Prove that if $ax=b$ has a solution, $R$ is a division ring

Question

Question:

Assume that $R$ is a ring which has more than $1$ members, such that for all $a,b\in R$, If $a\neq 0$, The equation $ax=b$ has a solution.

Prove that $R$ is a division ring.

I saw a huge(!!) proof here. I thought that it can be proved easier! But, I'm not sure if my way is correct...

Let $a$ be a non-zero member of $R$. $ax=a$ has a solution. Call that solution $1_R$.

Now, $ax=1_R$ should have a solution too! So, call that solution $a^{-1}$.

It seems so simple. Am i wrong?

Answer 1

You have defined $1_R$. But you haven't shown that $1_Ra=a$ nor that $b1_R=b$ or $1_Rb=b$ for any other element $b$.

Now if you really do have a unit element, you have $a^{-1}$ with $aa^{-1}=1_R$, but you still need $a^{-1}a=1_R$.