A nonzero unital ring $D$ in which every nonzero element is invertible is called a division ring.

My question :

Are the two following equations equals?

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

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

In order for $D$ to be division ring, should we have a two-sided idea? Is one-sided ideal enough? I mean, from Equation 2 it can be concluded that $D$ is a divisible ring?.


Firstly, no, because "1" can be vacuously satisfied by $\{0\}$ and that is precluded in "2".

We could try assuming $D^2\neq \{0\}$ and proving if the first condition is equivalent to $D$ having no right ideals other than $\{0\}$ and $D$. Then the answer is yes.

In that case, $1\implies 2$ is very easy. Suppose you had a proper nonzero right ideal. Pick a nonzero element $a$ in it, and an element $b$ outside of it. There is an $x$ such that $ax=b$: can you see the contradiction?

For $2\implies 1$, suppose $a$ is any nonzero element of $D$. By hypothesis $aR\neq\{0\}$ is a right ideal, so it can only be $R$. That means left multiplication by $a$ is onto $R$. Therefore for any $b\in R$, you can find $x$ such that $ax=b$.

  • $\begingroup$ Do you mean from Equation 2 it can not be concluded that $D$ is a divisible ring?. $\endgroup$ – Jak Nov 3 '17 at 6:02
  • $\begingroup$ @Jak I mean that as written, the zero ring satisfies 1 but not 2. $\endgroup$ – rschwieb Nov 3 '17 at 10:24

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