Theorem: a finite integral domain is a field
proof:
Let D be a finite integral domain with unity 1. Let a be any non-zero element of D.
If a=1, a is its own inverse and the proof concludes.
Suppose, then, a>1.
Since D is finite, the order of D is n. Then, $D =\left \{ e=0, a^{1},a^{2},\cdot \cdot \cdot a^{n-1} \right \}$.
Then, $\exists i,j \in \mathbb{Z}^{+}, i>j: a^{i}=a^{j}$
This implies $a^{i-j}=1$.
Now, I cannot understand the connection made from here(in bold):
Since $a\neq 1, i-j>1$.
What does the fact that $a\neq 1$ has anything to do with i-j>1?
And we have shown that $a^{i-j-1}$ is the inverse of a.
Explanation is appreciated. Thanks in advance.
Best Answer
For any number other than $1$, the difference $(i - j)$ will be greater than 1. We can prove this by showing that:
Let $j - i = 1$. Hence, $$ a^{(i - j)} = 1 \ \ \text{by assumption}\\ a^1 = 1 \\ a = 1 $$
So, If $a \neq 1$, $i - j > 1$ (since $i > j$ and $i - j \neq 1$)