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A ring is a structure of abstract algebra, similar to a group or a field. A ring $R$ is a set of elements closed under two operations, usually called multiplication and addition and denoted $\cdot$ and $+$, for which

In other words, the following properties hold for all $a,b,c$ in $R$:

  • $(a+b) + c = a+(b+c)$ (associativity of addition);
  • $a+b = b+a$ (commutativity of addition);
  • For some $0\in R$, $0+a=a+0=a$ (existance of additive identity);
  • There exists some $-a\in R$ for which $a+ (-a) = (-a)+a = 0$ (existance of additive inverses);
  • $(ab)c = a(bc)$ (associativity of multiplication);
  • For some $1\in R$, $1a=a1=a$ (existance of multiplicative identity)
  • $a(b+c)= ab+ac \\ (b+c)a = ba + ca = ab+ac$ (double distributivity of multiplication over addition).
  • $a(b-c)= ab-ac \\ (b-c)a = ba - ca$ (double distributivity of multiplication over subtraction).

Note especially that multiplicative inverses need not exist and that multiplication need not be commutative.

The elements of $R$ under addition is called the additive group of $R$; it is sometimes denoted $R^+$. (However, this can sometimes lead to confusion when $R$ is also an ordered set.) The set of invertible elements of $R$ constitute a group under multiplication, denoted $R^*$. The elements of $R$ under the multiplicative law $(a,b) \mapsto ba$ (i.e., the opposite multiplicative law) and the same additive law constitute the opposite ring of $R$, which can be denoted $R^0$.

Let $a$ be an element of $R$. Then the mapping $x \mapsto ax$ of $R$ into $R$ is an endomorphism of the abelian group $R^+$. Since group homomorphisms map identities to identities, it follows that $a0 = 0$, for all $a$ in $R$, and similarly, $0a = 0$.


Let $x$ and $y$ be elements of a ring $R$. If there exists an element $a$ of $R$ such that $x=ay$, then $y$ is said to be a right divisor of $x$, and $x$ is said to be a left multiple of $y$. Left divisors and right multiples are defined similarly. When $R$ is commutative, we say simply that $y$ is a divisor of $x$, or $y$ divides $x$, or $x$ is a multiple of $y$.

Note that the relation "$y$ is a right divisor of $x$" is transitive, for if $x = ay$ and $y = bz$, then $x= (ab)z$. Furthermore, every element of $R$ is a right divisor of itself. Therefore $R$ has the (sometimes trivial) structure of a partially ordered set.

Under these definitions, every element of $R$ is a left and right divisor of 0. However, by abuse of language, we usually only call an element $x$ a left (or right) divisor of zero (or left, right zero divisors) if there is a non-zero element $y$ for which $xy=0$ (or $yx=0$). The left zero divisors are precisely those $x$ elements of $R$ for which left multiplication is not cancellable. For if $y,z$ are distinct elements of $R$ for which $xy=xz$, then $x(y-z)=0$.

Examples of Rings

The sets of integers ($\mathbb{Z}$), rational numbers ($\mathbb{Q}$), real numbers ($\mathbb{R}$), and complex numbers ($\mathbb{C}$) are all examples of commutative rings, as is the set of Gaussian integers ($\mathbb{Z}[i]$). Note that of these, the integers and Gaussian integers do not have inverses; the rest do, and therefore also constitute examples of fields. All these rings are infinite, as well.

Among the finite commutative rings are sets of integers mod $m$ ($\mathbb{Z}/m\mathbb{Z}$), for any integer $m$.

If $G$ is an abelian group, then the set of endomorphisms on $G$ form a ring, under the rules \[(f+g)(x) = f(x)+ g(x); \qquad fg = f\circ g .\]

Let $R$ be a ring. The set of polynomials in $R$ is also a ring.

Let $F$ be a field. The set of $n\times n$ matrices of $F$ constitute a ring. In fact, they are the endomorphism ring of the additive group $(F^+)^n$.

If $R,R'$ are rings, then Cartesian product $R_1 \times R_2$ is a ring under coordinatewise multiplication and addition; this is called the direct product of these rings.

Let $\mathcal{F}$ be the set of weak multiplicative functions mapping the positive integers into themselves. Then the elements of $\mathcal{F}$ form a pseudo-ring, with multiplication defined as Dirichlet convolution, i.e., \[(fg)(n) = \sum_{d\mid n} f(d)g(n/d) ,\] for \[((fg)h)(n) = (f(gh))(n) = \sum_{abc=n} f(a)f(b)f(c) .\] However, there is no multiplicative identity, so this is not a proper ring.

See also

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