# A commutative ring with identity is a field if and only it has no nonzero proper ideals [duplicate]

Obviously, if $F$ is a field, and $I$ is it's nonzero ideal, then it contains an invertible element of $F$(any nonzero element of $F$). Denote this element as $a$. Since $I$ is ideal, $aa^{-1} = 1 \in I$. Hence, $I = F$.

But I'm not sure how to prove that any commutative ring with identity without nonzero proper ideals is a field.

## marked as duplicate by rschwieb ring-theory 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 15 '16 at 13:42

• Please try the search feature first, next time. – rschwieb Feb 15 '16 at 13:43

If you have a commutative ring $R$ with identity, the only missing property to be a field is, that any element is invertible.
So let's assume $R$ is no field. You have some non-invertible element $r \neq 0$ and thus $rR$ is a proper ideal, since $1 \notin rR$.
If $\{0\}$ is the only proper ideal, is a maximal ideal and $R/\{0\}\approx R$ is a field.