From Finite Integral Domains to Finite Fields

By Susam Pal on 25 May 2025

In this article, we explore a few well-known results from abstract algebra pertaining to fields and integral domains. We ask ourselves whether every field is an integral domain, and whether every integral domain is a field. We begin with the definition of an integral domain, discuss a few established results, and then proceed to answer these questions. Familiarity with algebraic structures such as rings and fields is assumed.

Contents

Definition of Integral Domain

An integral domain is a commutative ring, with distinct additive and multiplicative identities, in which the product of any two non-zero elements is also non-zero.

Equivalently, an integral domain is a commutative ring, with distinct additive and multiplicative identities, such that if the product of two elements is zero, then one of the elements must be zero.

Using standard notation, we can write that a commutative ring \( R \) is an integral domain if \( 0 \ne 1 \) and for \( a, b \in R, \) \[ a \ne 0 \text{ and } b \ne 0 \implies a \cdot b \ne 0 \] or equivalently, \[ a \cdot b = 0 \implies a = 0 \text{ or } b = 0. \] There are many other alternative ways to define an integral domain which are all equivalent. In a ring \( R, \) a zero divisor is a non-zero element \( a \in R \) such that there exists a non-zero element \( b \in R \) with \( ab = 0. \) With this definition of a zero divisor, we can define an integral domain to be a unital commutative ring, with \( 0 \ne 1, \) that has no zero divisors.

Examples of Integral Domains

The ring of integers \( \mathbb{Z} \) is an integral domain since the product of two non-zero integers is non-zero. The field of rational numbers \( \mathbb{Q} \) is also an integral domain. The ring of polynomials in the indeterminate \( t \) with coefficients in an integral domain \( R, \) denoted \( R[t], \) is an integral domain as well.

The ring of integers modulo 5, denoted \( \mathbb{Z}_5 \) is an integral domain. However, the ring of integers modulo 6, denoted \( \mathbb{Z}_6, \) is not an integral domain since \( 2 \cdot 3 = 0 \) in \( \mathbb{Z}_6. \) In other words, \( \mathbb{Z}_6 \) has zero divisors, namely \( 2 \) and \( 3, \) so it is not an integral domain. In fact, the ring of integers modulo \( n, \) denoted \( \mathbb{Z}_n \) is an integral domain if and only if \( n \) is prime.

Known Results

For the sake of brevity, we assume the following known results.

Proposition 1. Let \( R \) be a ring. Then, for all \( a \in R, \) we have \[ a \cdot 0 = 0 \cdot a = 0. \]

Proposition 2. Let \( D \) be an integral domain. Then, for all \( a, b, c \in D \) such that \( a \ne 0, \) we have \[ a \cdot b = a \cdot c \implies b = c. \]

The second result is also known as the cancellation property of integral domains.

On Distinct Identities

We have been mentioning the distinctiveness of the additive and multiplicative identities as a property of an integral domain. Some texts express this more concisely by saying that an integral domain is a non-zero unital commutative ring without zero divisors, i.e., the zero ring \( \{ 0 \} \) is excluded from the definition.

This follows directly from Proposition 1. If \( 0 = 1 \in R , \) then for all \( r \in R, \) we get \[ r = r \cdot 1 = r \cdot 0 = 0 \] which means that every element of \( R \) is zero, i.e., \( R = \{ 0 \}. \) To summarise \[ 0 = 1 \in R \implies R = \{ 0 \} \] or equivalently, for a ring \( R \) with unity, \[ R \ne \{ 0 \} \implies 0 \ne 1. \] Further, if \( 0 \) and \( 1 \) are two distinct elements of \( R , \) then \( R \) has at least two elements, so for a ring \( R \) with unity, \[ 0 \ne 1 \implies R \ne \{ 0 \} \] Therefore, a ring with unity has distinct additive and multiplicative identities if and only if it is a non-zero ring. This is why an integral domain can also be defined as a non-zero unital commutative ring without zero divisors.

Every Field Is an Integral Domain

We now show that every field is indeed an integral domain. Let \( F \) be a field, and let \( a, b \in F \) such that \( ab = 0. \) There are two cases to consider: \( a = 0 \) and \( a \ne 0. \) If \( a = 0, \) we are done.

Now suppose \( a \ne 0. \) Then by the properties of fields, there exists a multiplicative inverse \( a^{-1} \in F \) such that \( a \cdot a^{-1} = 1. \) Then using the properties of fields, we get \[ b = b \cdot 1 = b \cdot (a \cdot a^{-1}) = (a \cdot b) \cdot a^{-1} = 0 \cdot a^{-1} = 0. \] The last equality follows from Proposition 1. We have shown that if \( ab = 0, \) then either \( a = 0 \) or \( b = 0. \) Therefore, if both \( a \ne 0 \) and \( b \ne 0, \) then it must be that \( ab \ne 0. \) Therefore \( F \) is an integral domain.

