Zero-product property

In algebra, the zero-product property states that the product of two nonzero elements is nonzero. In other words, it is the following assertion:

If $ab = 0$, then $a=0$ or $b=0$.

The zero-product property is also known as the rule of zero product, the null factor law or the nonexistence of nontrivial zero divisors. All of the number systems studied in elementary mathematics — the integers $\mathbb{Z}$, the rational numbers $\mathbb{Q}$, the real numbers $\mathbb{R}$, and the complex numbers $\mathbb{C}$ — satisfy the zero-product property. In general, a ring which satisfies the zero-product property is called a domain.

Algebraic context

Suppose $A$ is an algebraic structure. We might ask, does $A$ have the zero-product property? In order for this question to have meaning, $A$ must have both additive structure and multiplicative structure.[note 1] Usually one assumes that $A$ is a ring, though it could be something else, e.g., the nonnegative integers $\{0,1,2,\ldots\}$.

Note that if $A$ satisfies the zero-product property, and if $B$ is a subset of $A$, then $B$ also satisfies the zero product property: if $a$ and $b$ are elements of $B$ such that $ab=0$, then either $a=0$ or $b=0$ because $a$ and $b$ can also be considered as elements of $A$.

Examples

• A ring in which the zero-product property holds is called a domain. A commutative domain with a multiplicative identity element is called an integral domain. Any field is an integral domain; in fact, any subring of a field is an integral domain (as long as it contains 1). Similarly, any subring of a skew field is a domain. Thus, the zero-product property holds for any subring of a skew field.
• In the strictly skew field of quaternions, the zero-product property holds. This ring is not an integral domain, because the multiplication is not commutative.
• The set of nonnegative integers $\{0,1,2,\ldots\}$ is not a ring, but it does satisfy the zero-product property.

Non-examples

• Let $\mathbb{Z}_n$ denote the ring of integers modulo $n$. Then $\mathbb{Z}_6$ does not satisfy the zero product property: 2 and 3 are nonzero elements, yet $2 \cdot 3 \equiv 0 \pmod{6}$.
• In general, if $n$ is a composite number, then $\mathbb{Z}_n$ does not satisfy the zero-product property. Namely, if $n = qm$ where $0 < q,m < n$, then $m$ and $q$ are nonzero modulo $n$, yet $qm \equiv 0 \pmod{n}$.
• The ring $\mathbb{Z}^{2 \times 2}$ of 2 by 2 matrices with integer entries does not satisfy the zero-product property: if
$M = \begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix}$ and $N = \begin{pmatrix}0 & 1 \\ 0 & 1\end{pmatrix}$,
then
$MN = \begin{pmatrix}1 & -1 \\ 0 & 0\end{pmatrix} \begin{pmatrix}0 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} = 0$,
yet neither $M$ nor $N$ is zero.
• The ring of all functions $f: [0,1] \to \mathbb{R}$, from the unit interval to the real numbers, has nontrivial zero divisors: there are pairs of functions which are not identically equal to zero yet whose product is the zero function. In fact, it is not hard to construct, for any n ≥ 2, functions $f_1,\ldots,f_n$, none of which is identically zero, such that $f_i \, f_j$ is identically zero whenever $i \neq j$.
• The same is true even if we consider only continuous functions, or only even infinitely smooth functions.

Application to finding roots of polynomials

Suppose $P$ and $Q$ are univariate polynomials with real coefficients, and $x$ is a real number such that $P(x)Q(x) = 0$. (Actually, we may allow the coefficients and $x$ to come from any integral domain.) By the zero-product property, it follows that either $P(x) = 0$ or $Q(x) = 0$. In other words, the roots of $PQ$ are precisely the roots of $P$ together with the roots of $Q$.

Thus, one can use factorization to find the roots of a polynomial. For example, the polynomial $x^3 - 2x^2 - 5x + 6$ factorizes as $(x-3)(x-1)(x+2)$; hence, its roots are precisely 3, 1, and -2.

In general, suppose $R$ is an integral domain and $f$ is a monic univariate polynomial of degree $d \geq 1$ with coefficients in $R$. Suppose also that $f$ has $d$ distinct roots $r_1,\ldots,r_d \in R$. It follows (but we do not prove here) that $f$ factorizes as $f(x) = (x-r_1) \cdots (x-r_d)$. By the zero-product property, it follows that $r_1,\ldots,r_d$ are the only roots of $f$: any root of $f$ must be a root of $(x-r_i)$ for some $i$. In particular, $f$ has at most $d$ distinct roots.

If however $R$ is not an integral domain, then the conclusion need not hold. For example, the cubic polynomial $x^3 + 3x^2 + 2x$ has six roots in $\mathbb{Z}_6$ (though it has only three roots in $\mathbb{Z}$).