20 '15 at 10:57
  • @Timbuc It's not symmetric because you don't have $M$ is a unique maximal right ideal. – MathStarter Mar 14 '16 at 02:57
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    You say "Using a lemma $1-rj$is not left invertible.". Which is the exact statement of the lemma? @user136266 – Mary Star Apr 12 '16 at 02:51
  • @MaryStar yes I believe that this part of OP's argument is incorrect - if $J$ is the unique maximal left ideal then it follows that $J$ consists exactly of the elements of the ring that are not left-invertible, and hence such elements are closed under addition. So when OP says $rj \in J$ we have that $rj$ is not left-invertible, and hence actually the opposite of what OP claims with their lemma is true: $1-rj$ must be left-invertible, since if it weren't then $rj$ and $1-rj$ not being left-invertible would imply that $rj + (1-rj) = 1$ is not left-invertible. – Joe Stephen Jul 01 '22 at 22:14

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Suppose there's another right ideal $J'$ such that $J' \not \subseteq J$. Take $x \in J'$ such that $x \notin J$. Then $x$ is left invertible, so there exists $r \in R$ such that $rx =1$. But $xr \neq 1$. Thus both $xr $ and $1 -xr$ are not left invertible (by your lemma), thus $xr, 1 - xr \in J$, so $1 \in J$, contradiction. Therefore $J$ is the unique maximal right ideal.

MathStarter
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