# Show that every finite integral domain is a field [duplicate]

Show that every finite integral domain is a field.

I have shown that every nonzero element in a finite ring with an identity is either a zero divisor or a unit.

## 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(); } ); }); }); May 5 '17 at 10:53

• I was going to use the definition that a ring is a field if it is a commutative division ring. I will need to show commutativity. I will also need to show that for every nonzero a in R there is a $a^{-1}$ such that $a * a^{-1} = a^{-1} * a = 1$. This confuses me because I just showed that every element should be a zero divisor or a unit, but now I want there to be an $a^{-1}$. – Moe May 5 '17 at 10:06
Take $x\in R^*$. For any $k\in\mathbb{Z}$ $x^k\neq0$, because $R$ is integral domain. But $|R|=n$, $|R^*|=n-1$, so $|\{x^1,..,x^n\}|<n$. There exists $a,b\in \{1,,n\},\ a<b$ such that $x^a=x^b$, thus $x^{b-a}=1$ and $x$ is invertible because $x^{-1}=x^{b-a-1}$. If every nonzero element is invertible, then $R$ is a field.
• $R^*=R-\{0\}$. Usual definition of integral domain assume, that $R$ is nonzero, thus $R^*$ is nonempty. – Przemek May 5 '17 at 10:19