Let R be a ring such that the only right ideals of R are (0) and R. Prove that R is either a division ring or that R is a ring with a prime number of elements in which $\displaystyle ab=0$ for every $\displaystyle a,b \in R$
one of the most notorious problems in Herstein's book as far as i remember! although it's an easy exercise for me now, it gave me a very hard time when i was an innocent undergrad student!
case 1 $\displaystyle R^2 =(0)$: so $\displaystyle ab=0,$ for all $\displaystyle a,b \in R$ and therefore any subgroup of $\displaystyle (R,+)$ is a (right) ideal of $\displaystyle R.$ but $\displaystyle R$ has no non-trivial right ideal. thus $\displaystyle (R,+)$ has no non-trivial subgroup and hence it
has to be of prime order.
case 2 $\displaystyle R^2 \neq (0)$: we need a couple of observations first:
(1) let $\displaystyle I=\{x \in R: \ xR=(0) \}.$ clearly $\displaystyle I$ is a (right) ideal of $\displaystyle R.$ hence either $\displaystyle I=R$ or $\displaystyle I=(0).$ if $\displaystyle I=R,$ then $\displaystyle R^2=(0),$ which is a contradiction. thus $\displaystyle I=(0).$ so $\displaystyle xR \neq (0),$ for all $\displaystyle 0 \neq x \in R.$
therefore for any $\displaystyle 0 \neq x \in R,$ we have $\displaystyle xR=R.$
(2) now suppose $\displaystyle 0 \neq x \in R$ and let $\displaystyle J=\{y \in R: \ \ xy=0 \}.$ clearly $\displaystyle J$ is a right ideal of $\displaystyle R.$ if $\displaystyle J=R,$ then $\displaystyle R=xR=(0),$ which is nonsense. thus $\displaystyle J=(0),$ i.e. $\displaystyle xy=0$ implies that $\displaystyle x=0$ or $\displaystyle y=0.$
now let $\displaystyle 0 \neq a \in R.$ by (1): $\displaystyle aR=R.$ so there exists $\displaystyle r \in R$ such that $\displaystyle ar = a.$ thus $\displaystyle a(ra-a)=0,$ which gives us $\displaystyle ra=a$ by (2). the claim is that $\displaystyle r=1_R$: let $\displaystyle K=\{x \in R: \ rx=x \}.$ obviously $\displaystyle K$
is a right ideal of $\displaystyle R$ and $\displaystyle K \neq (0),$ because $\displaystyle a \in K.$ thus $\displaystyle K=R,$ i.e. $\displaystyle \forall x \in R: \ rx=x.$ if $\displaystyle x=0,$ then obviously $\displaystyle xr=rx=x=0.$ if $\displaystyle x \neq 0,$ then since $\displaystyle (xr - x)x=x(rx)-x^2=x^2-x^2=0,$ we
will get $\displaystyle xr=x$ by (2). therefore $\displaystyle r=1_R.$ now if $\displaystyle 0 \neq x \in R,$ then by (1): $\displaystyle xR=R$ and so $\displaystyle xy=1,$ for some $\displaystyle y \in R.$ similarly $\displaystyle yz=1,$ for some $\displaystyle z \in R$ and thus $\displaystyle z=xyz=x(yz)=x.$ so $\displaystyle xy=yx=1.$
this proves that $\displaystyle R$ is a division ring.
Hi,
thanks man! herstein problems are the best! Ok can u help me with ur email id so that any doubts i can directly email u in(Algebra)...My email is chandru.mcc@gmail.com and i wud like to have ur help
i'm afraid that would be impossible! i also don't recommand private messeging me because i have a bad habit of replying those messeges quite late and sometimes never!
the best thing to do is to ask your doubts and questions on the forum to increase the chance of getting help and also give other members an opportunity to help or learn.
that's the main idea behind math forums like MHF, isn't it?