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

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?

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.