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=0_{R} iff a=0_{R} or b=0_{R}.

#### 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.

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*0_{R}=0_{R}. So K={0_{R}}, 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=0_{R} iff r=0_{R}, 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}=1_{R}. Further, let s be an arbitrary element of R. Then n*r*u^{-1}*s=1_{R}*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 n^{m}=1(mod c). Let r=n^{m-1}*1_{R}. Note that r ∈ R because n is an integer and so we are simply adding 1_{R} to itself n^{m-1} times. Now n*r ∈ K, and n*r=n*n^{m-1}*1_{R}=1_{R}. 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*2^{y}|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.