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, \[ a \cdot b = 0 \implies a = 0 \text{ or } b = 0. \]
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 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.
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 \( a = 0, \) then indeed \( ab = 0 \) by Proposition 1.
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 and 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:
- Every field is an integral domain.
- Every finite integral domain is a field.
- Some infinite integral domains are not fields. A convenient example is the set of integers \( \mathbb{Z}. \)
- Some infinite integral domains are fields. Every infinite field, such as \( \mathbb{Q}, \) \( \mathbb{R}, \) or \( \mathbb{C}, \) is an example.
It is worth reiterating that the final 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.