-1
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by rschwieb abstract-algebra May 5 '17 at 10:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 math.stackexchange.com/questions/2266874/… $\endgroup$ – Dirk May 5 '17 at 10:01
  • $\begingroup$ An integral domain has no zero divisors. $\endgroup$ – florence May 5 '17 at 10:03
  • $\begingroup$ 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 $a^{-1}$. $\endgroup$ – Moe May 5 '17 at 10:06
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ what does R* mean? $\endgroup$ – Moe May 5 '17 at 10:11
  • $\begingroup$ $R^*=R-\{0\}$. Usual definition of integral domain assume, that $R$ is nonzero, thus $R^*$ is nonempty. $\endgroup$ – Przemek May 5 '17 at 10:19
  • $\begingroup$ Thank you! Your explanation was really helpful for understanding it! $\endgroup$ – Moe May 5 '17 at 10:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.