1. ## Ring

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 $ab=0$ for every $a,b \in R$

2. Originally Posted by Chandru1

Let R be a non-zero 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 $ab=0$

for every $a,b \in R$
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 $R^2 =(0)$: so $ab=0,$ for all $a,b \in R$ and therefore any subgroup of $(R,+)$ is a (right) ideal of $R.$ but $R$ has no non-trivial right ideal. thus $(R,+)$ has no non-trivial subgroup and hence it

has to be of prime order.

case 2 $R^2 \neq (0)$: we need a couple of observations first:

(1) let $I=\{x \in R: \ xR=(0) \}.$ clearly $I$ is a (right) ideal of $R.$ hence either $I=R$ or $I=(0).$ if $I=R,$ then $R^2=(0),$ which is a contradiction. thus $I=(0).$ so $xR \neq (0),$ for all $0 \neq x \in R.$

therefore for any $0 \neq x \in R,$ we have $xR=R.$

(2) now suppose $0 \neq x \in R$ and let $J=\{y \in R: \ \ xy=0 \}.$ clearly $J$ is a right ideal of $R.$ if $J=R,$ then $R=xR=(0),$ which is nonsense. thus $J=(0),$ i.e. $xy=0$ implies that $x=0$ or $y=0.$

now let $0 \neq a \in R.$ by (1): $aR=R.$ so there exists $r \in R$ such that $ar = a.$ thus $a(ra-a)=0,$ which gives us $ra=a$ by (2). the claim is that $r=1_R$: let $K=\{x \in R: \ rx=x \}.$ obviously $K$

is a right ideal of $R$ and $K \neq (0),$ because $a \in K.$ thus $K=R,$ i.e. $\forall x \in R: \ rx=x.$ if $x=0,$ then obviously $xr=rx=x=0.$ if $x \neq 0,$ then since $(xr - x)x=x(rx)-x^2=x^2-x^2=0,$ we

will get $xr=x$ by (2). therefore $r=1_R.$ now if $0 \neq x \in R,$ then by (1): $xR=R$ and so $xy=1,$ for some $y \in R.$ similarly $yz=1,$ for some $z \in R$ and thus $z=xyz=x(yz)=x.$ so $xy=yx=1.$

this proves that $R$ is a division ring.

3. ## thanks

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

4. Originally Posted by Chandru1

Ok can u help me with ur email id so that any doubts i can directly email u in(Algebra)
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?