# every finite integral domain is a field

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## Solution 1

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 $f_{\alpha}(\beta) = 1 \implies \alpha \beta = 1$.

## Solution 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).

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### Simon

Updated on July 20, 2021

"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)."