Proving Every Finite Integral Domain is a Field: A Comprehensive Guide

Integral domains are a fundamental concept in algebra, and understanding the properties of these domains is crucial for advanced mathematical studies. One of the most intriguing properties of integral domains is that they can be extended to fields under certain conditions, specifically, when they are finite. This article delves into the algebraic proof that every finite integral domain is a field, providing a comprehensive and detailed explanation.

What is an Integral Domain?

In modern terminology, an integral domain is a commutative ring with an identity element and no zero divisors. Formally, it is a set ( R ) with binary operations addition ( ) and multiplication (×) that satisfy the following properties:

Commutativity of addition: For all ( a, b in R ), ( a b b a ). Associativity of addition and multiplication: For all ( a, b, c in R ), ( (a b) c a (b c) ) and ( (ab)c a(bc) ). Existence of an additive identity: There exists an element 0 in ( R ) such that for all ( a in R ), ( a 0 a ). Existence of additive inverses: For each ( a in R ), there exists an element (-a in R ) such that ( a (-a) 0 ). No zero divisors: If ( a, b in R ) and ( ab 0 ), then either ( a 0 ) or ( b 0 ). Existence of a multiplicative identity: There exists an element 1 in ( R ) such that for all ( a in R ), ( a times 1 a ) and ( 1 times a a ).

Integral domains are important structures because they behave similarly to the integers in many ways, but they do not necessarily include a multiplicative identity or ensure commutativity.

Proving Every Finite Integral Domain is a Field

Given a finite integral domain ( R ), we need to prove that it is a field. A field is a commutative ring in which every nonzero element has a multiplicative inverse. The key is to show that for any nonzero element ( a in R ), there exists some ( b in R ) such that ( ab 1 ).

Algebraic Proof

Let ( 0 eq a in R ). Consider the map ( f_a: R to R ) defined by ( f_a(r) ar ). We need to show that this map is a bijection, which will imply that there exists an element ( b in R ) such that ( f_a(b) 1 ).

Injectivity: Suppose ( f_a(x) f_a(y) ). Then, ( ax ay ). Since ( R ) is an integral domain, we can cancel ( a ) (as ( a eq 0 ) and ( R ) has no zero divisors), and we get ( x y ). Hence, ( f_a ) is injective. Surjectivity: Since ( R ) is finite, and ( f_a ) is injective, it must also be surjective (because an injective function from a finite set to itself is also surjective).

Thus, there exists an element ( b in R ) such that ( f_a(b) 1 ). This means ( ab 1 ), and ( a ) has a multiplicative inverse. Since ( a ) was an arbitrary nonzero element of ( R ), every nonzero element of ( R ) has a multiplicative inverse, making ( R ) a field.

Generalizations and Relaxations

Now, let's consider some generalizations and relaxations:

Without Multiplicative Identity

Suppose we relax the requirement for ( R ) to have a multiplicative identity. In this case, we can still prove that a finite integral domain is a field.

Choose an element ( a in R ) and consider the set ( {ax mid x in R} ). Since ( R ) is finite, this set must contain a repeated element. Suppose ( ax ay ) for some ( x eq y ). Then, ( a(x - y) 0 ). Since ( R ) has no zero divisors and ( a eq 0 ), we have ( x - y 0 ), which is a contradiction. Therefore, the elements ( ax ) are distinct. Since there are as many distinct products ( ax ) as there are elements ( x in R ), and ( R ) is finite, there must be some element ( b in R ) such that ( ab 1 ).

To summarize, even without a multiplicative identity, if ( R ) is finite and an integral domain, it is still a field.

Non-Commutative Integral Domains

Finally, let's consider non-commutative integral domains. For non-commutative rings, the situation is more complex, but Wedderburn's Theorem provides a powerful result.

Wedderburn's Theorem states that every finite simple ring is a matrix ring over a finite division ring. A field is a commutative division ring, so if we can show that a finite simple ring without a multiplicative identity is a matrix ring over a finite division ring and then apply the classification of simple finite rings, we can conclude that it is a field. This theorem effectively extends the result to non-commutative domains as well.

Conclusion

In summary, every finite integral domain is a field under the standard interpretation of integral domains. Even if the requirement for a multiplicative identity is relaxed or the domain is non-commutative, the result still holds under the right conditions. This property of integral domains is fascinating and highlights the deep connections between algebraic structures.

Keywords

finite integral domain, algebraic proof, Wedderburn's Theorem