In the context of this course, a ring is not necessarily commutative, and does not necessarily have the multiplicative identity.

All of the left ideals in ring $R$ are trivial, and there are two elements whose product is nonzero. Prove that $R$ forms a skew field.

I understand that if I show that $1\in R$, then this problem reduce to a well-known exercise. So here is what I have tried:

Suppose $ab\ne 0$. From the fact that the left ideal of the left zero divisors of $b$ is either $(0)$ or $R$, and it cannot be $R$ since $a$ is not in it, we learn that $b$ doesn’t have nontrivial left zero divisors.

Now consider the left ideal generated by $b^2$ (denoted $(b^2)_l$): this is exactly $R$ itself. So $b\in (b^2)_l$, and we have $b=rb^2+nb^2$ for some $r\in R$ and some integer $n$ (here $nb^2$ means the sum of $n$ $b^2$s). Let $e=rb+nb$. By definition, $eb=b$. For all $u\in R$, $ueb=ub$, so $(ue-u)b=0$. But $b$ is nonzero and does not have nontrivial left zero divisors. This gives $ue=u$. By now, we have shown that there is a right identity in $R$.

After that, I would like to show that there is a left identity or elements in $R$ have right inverses, but I stuck at this point. Can anyone point out some useful ideas?

Thanks in advance.


Let $\ E_0 = \left\{x\in R\left\vert\, ex=0\right.\right\} $. Then $\ E_0\ $ is a left ideal of $\ R\ $, because if $\ r\in R\ $ and $\ x\in E_0\ $ then $\ rex = 0 = rx\ $, so $\ erx = e0=0\ $, and therefore $\ rx\in E_0\ $. But $\ e\not\in\ E_0\ $, so $\ E_0\ne R\ $ and must be $\ \left(0\right)\ $. But if $\ y\in R\ $ then $\ e\left(y-ey\right)=0\ $, so $\ y-ey\in E_0\ $, and therefore $\ y-ey=0\ $. Since this is true for any $\ y\in R\ $, $\ e\ $ is a left identity.


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