Show that a ring with only trivial right ideals is either a division ring or $|R|=p$ and $R^2=\{0\}$. I have trouble with the following problem: Let $R$ be a ring (it doesn't necessarily have multiplicative identity). If the only left Ideals of $R$ are $(0)$ and itself then $R$ is a division ring or $|R|=p$ for some prime $p$ and $ab=0$ for all $a,b\in R$. I tried to prove that $R$ is a domain, because if that's true then I have an easy way of proving the exercise, but I don't know if it's true that $R$ is a domain
 A: As per this solution, it is a division ring if $R^2\neq\{0\}$.
If $ab=0$ for every $a,b\in R$, then it is a commutative ring, and the ideals are exactly the abelian subgroups of the underlying group.
Then ask yourself what simple abelian groups look like.
A: Let's consider these two cases

*

*$R$ has multiplicative identity $\implies$ $R$ is division Ring (of course vice versa)

*There is no multiplicative identity,

Let $a \in R, a\neq 0$.  $aR$ is a right ideal, hence  $aR=R$ or $aR=(0)$
Case 2.1:
If  $aR=0$ and $a\neq0$, then  $Y=\{x| xR=(0)\}$ is an ideal (hence right ideal).
Since, $ a\in Y \implies Y=R$,
Thus, $bR=(0), \: \:\forall b \in R \implies R^2 = (0)$
Consider for any $a \in R, a\neq 0$,
$$H = \{e^{(+)}\} \cup \{ ka \; | \; k \in \mathbb{Z} - \{0\} \} ,$$
where $e^{(+)}$ is additive identity.
$H$ is a subgroup under addition and it is also an ideal. Since $H\neq (0)$, we have $H=R$
Now for any $a,b \in R, a\neq0, b\neq0 $, we have
$$b=k_1a \implies b = k_1 k_2b \implies nb = 0, \: n>0. $$
Then you can show that, $pa=0, \: \forall a \in R$ for some prime $p$. Thus $|R| = p$
Case 2.2:
If  $aR=R$, you can show that it contains multiplicative identity, hence contradiction.
