In this, page 48, Exercies in chapter 1, there is a following exercise.
Exercise 2. Let $R$ be a ring (with $1$) such that the only left ideals of $R$ are $0$ and $R.$ Show that $R$ must be a division ring; that is, if $R$ is simple as a left $R$-module, then $R$ is a division ring. If the hypothesis that $R$ has an identity is dropped, the result no longer holds. Give an example to show this. In fact, the type of example you give is unique.
I have written a little about it. Regard a ring $R$ as an $R$-module. If $R$ is simple, we will show that $R$ is a divison ring. Indeed, if $x\in R,$ then $R=Rx,$ and so there is $y\in R$ such that $yx=1.$ Since $y$ is an element of the simple $R$-module $R,$ we have $R=Ry.$ This implies $1=zy.$ Moreover, $z=z1=z(yx)=(zy)x=1x=x.$ Therefore $R$ is a division ring. Otherwise, if $R$ is a division ring, then we will show that $R$ is simple. Indeed, if $x\in R,$ then $Rx$ is a submodule of $R.$ It seems that $R=Rx.$ This implies $R$ is a division ring.
The rest of the problem leaves me wondering. In this book, the ring is defined by a ring, has certainly identity $1.$ But the author gives $1$ and finds an example about the division ring without $1.$ I think the definition of a division ring requires $1.$ Moreover, I can't find an example about it.
Other way, in exercises 4 (c) and 4 (d), I think that $\mathbf{Q}$ must be written $\mathbf{Q}[x],$ and all simple $R$-modules, where $R$ is a principal ideal domain, are isomorphic to $R/I$ where $I$ is an ideal generated by a irreducible element. I'm not sure about this.
Please take a look and help me.