Zero divisor

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Short description: Ring element that can be multiplied by a non-zero element to equal 0

In abstract algebra, an element a of a ring R is called a left zero divisor if there exists a nonzero x in R such that ax = 0,[1] or equivalently if the map from R to R that sends x to ax is not injective.[lower-alpha 1] Similarly, an element a of a ring is called a right zero divisor if there exists a nonzero y in R such that ya = 0. This is a partial case of divisibility in rings. An element that is a left or a right zero divisor is simply called a zero divisor.[2] An element a that is both a left and a right zero divisor is called a two-sided zero divisor (the nonzero x such that ax = 0 may be different from the nonzero y such that ya = 0). If the ring is commutative, then the left and right zero divisors are the same.

An element of a ring that is not a left zero divisor is called left regular or left cancellable. Similarly, an element of a ring that is not a right zero divisor is called right regular or right cancellable. An element of a ring that is left and right cancellable, and is hence not a zero divisor, is called regular or cancellable,[3] or a non-zero-divisor. A zero divisor that is nonzero is called a nonzero zero divisor or a nontrivial zero divisor. A nonzero ring with no nontrivial zero divisors is called a domain.


  • In the ring [math]\displaystyle{ \mathbb{Z}/4\mathbb{Z} }[/math], the residue class [math]\displaystyle{ \overline{2} }[/math] is a zero divisor since [math]\displaystyle{ \overline{2} \times \overline{2}=\overline{4}=\overline{0} }[/math].
  • The only zero divisor of the ring [math]\displaystyle{ \mathbb{Z} }[/math] of integers is [math]\displaystyle{ 0 }[/math].
  • A nilpotent element of a nonzero ring is always a two-sided zero divisor.
  • An idempotent element [math]\displaystyle{ e\ne 1 }[/math] of a ring is always a two-sided zero divisor, since [math]\displaystyle{ e(1-e)=0=(1-e)e }[/math].
  • The ring of [math]\displaystyle{ n \times n }[/math] matrices over a field has nonzero zero divisors if [math]\displaystyle{ n \geq 2 }[/math]. Examples of zero divisors in the ring of [math]\displaystyle{ 2\times 2 }[/math] matrices (over any nonzero ring) are shown here: [math]\displaystyle{ \begin{pmatrix}1&1\\2&2\end{pmatrix}\begin{pmatrix}1&1\\-1&-1\end{pmatrix}=\begin{pmatrix}-2&1\\-2&1\end{pmatrix}\begin{pmatrix}1&1\\2&2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix} , }[/math] [math]\displaystyle{ \begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix} =\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix} =\begin{pmatrix}0&0\\0&0\end{pmatrix}. }[/math]
  • A direct product of two or more nonzero rings always has nonzero zero divisors. For example, in [math]\displaystyle{ R_1 \times R_2 }[/math] with each [math]\displaystyle{ R_i }[/math] nonzero, [math]\displaystyle{ (1,0)(0,1) = (0,0) }[/math], so [math]\displaystyle{ (1,0) }[/math] is a zero divisor.
  • Let [math]\displaystyle{ K }[/math] be a field and [math]\displaystyle{ G }[/math] be a group. Suppose that [math]\displaystyle{ G }[/math] has an element [math]\displaystyle{ g }[/math] of finite order [math]\displaystyle{ n\gt 1 }[/math]. Then in the group ring [math]\displaystyle{ K[G] }[/math] one has [math]\displaystyle{ (1-g)(1+g+ \cdots +g^{n-1})=1-g^{n}=0 }[/math], with neither factor being zero, so [math]\displaystyle{ 1-g }[/math] is a nonzero zero divisor in [math]\displaystyle{ K[G] }[/math].

