MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a finite ring $R$ with identity. If every left ideal of $R$ is two-sided then is it true that any right ideal of $R$ is also two-sided !?

share|cite|improve this question
My gut says this must be true. The only examples of rings where things like this fail to be true (that I know of anyway) are those constructed by Jategaonkar using large cardinal axioms (discussed late in Chapter 5 of Lam's Lectures on Modules and Rings). So for finite rings I can't imagine examples like that. In fact, I'm guessing the full answer is in that book, and may be known for any Noetherian ring $R$. Some other places large cardinal axioms pop up in homological algebra are discussed here:… – David White Dec 30 '12 at 0:26
@David White: The problem is not true even for Artinian rings. K. Asano has constructed an Artianian ring such that every left ideal is two-sided but not every right ideal is two-sided. – user30230 Dec 30 '12 at 6:59
up vote 10 down vote accepted

Yes. For each $x\in R$ let $a(x)$ and $b(x)$ be the number of $y$ such that $xy=0$ and the number of $y$ such that $yx=0$, respectively. These are also the orders of the groups $R/xR$ and $R/Rx$.

For every $x$ the left ideal $Rx$ is a right ideal containing $x$ and therefore contains $xR$, whence $b(x)\le a(x)$. On the other hand the sum of $a(x)$ over all $x$ is clearly the same as the sum of $b(x)$ (both are the number of pairs $(x,y)$ such that $xy=0$). Therefore $a(x)=b(x)$ and $Rx=xR$. This shows that $xR$ is a left ideal for every $x$, and it follows that every right ideal is a left ideal.

share|cite|improve this answer
@Tom: I am afraid the argument is not right. How do you conclude that the sum of all $a(x)$'s and the sum of all $b(x)$'s are equal ?! – user30230 Apr 22 '13 at 18:36
I gave the reason. – Tom Goodwillie Apr 23 '13 at 0:00
Let $C$ be the set of all ordered pairs $(x,y)\in R\times R$ such that $xy=0$. Then $|C|$ is the sum, over all $x\in R$, of $|A(x)|$. On the other hand $C$ is also the sum, over all $y\in R$, of $|B(y)|$. – Tom Goodwillie Apr 23 '13 at 11:32
@Tom: I see. Thank You. – user30230 Apr 27 '13 at 13:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.