Integral domain explained

In mathematics, an integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero. Integral domains are generalizations of the ring of integers and provide a natural setting for studying divisibility. In an integral domain, every nonzero element a has the cancellation property, that is, if, an equality implies .

"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:

Definition

An integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero. Equivalently:

Examples

\Z

of all integers.

\R

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

\Z

provides an example of a non-Artinian infinite integral domain that is not a field, possessing infinite descending sequences of ideals such as:

\Z\supset2\Z\supset … \supset2n\Z\supset2n+1\Z\supset …

\Z[x]

of all polynomials in one variable with integer coefficients is an integral domain; so is the ring

\Complex[x1,\ldots,xn]

of all polynomials in n-variables with complex coefficients.

\Complex[x,y]/(y2-x(x-1)(x-2))

corresponding to a plane elliptic curve is an integral domain. Integrality can be checked by showing

y2-x(x-1)(x-2)

is an irreducible polynomial.

\Z[x]/(x2-n)\cong\Z[\sqrt{n}]

is an integral domain for any non-square integer

n

. If

n>0

, then this ring is always a subring of

\R

, otherwise, it is a subring of

\Complex.

\Zp

is an integral domain.

U

is a connected open subset of the complex plane

\Complex

, then the ring

l{H}(U)

consisting of all holomorphic functions is an integral domain. The same is true for rings of analytic functions on connected open subsets of analytic manifolds.

Non-examples

The following rings are not integral domains.

0=1

).

\Z/m\Z

when m is a composite number. Indeed, choose a proper factorization

m=xy

(meaning that

x

and

y

are not equal to

1

or

m

). Then

x\not\equiv0\bmod{m}

and

y\not\equiv0\bmod{m}

, but

xy\equiv0\bmod{m}

.

R x S

, one has

(1,0) ⋅ (0,1)=(0,0)

.

\Z[x]/(x2-n2)

for any

n\inZ

. The images of

x+n

and

x-n

are nonzero, while their product is 0 in this ring.

M

and

N

are matrices such that the image of

N

is contained in the kernel of

M

, then

MN=0

. For example, this happens for

M=N=(\begin{smallmatrix}0&1\ 0&0\end{smallmatrix})

.

k[x1,\ldots,xn]/(fg)

for any field

k

and any non-constant polynomials

f,g\ink[x1,\ldots,xn]

. The images of and in this quotient ring are nonzero elements whose product is 0. This argument shows, equivalently, that

(fg)

is not a prime ideal. The geometric interpretation of this result is that the zeros of 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 is a power of an irreducible polynomial, which defines the same algebraic set.

f(x)=\begin{cases}1-2x&x\in\left[0,\tfrac{1}{2}\right]\ 0&x\in\left[\tfrac{1}{2},1\right]\end{cases}    g(x)=\begin{cases}0&x\in\left[0,\tfrac{1}{2}\right]\ 2x-1&x\in\left[\tfrac{1}{2},1\right]\end{cases}

Neither

f

nor

g

is everywhere zero, but

fg

is.

\Complex ⊗ \R\Complex

. This ring has two non-trivial idempotents,

e1=\tfrac{1}{2}(1 ⊗ 1)-\tfrac{1}{2}(i ⊗ i)

and

e2=\tfrac{1}{2}(1 ⊗ 1)+\tfrac{1}{2}(i ⊗ i)

. They are orthogonal, meaning that

e1e2=0

, and hence

\Complex ⊗ \R\Complex

is not a domain. In fact, there is an isomorphism

\Complex x \Complex\to\Complex ⊗ \R\Complex

defined by

(z,w)\mapstoz ⋅ e1+w ⋅ e2

. Its inverse is defined by

z ⊗ w\mapsto(zw,z\overline{w})

. This example shows that a fiber product of irreducible affine schemes need not be irreducible.

Divisibility, prime elements, and irreducible elements

See also: Divisibility (ring theory). In this section, R is an integral domain.

Given elements a and b of R, one says that a divides b, 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 .

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 for some unit u.

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

\Z,

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

\Z\left[\sqrt{-5}\right]

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

a2+5b2=3

has no integer solutions), but not prime (since 3 divides

\left(2+\sqrt{-5}\right)\left(2-\sqrt{-5}\right)

without dividing either factor). In a unique factorization domain (or more generally, a GCD domain), an irreducible element is a prime element.

While unique factorization does not hold in

\Z\left[\sqrt{-5}\right]

, there is unique factorization of ideals. See Lasker–Noether theorem.

Properties

Field of fractions

See main article: Field of fractions.

The field of fractions K of an integral domain R is the set of fractions a/b with a and b in R and 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

\Z

is the field of rational numbers

\Q.

The field of fractions of a field is isomorphic to the field itself.

Algebraic geometry

Integral domains are characterized by the condition that they are reduced (that is implies) 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.

This translates, in algebraic geometry, into the fact that the coordinate ring of an affine algebraic set is an integral domain if and only if the algebraic set is an algebraic variety.

More generally, a commutative ring is an integral domain if and only if its spectrum is an integral affine scheme.

Characteristic and homomorphisms

The characteristic of an integral domain is either 0 or a prime number.

If R is an integral domain of prime characteristic p, then the Frobenius endomorphism is injective.

See also

References

External links

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