# How do I prove these two definitions for a local ring are equivalent?

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.

• I don't think it's a full answer, but this question may be helpful: math.stackexchange.com/questions/1243171/… – Henry Swanson Jul 11 '17 at 15:52
• @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.. – Rubertos Jul 11 '17 at 15:55
• 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? – Daniel Schepler Jul 11 '17 at 15:55
• 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. – user45765 Jul 11 '17 at 15:59
• 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. – Henry Swanson Jul 11 '17 at 16:01

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.

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 $J$.