# Trivial (right or left) ideals of R imply that R is a division ring [duplicate]

It is clear that a division ring is simple, having only trivial right ideals. It is also clear that if a ring, R, is unital and finite, having only trivial ideals, then the ring is a division ring.

However, is it true that a unital ring with only trivial right ideals is a division ring, with its order being irrelevant?

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 17 '15 at 3:57

• Two-sided or one-sided? – whacka Feb 16 '15 at 23:38
• one-sided ideals. – Anerky Feb 16 '15 at 23:41
• – rschwieb Feb 17 '15 at 4:01

Suppose $aR=R$ for all nonzero $a\in R$, and $R$ is unital. Fix $a\in R$; we seek an inverse. By hypothesis we know $1\in aR$ so $1=ab$ for some $b\in R$. And similarly $bR=R$ so $1=bc$ for some element $c\in R$. But then we have $a=abc=c$, so $1=ab$ and $1=ba$ and so arbitrary $a\in R$ have two-sided inverses, and hence $R$ is a division ring.
A symmetrical argument holds if we suppose $R=Ra$ for all nonzero $a\in R$.
A unital ring $R$ may fail to have two-sided ideals besides $0$ and $R$ and still fail to be a division ring due to the presence of zero divisors. The textbook examples are the matrix rings $M_n(k)$ (for any integer $n>1$) over any scalar field $k$ (see this question). Indeed if $k$ is finite then so is $M_n(k)$!