# Showing that a divisor of zero in a commutative ring with unity can have no multiplicative inverse" [duplicate]

A divisor of zero in a commutative ring with unity can have no multiplicative inverse.

I don't understand why this statement is true.

So for $a,b$ in the ring, $ab=0$ by zero divisor.

How can we guarantee that there does not exist $c$ such that $ac=1$?

## 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(); } ); }); }); Apr 8 '15 at 9:57

Hint Suppose $a$ has a multiplicative inverse, and multiply both sides of $ab = 0$ by $a^{-1}$.
• Yes, by hypothesis $b \neq 0$ but the assumption that $a$ has a multiplicative inverse lead to the conclusion that $b = 0$, which is indeed a contradiction. – Travis Apr 8 '15 at 3:57