Commutative Rings, Domains, Units and Subrings

A commutative ring1 R is a set with two binary operations, addition and multiplication, such that (i) R is an abelian group under addition; (ii) (commutativity) ab = ba for all a, b ∈R; (iii) (associativity) a(bc) = (ab)c for every a, b, c ∈R; 1This term was probably coined by D. One of the meanings of the word ring, in German as in English, is collection, as in the phrase “a ring of thieves.” (It has also been suggested that polynomial finite-generation used this term because, for a ring of algebraic integers, an appropriate power of each element “cycles back” to being a linear combination of lower powers.) First Properties (iv) there is an element 1 ∈R with 1a = a for every a ∈R;2 (v) (distributivity) a(b + c) = ab + ac for every a, b, c ∈R. The element 1 in a ring R has several names; it is called one, the unit of R, or the identity in R. Addition and multiplication in a commutative ring R are binary operations, so there are functions α : R × R →R with α(r,r′) = r + r′ ∈R and µ : R × R →R with µ(r,r′) = rr′ ∈R for all r,r′ ∈R. The law of substitution holds here, as it does for any operation: If r = r′ and s = s′, then r + s = r′ + s′ and rs = r′s′.

A subset S of a commutative ring R is a subring of R if (i) 1 ∈S;4 (ii) if a, b ∈S, then a −b ∈S; (iii) if a, b ∈S, then ab ∈S. Notation. In contrast to the usage H ≤G for a subgroup, the tradition in ring theory is to write S ⊆R for a subring. We shall also write S ⊊R to denote a proper subring; that is, S ⊆R and S ̸= R.

Let a and b be elements of a commutative ring R. Then a divides b in R (or a is a divisor of b or b is a multiple of a), denoted by a | b, if there exists an element c ∈R with b = ca. As an extreme example, if 0 | a, then a = 0 · b for some b ∈R. Since 0 · b = 0, however, we must have a = 0. Thus, 0 | a if and only if a = 0. Notice that whether a | b depends not only on the elements a and b but on the ambient ring R as well. For example, 3 does divide 2 in Q, for 2 = 3 × 2 3, and 2 3 ∈Q; on the other hand, 3 does not divide 2 in Z, because there is no integer c with 3c = 2.

An element u in a commutative ring R is called a unit if u | 1 in R, that is, if there exists v ∈R with uv = 1; the element v is called the inverse of u and v is often denoted by u−1. Units are of interest because we can always divide by them: If a ∈R and u is a unit in R (so there is v ∈R with uv = 1), then a = u(va) is a factorization of a in R, for va ∈R; thus, it is reasonable to define the quotient a/u as va = u−1a. Given elements a and b, whether a | b depends not only on these elements but also on the ambient ring R; similarly, whether an element u ∈R is a unit also depends on the ambient ring R (for it is a question whether u | 1 in R). For example, the number 2 is a unit in Q, for 1 2 lies in Q and 2 × 1 2 = 1, but 2 is not a unit in Z, because there is no integer v with 2v = 1. In fact, the only units in Z are 1 and −1.

A field5 F is a commutative ring in which 1 ̸= 0 and every nonzero element a is a unit; that is, there is a−1 ∈F with a−1a = 1. The first examples of fields are Q, R, and C. The definition of field can be restated in terms of the group of units; a commutative ring R is a field if and only if U(R) = R×, the nonzero elements of R. To say this another way, R is a field if and only if R× is a multiplicative group [note that U(R×) ̸= ∅because we are assuming that 1 ̸= 0].

The field F constructed from R in Theorem 3.13 is called the fraction field of R; we denote it by Frac(R), and we denote [a, b] ∈Frac(R) by a/b; in particular, the elements [a, 1] of R′ are denoted by a/1 or, more simply, by a. Notice that the fraction field of Z is Q; that is, Frac(Z) = Q.

A sequence σ = (s0, s1, . . . , si, . . . ) in a commutative ring R is called a polynomial if there is some integer m ≥0 with si = 0 for all i > m; that is, σ = (s0, s1, . . . , sm, 0, 0, . . . ). A polynomial has only finitely many nonzero coefficients. The zero polynomial, denoted by σ = 0, is the sequence σ = (0, 0, 0, . . . ).

If σ = (s0, s1, . . . , sn, 0, 0, . . . ) ̸= 0 is a polynomial, then there is sn ̸= 0 with si = 0 for all i > n. We call sn the leading coefficient of σ, we call n the degree of σ, and we denote the degree n by deg(σ). The zero polynomial 0 does not have a degree because it has no nonzero coefficients. Some authors define deg(0) = −∞, and this is sometimes convenient, for −∞< n for every integer n. On the other hand, we choose not to assign a degree to 0 because it is often a genuinely different case that must be dealt with separately. Notation. If R is a commutative ring, then the set of all polynomials with coefficients in R is denoted by R[x].

