A ring R is called an integral domain if whenever a,b in R satisfy ab=0 we have either a=0 or b=0.

Examples

• Any field is an integral domain. So this includes Q,R, C, and Zp, the rational numbers, the real numbers, the complex numbers, and the integers modulo a prime p.
• Any subring of a field is an integral domain. So this includes the ring of integers Z and the Gaussian integers Z[i].
• The ring k[x] of polynomials over another integral domain k form another integral domain (proved by looking at leading terms). So this gives lots more examples, like Z[x],C[x] or Q[x,y]
• The ring Zm of integers modulo m for m> 3 composite is not an integral domain.
• The ring of 2x2 matrices with complex number coefficients is not an integral domain.
Books give different definitions for an integral domain.  Namely:

A commutative ring D that has an additional axiom, the cancellation law, which is:
for any a, b, c in a commutative ring D, ab = ac and a != 0 implies b = c.    (1)
And as Noether points out:
A commutative ring D that has the property: for any a, b in D, ab = 0 implies a = 0 or b = 0.    (2)
The definitions are equivalent, and it is easy to show.  (Upon request, I'll post the proof that says they are equivalent definitions.)  The first definition emphasizes the multiplicative cancellation law, while the second emphasizes a domain's usefulness in finding roots.

The definitions given above are not wrong per se, however the books used in my undergraduate studies used a slightly more restrictive version. The Characterization of Integral Domains depends on one important feature that is missing in the definitions given above. ariels notes that it is (fairly) common to assume that a ring has an identity and that a ring without an identity will be called a rng.

#### Definition

A commutative ring R with identity is called an integral domain if for all a and b ∈ R, ab=0R iff a=0R or b=0R.

#### Theorem: characterization

Let R be an integral domain (but not a field), let n be an integer greater than 1, and let K be a subset of R defined by K={n*r|r ∈ R}. Then
(1) n is relatively prime to the characteristic of R iff K is an ideal in R;
(2) if u ∈ R is a unit and u ∈ K, then K=R;
(3) if the characteristic of R is finite and relatively prime to n, then K=R.

#### Proof

Note that when I use "characteristic" I am using my definition as stated in that node. Also note that characteristic infinity is assumed to be relatively prime to all positive integers.

(1a) We try the contrapositive; assume that n is not relatively prime to the characteristic of R. Then the char of R must be finite and we will call it c. c must be a prime number, and since c and n are not relatively prime, c divides n. This means that there exists an integer x such that c*x=n. Let r be an element of R. Then n*r=c*x*r=x*(c*r)=x*0R=0R. So K={0R}, which is not a ring (and thus not an ideal). By contrapositive, K is an ideal implies that n is relatively prime to the char of R.

(1b) Now we use the direct method. Let n be relatively prime to the char of R. Then n*r=0R iff r=0R, so let a, b ∈ K and r ∈ R. There exist s, t ∈ R such that a=n*s and b=n*t, so a*b=n*s*n*t, which is in K, and a-b=n*a-n*b=n*(a-b), which is also in K. Further, a*r=n*s*r, which is also in K. So K is an ideal in R. QED (1)

(2) Let R and K be defined as above and let u be a unit in R such that u is in K. Then there exists r ∈ R such that n*r=u. u is a unit, so u-1 (u inverse) is in R. Then n*r*u-1=1R. Further, let s be an arbitrary element of R. Then n*r*u-1*s=1R*s=s. s was arbitrary, so K=R. QED (2)

(3) Let R have finite char c and let gcd(c,n)=1. Then n has a multiplicative order mod c, call it m. So nm=1(mod c). Let r=nm-1*1R. Note that r ∈ R because n is an integer and so we are simply adding 1R to itself nm-1 times. Now n*r ∈ K, and n*r=n*nm-1*1R=1R. So there is a unit in K, and K=R by (2). QED (3)

Note that there is a logically appealing but false statement related to (3): K=R implies that the char of R is finite. As an example, consider R={x*2y|x,y ∈ Z}, which is fairly easily identified as an integral domain. So this is the set of integers multiplied by all powers of 2. Suppose n=2. In our given R here, 2*1/2=1 is a unit in R, so K={n*r|r in R}=R for n equal to (any positive integer power of) 2. But we also have that the char of R is infinite. Thus K=R has no say in the characteristic of R.

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