One-sided zero-divisor

  • Consider the ring of (formal) matrices [math]\displaystyle{ \begin{pmatrix}x&y\\0&z\end{pmatrix} }[/math] with [math]\displaystyle{ x,z\in\mathbb{Z} }[/math] and [math]\displaystyle{ y\in\mathbb{Z}/2\mathbb{Z} }[/math]. Then [math]\displaystyle{ \begin{pmatrix}x&y\\0&z\end{pmatrix}\begin{pmatrix}a&b\\0&c\end{pmatrix}=\begin{pmatrix}xa&xb+yc\\0&zc\end{pmatrix} }[/math] and [math]\displaystyle{ \begin{pmatrix}a&b\\0&c\end{pmatrix}\begin{pmatrix}x&y\\0&z\end{pmatrix}=\begin{pmatrix}xa&ya+zb\\0&zc\end{pmatrix} }[/math]. If [math]\displaystyle{ x\ne0\ne z }[/math], then [math]\displaystyle{ \begin{pmatrix}x&y\\0&z\end{pmatrix} }[/math] is a left zero divisor if and only if [math]\displaystyle{ x }[/math] is even, since [math]\displaystyle{ \begin{pmatrix}x&y\\0&z\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&x\\0&0\end{pmatrix} }[/math], and it is a right zero divisor if and only if [math]\displaystyle{ z }[/math] is even for similar reasons. If either of [math]\displaystyle{ x,z }[/math] is [math]\displaystyle{ 0 }[/math], then it is a two-sided zero-divisor.
  • Here is another example of a ring with an element that is a zero divisor on one side only. Let [math]\displaystyle{ S }[/math] be the set of all sequences of integers [math]\displaystyle{ (a_1,a_2,a_3,...) }[/math]. Take for the ring all additive maps from [math]\displaystyle{ S }[/math] to [math]\displaystyle{ S }[/math], with pointwise addition and composition as the ring operations. (That is, our ring is [math]\displaystyle{ \mathrm{End}(S) }[/math], the endomorphism ring of the additive group [math]\displaystyle{ S }[/math].) Three examples of elements of this ring are the right shift [math]\displaystyle{ R(a_1,a_2,a_3,...)=(0,a_1,a_2,...) }[/math], the left shift [math]\displaystyle{ L(a_1,a_2,a_3,...)=(a_2,a_3,a_4,...) }[/math], and the projection map onto the first factor [math]\displaystyle{ P(a_1,a_2,a_3,...)=(a_1,0,0,...) }[/math]. All three of these additive maps are not zero, and the composites [math]\displaystyle{ LP }[/math] and [math]\displaystyle{ PR }[/math] are both zero, so [math]\displaystyle{ L }[/math] is a left zero divisor and [math]\displaystyle{ R }[/math] is a right zero divisor in the ring of additive maps from [math]\displaystyle{ S }[/math] to [math]\displaystyle{ S }[/math]. However, [math]\displaystyle{ L }[/math] is not a right zero divisor and [math]\displaystyle{ R }[/math] is not a left zero divisor: the composite [math]\displaystyle{ LR }[/math] is the identity. [math]\displaystyle{ RL }[/math] is a two-sided zero-divisor since [math]\displaystyle{ RLP=0=PRL }[/math], while [math]\displaystyle{ LR=1 }[/math] is not in any direction.



  • In the ring of n-by-n matrices over a field, the left and right zero divisors coincide; they are precisely the singular matrices. In the ring of n-by-n matrices over an integral domain, the zero divisors are precisely the matrices with determinant zero.
  • Left or right zero divisors can never be units, because if a is invertible and ax = 0 for some nonzero x, then 0 = a−10 = a−1ax = x, a contradiction.
  • An element is cancellable on the side on which it is regular. That is, if a is a left regular, ax = ay implies that x = y, and similarly for right regular.

Zero as a zero divisor

There is no need for a separate convention for the case a = 0, because the definition applies also in this case:

  • If R is a ring other than the zero ring, then 0 is a (two-sided) zero divisor, because any nonzero element x satisfies 0x = 0 = x0.
  • If R is the zero ring, in which 0 = 1, then 0 is not a zero divisor, because there is no nonzero element that when multiplied by 0 yields 0.

Some references include or exclude 0 as a zero divisor in all rings by convention, but they then suffer from having to introduce exceptions in statements such as the following:

  • In a commutative ring R, the set of non-zero-divisors is a multiplicative set in R. (This, in turn, is important for the definition of the total quotient ring.) The same is true of the set of non-left-zero-divisors and the set of non-right-zero-divisors in an arbitrary ring, commutative or not.
  • In a commutative noetherian ring R, the set of zero divisors is the union of the associated prime ideals of R.

Zero divisor on a module

Let R be a commutative ring, let M be an R-module, and let a be an element of R. One says that a is M-regular if the "multiplication by a" map [math]\displaystyle{ M \,\stackrel{a}\to\, M }[/math] is injective, and that a is a zero divisor on M otherwise.[4] The set of M-regular elements is a multiplicative set in R.[4]

Specializing the definitions of "M-regular" and "zero divisor on M" to the case M = R recovers the definitions of "regular" and "zero divisor" given earlier in this article.

See also


  1. Since the map is not injective, we have ax = ay, in which x differs from y, and thus a(xy) = 0.


  1. N. Bourbaki (1989), Algebra I, Chapters 1–3, Springer-Verlag, p. 98 
  2. Charles Lanski (2005), Concepts in Abstract Algebra, American Mathematical Soc., p. 342 
  3. Nicolas Bourbaki (1998). Algebra I. Springer Science+Business Media. p. 15. 
  4. 4.0 4.1 Hideyuki Matsumura (1980), Commutative algebra, 2nd edition, The Benjamin/Cummings Publishing Company, Inc., p. 12 

Further reading