Zero-product property

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 ${\displaystyle p}$ is a prime number, then the ring of integers modulo ${\displaystyle p}$ 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 ${\displaystyle \{0,1,2,\dots \}}$ is not a ring (being instead a semiring), but it does satisfy the zero-product property.

Non-examples

• Let ${\displaystyle {\mathbb{Z}}_n}$ denote the ring of integers modulo ${\displaystyle n}$. Then ${\displaystyle {\mathbb{Z}}_6}$ does not satisfy the zero product property: 2 and 3 are nonzero elements, yet ${\displaystyle 2\cdot 3\equiv 0(\mathrm{mod}6)}$.
• In general, if ${\displaystyle n}$ is a composite number, then ${\displaystyle {\mathbb{Z}}_n}$ does not satisfy the zero-product property. Namely, if ${\displaystyle n=qm}$ where ${\displaystyle 0<q,m<n}$, then ${\displaystyle m}$ and ${\displaystyle q}$ are nonzero modulo ${\displaystyle n}$, yet ${\displaystyle qm\equiv 0(\mathrm{mod}n)}$.
• The ring ${\displaystyle {\mathbb{Z}}^{2\times 2}}$ of 2×2 matrices with integer entries does not satisfy the zero-product property: if $${\displaystyle M=\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)}$$ and ${\displaystyle N=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right),}$ then $${\displaystyle MN=\left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)=0,}$$ yet neither ${\displaystyle M}$ nor ${\displaystyle N}$ is zero.
• The ring of all functions ${\displaystyle f:[0,1]\to \mathbb{R}}$, 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 ${\displaystyle f_1,\dots ,f_n}$, none of which is identically zero, such that ${\displaystyle f_i{f}_j}$ is identically zero whenever ${\displaystyle i\ne j}$.
• The same is true even if we consider only continuous functions, or only even infinitely smooth functions. On the other hand, analytic functions have the zero-product property.

Application to finding roots of polynomials

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

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

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

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