Zero-product property

In the branch of mathematics called 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 either a = 0 or b = 0.

The zero-product property is also known as the rule of zero product or nonexistence of 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. 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.

• If p is a prime number, then the ring of integers modulo p has the zero-product property (in fact, it is a field).

• The Gaussian integers are an integral domain because they are a subring of the complex numbers.

• In the 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 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 k-times differentiable 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 x³ − 2x² − 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 x³ + 3x² + 2x has six roots in $\mathbb{Z}_6$ (though it has only three roots in $\mathbb{Z}$).

See also

• Fundamental theorem of algebra
• Integral domain and domain
• Prime ideal
• Zero divisor

References

• David S. Dummit and Richard M. Foote, Abstract Algebra (3d ed.), Wiley, 2003, ISBN 0-471-43334-9.

External links

• PlanetMath: Zero rule of product