Infinite Integral Domains

Now we arrive at the next natural question. Is every integral domain a field?

The ring of integers \( \mathbb{Z} \) is an integral domain but it is not a field since \( 2 \in \mathbb{Z}, \) but \( 2^{-1} \notin \mathbb{Z}. \) Therefore, \( \mathbb{Z} \) is an example of an infinite integral domain that is not a field.

Next we ask ourselves: Is every infinite integral domain not a field? Not quite! Some infinite integral domains are, in fact, fields. This follows directly from the result in the previous section. Every field is an integral domain, and there are plenty of infinite fields, so they must all be integral domains too. Consider the field of rational numbers \( \mathbb{Q} \) or the field of complex numbers \( \mathbb{C}. \) Since these are fields, they are also integral domains. So, clearly, there are infinite integral domains that are also fields.

Every Finite Integral Domain Is a Field

We will now turn our attention to finite integral domains. Is every finite integral domain a field? Yes! This can be shown as follows.

Let \( D \) be a finite integral domain. Let \( a \in D \) with \( a \ne 0. \) Consider the set \[ A = \{ a, a^2, a^3, \dots \}. \] Since a ring is closed under multiplication, every element of \( A \) belongs to \( D, \) so \( A \subseteq D. \) Since \( D \) is finite, \( A \) is finite too. Therefore, by the pigeonhole principle, there exist integers \( m \gt n \ge 0 \) such that \[ a^m = a^n \] This equation can be rewritten as \[ a \cdot a^{m - n - 1} \cdot a^n = 1 \cdot a^n. \] Since \( a \) is a non-zero element of an integral domain, it follows that \( a^n \ne 0. \) Therefore we can use Proposition 2 (the cancellation property of integral domains) to get \[ a \cdot a^{m - n - 1} = 1. \] Since a ring is closed under multiplication and since \( m - n - 1 \ge 0, \) it follows that \( a^{m - n - 1} \in D. \) Thus every non-zero element \( a \in D \) has a multiplicative inverse in \( D . \) This establishes the multiplicative inverse property of a field.

Since an integral domain has distinct additive and multiplicative identities, it satisfies two additional field properties: the existence of an additive identity and a distinct multiplicative identity.

Finally, the remaining field properties are inherited from the ring structure, i.e., associativity and commutativity of addition and multiplication, the existence of additive inverses, and the distributivity of multiplication over addition all hold in \( D, \) since they hold in any ring. Thus, \( D \) satisfies all the field properties. Therefore \( D \) is a field.

Alternate Proof

The proof in the previous section presents what I initially came up with while working through these concepts and proving these results for myself. However, I later found that there is another proof that is quite popular in the literature. This alternate proof differs in one key aspect: it does not invoke the cancellation property of integral domains stated in Proposition 2. Let us examine this alternate proof.

As before, we consider the set \( A = \{ a, a^2, a^3, \dots \} \subseteq D, \) where \( a \in D \) and \( a \ne 0, \) and we obtain the equation \[ a^m = a^n \] for some integers \( m \gt n \ge 0. \) As before, we use the fact that \( a \) is an element of an integral domain to conclude that \( a^n \ne 0. \) Now adding the additive inverse of \( a^n \) to both sides we get \[ a^m - a^n = 0. \] Using the distributivity property of rings, we get \[ a^n (a^{m - n} - 1) = 0 \] Since a ring is closed under addition and multiplication, both \( a^n \) and \( a^{m - n} - 1 \) belong to \( D. \) As \( D \) is an integral domain and \( a^n \ne 0, \) we conclude that \( a^{m - n} - 1 = 0. \) Therefore \( a^{m - n} = 1. \) Since \( m - n \ge 1, \) we can write: \[ a \cdot a^{m - n - 1} = 1. \] Therefore every non-zero element \( a \in D \) has a multiplicative inverse in \( D. \) The remaining properties of a field are established in the same manner as in the previous section. Hence, if \( D \) is a finite integral domain, then it is also a field.

Conclusion

We now summarise all the results here before concluding the article:

It is worth reiterating here that the fourth result in the summary above follows from the fact that every field is an integral domain. These results reveal how structure and size interact in algebraic systems. It is interesting how simply being finite guarantees that an integral domain is a field.

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