Suppose ${\displaystyle A}$  is an algebraic structure. We might ask, does ${\displaystyle A}$  have the zero-product property? In order for this question to have meaning, ${\displaystyle A}$  must have both additive structure and multiplicative structure.^[2] Usually one assumes that ${\displaystyle A}$  is a ring, though it could be something else, e.g. the set of nonnegative integers ${\displaystyle \{0,1,2,\ldots \}}$  with ordinary addition and multiplication, which is only a (commutative) semiring.

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

• 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 ${\displaystyle \{0,1,2,\ldots \}}$  is not a ring (being instead a semiring), but it does satisfy the zero-product property.

• 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 ${\displaystyle 2\cdot 3\equiv 0{\pmod {6}}}$ .
• 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 ${\displaystyle qm\equiv 0{\pmod {n}}}$ .
• The ring ${\displaystyle \mathbb {Z} ^{2\times 2}}$  of 2×2 matrices with integer entries does not satisfy the zero-product property: if
  $${\displaystyle M={\begin{pmatrix}1&-1\\0&0\end{pmatrix}}}$$

   
  and ${\displaystyle N={\begin{pmatrix}0&1\\0&1\end{pmatrix}},}$  then
  $${\displaystyle MN={\begin{pmatrix}1&-1\\0&0\end{pmatrix}}{\begin{pmatrix}0&1\\0&1\end{pmatrix}}={\begin{pmatrix}0&0\\0&0\end{pmatrix}}=0,}$$

   
  yet neither ${\displaystyle M}$  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 ${\displaystyle f_{1},\ldots ,f_{n}}$ , none of which is identically zero, such that ${\displaystyle f_{i}\,f_{j}}$  is identically zero whenever ${\displaystyle i\neq j}$ .
• 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.

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 ${\displaystyle d\geq 1}$  with coefficients in ${\displaystyle R}$ . Suppose also that ${\displaystyle f}$  has ${\displaystyle d}$  distinct roots ${\displaystyle r_{1},\ldots ,r_{d}\in R}$ . 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 ${\displaystyle r_{1},\ldots ,r_{d}}$  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 ${\displaystyle \mathbb {Z} }$ ).

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

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

• PlanetMath: Zero rule of product