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Algebraic structure with two binary operations
Commutative ring with no zero divisors other than zero
"Integral domain" is defined almost universally as above, but there is some variation. This article follows the convention that rings have a multiplicative identity, generally denoted 1, but some authors do not follow this, by not requiring integral domains to have a multiplicative identity. Noncommutative integral domains are sometimes admitted. This article, however, follows the much more usual convention of reserving the term "integral domain" for the commutative case and using "domain" for the general case including noncommutative rings.
Some sources, notably Lang, use the term entire ring for integral domain.
Some specific kinds of integral domains are given with the following chain of class inclusions:
An integral domain is basically defined as a nonzerocommutative ring in which the product of any two nonzero elements is nonzero. This definition may be reformulated in a number of equivalent definitions :
An integral domain is a nonzero commutative ring with no nonzero zero divisors.
An integral domain is a nonzero commutative ring for which every non-zero element is cancellable under multiplication.
An integral domain is a ring for which the set of nonzero elements is a commutative monoid under multiplication (because a monoid must be closed under multiplication).
An integral domain is a nonzero commutative ring in which for every nonzero element r, the function that maps each element x of the ring to the product xr is injective. Elements r with this property are called regular, so it is equivalent to require that every nonzero element of the ring be regular.
A fundamental property of integral domains is that every subring of a field is an integral domain, and that, conversely, given any integral domain, one may construct a field that contains it as a subring, the field of fractions. This characterization may be viewed as a further equivalent definition:
An integral domain is a ring that is (isomorphic to) a subring of a field.
The archetypical example is the ring of all integers.
Every field is an integral domain. For example, the field of all real numbers is an integral domain. Conversely, every Artinian integral domain is a field. In particular, all finite integral domains are finite fields (more generally, by Wedderburn's little theorem, finite domains are finite fields). The ring of integers provides an example of a non-Artinian infinite integral domain that is not a field, possessing infinite descending sequences of ideals such as:
Rings of polynomials are integral domains if the coefficients come from an integral domain. For instance, the ring of all polynomials in one variable with integer coefficients is an integral domain; so is the ring of all polynomials in n-variables with complex coefficients.
The previous example can be further exploited by taking quotients from prime ideals. For example, the ring corresponding to a plane elliptic curve is an integral domain. Integrality can be checked by showing is an irreducible polynomial.
The ring is an integral domain for any non-square integer . If , then this ring is always a subring of , otherwise, it is a subring of
A product of two nonzero commutative rings. In such a product , one has .
When is a square, the ring is not an integral domain. Write , and note that there is a factorization in . By the Chinese remainder theorem, there is an isomorphism
The ring of n × nmatrices over any nonzero ring when n >= 2. If and are matrices such that the image of is contained in the kernel of , then . For example, this happens for .
The quotient ring for any field and any non-constant polynomials . The images of f and g in this quotient ring are nonzero elements whose product is 0. This argument shows, equivalently, that is not a prime ideal. The geometric interpretation of this result is that the zeros of fg form an affine algebraic set that is not irreducible (that is, not an algebraic variety) in general. The only case where this algebraic set may be irreducible is when fg is a power of an irreducible polynomial, which defines the same algebraic set.
The tensor product. This ring has two non-trivial idempotents, and . They are orthogonal, meaning that , and hence is not a domain. In fact, there is an isomorphism defined by . Its inverse is defined by . This example shows that a fiber product of irreducible affine schemes need not be irreducible.
Divisibility, prime elements, and irreducible elements
In this section, R is an integral domain.
Given elements a and b of R, one says that adividesb, or that a is a divisor of b, or that b is a multiple of a, if there exists an element x in R such that ax = b.
The units of R are the elements that divide 1; these are precisely the invertible elements in R. Units divide all other elements.
If a divides b and b divides a, then a and b are associated elements or associates. Equivalently, a and b are associates if a = ub for some unitu.
An irreducible element is a nonzero non-unit that cannot be written as a product of two non-units.
A nonzero non-unit p is a prime element if, whenever p divides a product ab, then p divides a or p divides b. Equivalently, an element p is prime if and only if the principal ideal (p) is a nonzero prime ideal.
Both notions of irreducible elements and prime elements generalize the ordinary definition of prime numbers in the ring if one considers as prime the negative primes.
Every prime element is irreducible. The converse is not true in general: for example, in the quadratic integer ring the element 3 is irreducible (if it factored nontrivially, the factors would each have to have norm 3, but there are no norm 3 elements since has no integer solutions), but not prime (since 3 divides without dividing either factor). In a unique factorization domain (or more generally, a GCD domain), an irreducible element is a prime element.
Let R be an integral domain. Then the polynomial rings over R (in any number of indeterminates) are integral domains. This is in particular the case if R is a field.
The cancellation property holds in any integral domain: for any a, b, and c in an integral domain, if a ? 0 and ab = ac then b = c. Another way to state this is that the function x ? ax is injective for any nonzero a in the domain.
The cancellation property holds for ideals in any integral domain: if xI = xJ, then either x is zero or I = J.
An integral domain is equal to the intersection of its localizations at maximal ideals.
If are integral domains over an algebraically closed field k, then is an integral domain. This is a consequence of Hilbert's nullstellensatz,[note 1] and, in algebraic geometry, it implies the statement that the coordinate ring of the product of two affine algebraic varieties over an algebraically closed field is again an integral domain.
Field of fractions
The field of fractionsK of an integral domain R is the set of fractions a/b with a and b in R and b ? 0 modulo an appropriate equivalence relation, equipped with the usual addition and multiplication operations. It is "the smallest field containing R " in the sense that there is an injective ring homomorphism such that any injective ring homomorphism from R to a field factors through K. The field of fractions of the ring of integers is the field of rational numbers The field of fractions of a field is isomorphic to the field itself.
Integral domains are characterized by the condition that they are reduced (that is x2 = 0 implies x = 0) and irreducible (that is there is only one minimal prime ideal). The former condition ensures that the nilradical of the ring is zero, so that the intersection of all the ring's minimal primes is zero. The latter condition is that the ring have only one minimal prime. It follows that the unique minimal prime ideal of a reduced and irreducible ring is the zero ideal, so such rings are integral domains. The converse is clear: an integral domain has no nonzero nilpotent elements, and the zero ideal is the unique minimal prime ideal.
^Proof: First assume A is finitely generated as a k-algebra and pick a -basis of . Suppose (only finitely many are nonzero). For each maximal ideal of , consider the ring homomorphism . Then the image is and thus either or and, by linear independence, for all or for all . Since is arbitrary, we have the intersection of all maximal ideals where the last equality is by the Nullstellensatz. Since is a prime ideal, this implies either or is the zero ideal; i.e., either are all zero or are all zero. Finally, is an inductive limit of finitely generated k-algebras that are integral domains and thus, using the previous property, is an integral domain.