zero-divisor

A zero-divisor is something that, like zero itself, can be multiplied by something nonzero to produce zero as a product.

Let $R$ be a commutative ring (or any ring).

An element $x$ of $R$ is a **non-zero-divisor** if, whenever $x y = 0$ or $y x = 0$, then $y = 0$. An element $x$ is a **zero-divisor** if there exists $y \ne 0$ such that $x y = 0$ or $y x = 0$.

In constructive mathematics, we want $\ne$ to be a tight apartness relation on $R$ in the definition of zero-divisor. We also say that $x$ is a **strong non-zero-divisor** if, whenever $y \ne 0$, then $x y \ne 0$ and $y x \ne 0$. (The notion of (weak) non-zero-divisor makes sense even without any apartness relation.)

If $R$ is (or may be) non-commutative, then we may distinguish **left** and **right** (non)-zero-divisors in the usual way.

Note that, in a ring, an element is a non-zero-divisor if and only if the operation of multiplication by that element is injective. This is probably the right definition of zero-divisor to use in a rig, even though then it no longer literally has anything to do with being a divisor of zero.

By this definition, zero itself is a zero-divisor if and only if $R$ is non-trivial. (Some authorities will differ on this point, but if you think about it, this is clearly the correct definition, by the same principle that the trivial ring is not a field, $1$ is not a prime number, etc. See too simple to be simple.)

An integral domain is precisely a commutative ring in which zero is the unique zero-divisor (or constructively, in which the strong non-zero-divisors are precisely the strong non-zero elements, that is those elements $x$ such that $x \ne 0$).

The non-zero-divisors of any rig $R$ form a monoid under multiplication, which may be denoted $R^{\times}$. Note that if $R$ happens to be a field, then this $R^{\times}$ agrees with the usual notation $R^{\times}$ for the group of invertible elements of $R$, but $R^{\times}$ is not a group in general. (We may use $R^{\div}$ or $R^*$ for the group of invertible elements.)

If $I$ is any ideal of $R$, then we can generalise from a zero-divisor to an $I$-divisor. In a way, this is nothing new; $x$ is an $I$-divisor in $R$ if and only if $[x]$ is a zero-divisor in $R/I$. Ultimately, this is related to the notion of divisor in algebraic geometry.

Revised on June 30, 2010 05:25:32
by Toby Bartels
(75.88.78.90)