0

I am trying to understand a proof that every finite integral domain is a field, and in part it states:

"Consider $a, a^2, a^3,\dots$. Since there are only finitely many elements we must have $a^m = a^n$ for some $m < n$ (say)."

Can someone let me know what that means please?

J. W. Tanner
  • 60,406
Simon
  • 73
  • You don't know what $a^n$ means? It means $\underbrace{a\cdot a\cdots a}_n$. – anon Aug 21 '14 at 07:42
  • It means the sequence $a,a^2,a^3,\dots$, as elements of your ring, cannot go on unrepeated forever since the domain is finite. – Troy Woo Aug 21 '14 at 07:46
  • Thanks for your responses. I still don't see why a^m = a^n for some m < n though? – Simon Aug 21 '14 at 08:08
  • If the set$ {a^k} $contains infinitely many elements then it cannot be a subset of the ring ( which has only finitely many elements). This means some power of $a$ must equal another power of $a$. – Anurag A Aug 21 '14 at 08:11
  • Related: http://math.stackexchange.com/questions/62548 – Watson Jan 19 '17 at 13:04

2 Answers2

4

Let D be an integral domain. Then if $a$ is a non-zero element in D, then $a^{2}$ is also an element of D and so is $a^{3}$ and so are all the powers of $a$. If the powers are distinct, then you will have an infinite number of elements in D, which is not possible because D is finite and hence the powers of $a$ cannot all be distinct, which means that for some $k \neq m$, $a^{k} =a^{m}$.

Here, is another way of proving that D must be a field:

Choose $\alpha \neq 0 \in D$. What we must show is that $\alpha$ is invertible, that is, we must find a $\beta \in D$ such that $\alpha \beta = 1$.

Consider the map: $f_{\alpha}: D \rightarrow D: f_{\alpha}(x) = \alpha x$. $f_{\alpha}(x) = f_{\alpha}(y) \iff 0 = \alpha x - \alpha y = \alpha (x-y)$ and since there are no zero divisors in an integral domain and $\alpha \neq 0$, $x - y = 0$ or $x = y$. So $f_{\alpha}$ is one-to-one. Since $D$ is finite, a one-to-one map must be onto. Since $f_{\alpha}$ is onto, there exists a $\beta \in D$ so that $f_{\alpha}(\beta) = 1 \implies \alpha \beta = 1$.

Daniel Akech Thiong
  • 4,061
2

Elements $a,a^2,a^3,a^4…. $ are all in the integral domain since it is closed under multiplication.
But the Integral domain is finite.
Hence all powers of element a start repeating after a certain stage.
Thus we've have $a^m=a^n$ for some m < n (say).

Nerdy
  • 744
  • 8
  • 16