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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).

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    $\begingroup$ 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). $\endgroup$
    – coiso
    Commented Dec 1, 2023 at 0:27
  • $\begingroup$ @coiso Please post answers as answers, not inside of comments. $\endgroup$ Commented Dec 1, 2023 at 0:32
  • $\begingroup$ @Ignacio The trivial ring is not an integral domain - you forgot that in the definition. $\endgroup$ Commented Dec 1, 2023 at 0:32
  • $\begingroup$ In step 5, I believe you mean $i-j-1\geq 0$ $\endgroup$
    – chi
    Commented Dec 1, 2023 at 10:41

1 Answer 1

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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 (this Wikipedia article includes the usual proof).

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  • $\begingroup$ The term "abelian" is specific to groups, not ring multiplication. --- This is not completely correct. The term "abelian" is used for Lie algebras, von Neumann algebras and other types of operator algebras. Even more confusingly, some authors have defined "abelian rings" by the property that idempotent elements are central. $\endgroup$ Commented Dec 2, 2023 at 3:57

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