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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.

Jyrki Lahtonen
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Moe
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  • Well, you have asked it at least... Now, what do you want to do? What is the definition of a integral domain you want to use, which definition of a field comes to mind?

    Oh, and btw, partially dublicate of https://math.stackexchange.com/questions/2266874/proving-that-elements-in-a-finite-ring-are-zero-divisors-or-units-and-deducing-t

     – Dirk May 05 '17 at 10:01
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    An integral domain has no zero divisors. – florence May 05 '17 at 10:03
  • 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 05 '17 at 10:06

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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.

Przemek
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