I have the following question I am trying to solve:
Let $R$ be a ring such that the only right ideals of $R$ are $(0)$ and $R$. Prove that either $R$ is a division ring or that $R$ is a ring with a prime number of elements in which $ab=0$ for every $a,b \in R$.
My progress:
We assume that $R\neq \{0\}$.
We first prove the following result: If $R$ has an identity element, then $R$ is a division ring.
Let $a\neq 0\in R$ be arbitrary. We proceed by observing that $aR$ is a right ideal of $R$ and since $a=a.1\neq 0\in aR$, we must have $aR=R$. Since $1\in R$, this shows that $\exists{x},\space ax=1.$ Now we have, similarly, $xR=R$ and thus$\exists y, \space xy=1.$
We have then, $a=a.1=a(xy)=(ax)y=1.y=y.$ Thus $xa=ax=1.$ As $a\neq0$ was arbitrarily chosen, this shows that any non-zero element has an inverse.
So, if $\alpha$ and $\beta$ are two non-zero elements in $R$, we have then $\alpha \beta \neq 0.$ For if $\alpha \beta = 0,$ then $ 1= (\alpha^{-1}\alpha)(\beta^{-1}\beta)=\alpha^{-1}(\alpha \beta)\beta^{-1}=0$, which is not the case as $R \neq \{0\}$. Thus, $R$ is a division ring.
Then we prove the following result: If $\forall a,b \in R \space ab=0$, then $R$ is of prime order.
Let $R$ be of infinite order. Now we know that $(R,+)$ is an abelian group. If a non-zero element of $R$ is of finite order, i.e., $\exists a \in R, \space na=0$ for some $n \in \Bbb N$, we consider the group generated by $a$, $<a>$. Clearly $0 \in <a>,$ and $<a> \neq \{0\}, R.$ Now, if $x \in <a>, r\in R$ are arbitrary, then $xr=0 \in <a>$. Thus $<a>$ is a right ideal such that $<a> \neq \{0\}, R$, which cannot happen and hence no non-zero element of $R$ has a finite order. In that case, for some non-zero element $b \in R$, we consider $<2b>$. By a similar argument, we can show that $<2b>\neq \{0\}, R$ is a right ideal of $R$. Thus $R$ can not be of infinite order.
Now, let $|R|=m=rs$ where $1 < r \leq s < m.$ If some non-zero $a \in R$, we have $|a|<m$, then we consider $<a>$. Using arguments similar to those in the last paragraph, we can show that $<a>$ is a right ideal and $<a>\neq \{0\}, R$. Thus no non-zero element of $R$ has order less than $m$. But then, this cannot happen as if $a\neq 0 \in R$ has order $m$, then $ra\neq 0$ has order $s<m.$ Thus $|R|$ cannot be composite also. Thus the result holds.
So, if we can show that if $R$ is not a division ring, then $\forall a,b \in R, \space ab=0$, then we would be through.
We note that for any $a\in R$, $r(a):=\{x\in R | ax=0\}$ is a right ideal of $R$. So, if we can prove that $\forall a \in R, r(a)=R$, we would be done. Clearly, $r(0)=R$. So let $a\neq 0$ and $r(a)=\{0\}$. This means that $ax=0\implies x=0$. As $a\neq 0$, we have $a.a\neq 0 \in aR$. Thus, $aR=R$. Hence $\exists x_{0}, \space ax_{0}=a.$ We then have $$(ax_{0})a=a.a\implies a(x_{0}a)=a.a\implies a(x_{0}a-a)=0\implies x_{0}a=a.$$ Thus, $x_{0}a=ax_{0}=a.$ Now let $b\in R$ be arbitrary. Then, as $aR=R$, $\exists c, \space b=ac.$ We then observe that $x_{0}b=x_{0}(ac)=(x_{0}a)c=ac=b.$
Now, if we can prove that $bx_{0}=b$ also, then we can see that $x_{0}$ would then act as an identity element of $R$, which would then mean that $R$ is a division ring due to our first result, contrary to what we assumed. Thus, then we can conclude that $r(a)=\{0\}$ would be impossible from where we can tell that $r(a)=R \space \forall a \in R$, which would then lead us to our result.
This is where I got stuck. I can't prove $bx_{0}=b$. So, any hints on how to proceed from here would be welcome.