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Let $R$ be a non-trivial ring with unity. We say $R$ is a local ring if there exists a unique maximal left-ideal.

Let $R$ be a local ring. Then, how do I prove that there exists a unique maximal right-ideal?

I know that the maximal left-ideal is the Jacobson radical and it is the intersection of all the maximal right-ideals. However, I am not sure if this helps to prove the problem.

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  • $\begingroup$ I don't think it's a full answer, but this question may be helpful: math.stackexchange.com/questions/1243171/… $\endgroup$ – Henry Swanson Jul 11 '17 at 15:52
  • $\begingroup$ @HenrySwanson Well, but I know that the Jacobson radical is a two-sided ideal in general (That question merely shows that Jacobson radical is a two-sided ideal (when it is defined as the intersection of all maximal left-ideals) with stronger assumption) I am not sure how it is helpful.. $\endgroup$ – Rubertos Jul 11 '17 at 15:55
  • $\begingroup$ In the commutative case at least, being a local ring is equivalent to the set of non-units being closed under addition (then the maximal ideal is the set of non-units). Maybe the same thing happens in the noncommutative case? $\endgroup$ – Daniel Schepler Jul 11 '17 at 15:55
  • $\begingroup$ You can start proof by contradiction. Any element of the form $1+rs$ for all $s\in R$ is right invertible for $r\in J_R$ which rules out all the elements that are not contained in some right max ideal. $\endgroup$ – user45765 Jul 11 '17 at 15:59
  • $\begingroup$ So, if I understand your post, you know that the unique maximal left ideal (say $J$) is the same as the intersection of all maximal right ideals. But you know that $J$ is a right ideal, being the intersection of right ideals. I might be missing something, but I think you can still apply the posted answer from there. $\endgroup$ – Henry Swanson Jul 11 '17 at 16:01
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There are at least two 'symmetric' characterizations of local rings available:

  • For all $x\in R$, at least one of $x$ and $1-x$ is a unit
  • The set of nonunits is closed under addition.

You should be able to establish the equivalence of at least one of these with your definition, and then it is automatic.

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One way to start your proof is to suppose $K$ is some maximal right ideal. You know that $K$ contains $J$, since $J$ is the intersection of all the right ideals. (Incidentally, I'm assuming you know that there is at least one right ideal, since $J$ is the intersection of them and $J$ is non-degenerate). Now, consider any element $k \in K$. Then show that $Nk+J:=\{nk+j:n\in N, j\in J\}$ is a subgroup of the additive group of $R$, where $nk=k+k+...+k$ if $n>=0$, and $(-n)k=n(-k)$. Then show that $Nk+J$ is a left ideal of $R$. Lastly, draw some conclusions about $k$ and $K$ in relation to $J$.

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