Maximum term method

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 zero divisors. All of the number systems studied in elementary mathematics — the integers $\mathbb{Z}$, the rational numbers $\mathbb{Q}$, the real numbers $ℝ$, and the complex numbers $ℂ$ — 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 $\{0,1,2,\dots \}$.

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 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 $\{0,1,2,\dots \}$ 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 $2\cdot 3\equiv 0(\mathrm{mod}6)$.

• 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 $qm\equiv 0(\mathrm{mod}n)$.

• The ring ${\mathbb{Z}}^{2\times 2}$ of 2 by 2 matrices with integer entries does not satisfy the zero-product property: if

$M=\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)$,

then

$MN=\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)=0$,

yet neither $M$ nor $N$ is zero.

• The ring of all functions $f:[0,1]\to ℝ$, 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 $f_1,\dots ,f_n$, none of which is identically zero, such that $f_i{f}_j$ is identically zero whenever $i\ne j$.

• 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 $d\ge 1$ with coefficients in $R$. Suppose also that $f$ has $d$ distinct roots $r_1,\dots ,r_d\in R$. 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 $r_1,\dots ,r_d$ 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 $\mathbb{Z}$).

See also

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

Notes

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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