Let R be a commutative ring. (i) 0 · a = 0 for every a ∈R. (ii) If 1 = 0, then R consists of the single element 0. In this case, R is called the zero ring.3 (iii) If −a is the additive inverse of a, then (−1)(−a) = a. (iv) (−1)a = −a for every a ∈R. (v) If n ∈N and n1 = 0, then na = 0 for all a ∈R. (vi) The binomial theorem holds: If a, b ∈R, then (a + b)n = n r=0 n r arbn−r.

A nonzero commutative ring R is a domain if and only if the product of any two nonzero elements of R is nonzero.

Let R be a domain, and let a, b ∈R be nonzero. Then a | b and b | a if and only if b = ua for some unit u ∈R.

If a is an integer, then [a] is a unit in Im if and only if a and m are relatively prime. In fact, if sa + tm = 1, then [a]−1 = [s].

Every subring of a domain is itself a domain. Since fields are domains, it follows that every subring of a field is a domain. The converse of this exercise is true, and it is much more interesting: Every domain is a subring of a field. 5The derivation of the mathematical usage of the English term field (first used by E. First Properties Given four elements a, b, c, and d in a field F with b ̸= 0 and d ̸= 0, assume that ab−1 = cd−1. Multiply both sides by bd to obtain ad = bc. In other words, were ab−1 written as a/b, then we have just shown that a/b = c/d implies ad = bc; that is, “cross-multiplication” is valid. Conversely, if ad = bc and both b and d are nonzero, then multiplication by b−1d−1 gives ab−1 = cd−1, that is, a/b = c/d. The proof of the next theorem is a straightforward generalization of the usual construction of the field of rational numbers Q from the domain of integers Z.

If R is a domain, then there is a field F containing R as a subring. Moreover, F can be chosen so that, for each f ∈F, there are a, b ∈R with b ̸= 0 and f = ab−1.

If R is a commutative ring, then R[x] is a commutative ring that contains R as a subring.

(i) Z, Q, R, and C are commutative rings with the usual addition and multiplication (the ring axioms are verified in courses in the foundations of mathematics). (ii) Im, the integers mod m, is a commutative ring. (iii) Let Z[i] be the set of all complex numbers of the form a + bi, where a, b ∈Z and i2 = −1. Z[i] is called the ring of quadratic complex integer ring. (iv) Consider the set R of all real numbers x of the form x = a + bω, where a, b ∈Q and ω = 3√ 2. However, if R is closed under multiplication, then ω2 ∈R, and there are rationals a and b with ω2 = a + bω. Multiplying both sides by ω and by b gives the equations 2 = aω + bω2 bω2 = ab + b2ω. Hence, 2 −aω = ab + b2ω, and so 2 −ab = (b2 + a)ω. If b2 + a ̸= 0, then ω = 3√ 2 is rational; if b2 + a = 0, then this coupled with 2 −ab = 0 yields 2 = (−b)3. Thus, either case forces 3√ 2 rational, and this contradiction shows that R is not a commutative ring. ◀ 2Some authors do not demand that commutative rings have 1. For them, the set of all even integers is a commutative ring, but we do not recognize it as such. Remark. There are noncommutative rings; that is, sets having an addition and a multiplication satisfying all the axioms of a commutative ring except the axiom: ab = ba. ◀ Here are some elementary results.

(i) Let F(R) be the set of all the functions R →R equipped with the operations of pointwise addition and pointwise multiplication: Given f, g ∈F(R), define functions f + g and f g by f + g : a ↦f (a) + g(a) and f g : a ↦f (a)g(a) (notice that f g is not their composite). We claim that F(R) with these operations is a commutative ring. Verification of the axioms is left to the reader with the following hint: The zero element in F(R) is the constant function z with value 0 [that is, z(a) = 0 for all a ∈R] and the unit is the constant function ε with ε(a) = 1 for all a ∈R. We now show that F(R) is not a domain. x x y f g y Figure 3.1 Define f and g as drawn in Figure 3.1: f (a) = a if a ≤0 if a ≥0; g(a) = if a ≤0 a if a ≥0. Clearly, neither f nor g is zero (i.e., f ̸= z and g ̸= z). On the other hand, for each a ∈R, f g : a ↦f (a)g(a) = 0, because at least one of the factors f (a) or g(a) is the number zero. Therefore, f g = z, by Proposition 1.43, and F(R) is not a domain. First Properties (ii) All differentiable functions f : R →R form a subring of F(R). The identity ε is a constant function, hence is differentiable, while the sum and product of differentiable functions are also differentiable. Hence, the differentiable functions form a commutative ring. ◀ Many theorems of ordinary arithmetic, that is, properties of the commutative ring Z, hold in more generality. We now generalize some familiar definitions from Z to arbitrary commutative rings.