2
$\begingroup$

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?.

$\endgroup$
1
$\begingroup$

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 $ax=b$.

$\endgroup$
  • $\begingroup$ 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

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.