# 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 or 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 $\mathbb{Z}$).