Zero-product property

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