Suppose $R $ is a non zero ring.

How can we prove the following conditions are equivalent?

In Proof 2 to 3, how can we prove from $ a R = R $ that the ring has identity?

1: For all $a, b \in R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ .

2: $ R^{2} \neq 0$ and $ R$ has no right ideals other than $ 0$ and $R$.

3: $ R$ is a division ring.

  • $\begingroup$ To your "2 to 3" question, $a\in aR$ should give you some info. It just remains to show that whichever $x$ has $ax=a$ works for all other elements as well. $\endgroup$ – Arthur Nov 4 '17 at 17:46

The set of all elements $x$ such that $xR=0$ forms a right ideal of $R$.

By hypothesis it can't be all of $R$, so it is just $\{0\}$.

In other words, $xR=R$ for nonzero $x$.

Let $x$ be a nonzero element. Two observations:

  1. The right annihilator of $x$ is a right ideal, and it can't be $R$, so it is zero. Therefore $x$ is left cancelable. (Notice this is true for any nonzero $x$.)

  2. $xe=x$ for some $e$, and furthermore $xee=xe=x$. Canceling $x$ we find that $e^2=e$.

Using the $e$ above, we see that $ey=e^2y$ implies $y=ey$ for any $y\in R$, so $e$ is a left identity.

Finally, there's probably a shorter trick, but this one is the first that occurred to me: $(y-ye)^2=y^2-y^2e-yey+yeye=y^2-y^2e-y^2+y^2e=0=(y-ye)0$. If $y-ye\neq0$, we could cancel it on the left, obtaining $y-ye=0$, a contradiction. Therefore $y-ye=0$, and $e$ is a right identity too.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.