# Every finite integral domain is a field (why is it commutative?)

Okay, a finite integral domain is a finite ring $$D$$ such that for every $$a, b \in D, ab = 0$$ iff $$a = 0$$ or $$b = 0$$. We want that $$D$$ is a field, meaning it has a unit element ($$1$$), every element is invertible and it is abelian.

First, I prove that every element is invertible:

1. Assume $$|D| = n$$
2. Take $$a \in D, a \neq 0$$, then multiply it by itself $$n + 1$$ times. By the pidgeonhole principle, there exists at least one element in $$\{a, a^2, a^3, ..., a^{n+1}\}$$ such that $$a^i = a^j$$ with $$i > j$$
3. Now, take $$a^i - a^j = 0$$ (we can do it because $$D$$ is a ring) and get $$a^j(a^{i-j}-1)=0$$
4. Since $$D$$ is an integral domain, either $$a^j=0$$ or $$(a^{i-j}-1)=0$$, but $$a^j \neq 0$$ because $$D$$ is an integral domain, therefore $$(a^{i-j}-1)=0$$, so $$a^{i-j}=1$$.
5. $$a^{i-j}=1$$ means $$a·a^{i-j-1}=1$$ (Notice $$i-j-1\geq0$$ because $$i-j>0$$) so $$a$$ is invertible.

Notice from step 4 we also know $$1 \in D$$ as $$D$$ is a ring and, therefore, the product operation of two elements of $$D$$ stays in $$D$$.

Now, I am stuck at how do you prove it is abelian by the product? Everywhere I look people just assume it is abelian, but on my book it says it needn't be (for example, the quaternions).

• An integral domain is commutative by definition, or at least the standard definition. (The term "abelian" is specific to groups, not ring multiplication.) You're thinking of what's generally called a finite division ring. Then the result you speak of is Wedderburn's Little Theorem (the proof is on Wikipedia - link). Dec 1, 2023 at 0:27
• In step 5, I believe you mean $i-j-1\geq 0$