Zero divisor

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.[a] 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 (respectively, not a right zero divisor) is called left regular or left cancellable (respectively, 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 non-zero ring with no nontrivial zero divisors is called a domain.

Examples

  • In the ring Z / 4 Z {\displaystyle \mathbb {Z} /4\mathbb {Z} } , the residue class 2 ¯ {\displaystyle {\overline {2}}} is a zero divisor since 2 ¯ × 2 ¯ = 4 ¯ = 0 ¯ {\displaystyle {\overline {2}}\times {\overline {2}}={\overline {4}}={\overline {0}}} .
  • The only zero divisor of the ring Z {\displaystyle \mathbb {Z} } of integers is 0 {\displaystyle 0} .
  • A nilpotent element of a nonzero ring is always a two-sided zero divisor.
  • An idempotent element e 1 {\displaystyle e\neq 1} of a ring is always a two-sided zero divisor, since e ( 1 e ) = 0 = ( 1 e ) e {\displaystyle e(1-e)=0=(1-e)e} .
  • The ring of n × n matrices over a field has nonzero zero divisors if n ≥ 2. Examples of zero divisors in the ring of 2 × 2 matrices (over any nonzero ring) are shown here:

( 1 1 2 2 ) ( 1 1 1 1 ) = ( 2 1 2 1 ) ( 1 1 2 2 ) = ( 0 0 0 0 ) , {\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}},}
( 1 0 0 0 ) ( 0 0 0 1 ) = ( 0 0 0 1 ) ( 1 0 0 0 ) = ( 0 0 0 0 ) . {\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}}.}

  • A direct product of two or more nonzero rings always has nonzero zero divisors. For example, in R 1 × R 2 {\displaystyle R_{1}\times R_{2}} with each R i {\displaystyle R_{i}} nonzero, ( 1 , 0 ) ( 0 , 1 ) = ( 0 , 0 ) {\displaystyle (1,0)(0,1)=(0,0)} , so ( 1 , 0 ) {\displaystyle (1,0)} is a zero divisor.
  • Let K {\displaystyle K} be a field and G {\displaystyle G} be a group. Suppose that G {\displaystyle G} has an element g {\displaystyle g} of finite order n > 1 {\displaystyle n>1} . Then in the group ring K [ G ] {\displaystyle K[G]} one has ( 1 g ) ( 1 + g + + g n 1 ) = 1 g n = 0 {\displaystyle (1-g)(1+g+\cdots +g^{n-1})=1-g^{n}=0} , with neither factor being zero, so 1 g {\displaystyle 1-g} is a nonzero zero divisor in K [ G ] {\displaystyle K[G]} .

One-sided zero-divisor

  • Consider the ring of (formal) matrices ( x y 0 z ) {\displaystyle {\begin{pmatrix}x&y\\0&z\end{pmatrix}}} with x , z Z {\displaystyle x,z\in \mathbb {Z} } and y Z / 2 Z {\displaystyle y\in \mathbb {Z} /2\mathbb {Z} } . Then ( x y 0 z ) ( a b 0 c ) = ( x a x b + y c 0 z c ) {\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}}} and ( a b 0 c ) ( x y 0 z ) = ( x a y a + z b 0 z c ) {\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}}} . If x 0 z {\displaystyle x\neq 0\neq z} , then ( x y 0 z ) {\displaystyle {\begin{pmatrix}x&y\\0&z\end{pmatrix}}} is a left zero divisor if and only if x {\displaystyle x} is even, since ( x y 0 z ) ( 0 1 0 0 ) = ( 0 x 0 0 ) {\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}}} , and it is a right zero divisor if and only if z {\displaystyle z} is even for similar reasons. If either of x , z {\displaystyle x,z} is 0 {\displaystyle 0} , 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 S {\displaystyle S} be the set of all sequences of integers ( a 1 , a 2 , a 3 , . . . ) {\displaystyle (a_{1},a_{2},a_{3},...)} . Take for the ring all additive maps from S {\displaystyle S} to S {\displaystyle S} , with pointwise addition and composition as the ring operations. (That is, our ring is E n d ( S ) {\displaystyle \mathrm {End} (S)} , the endomorphism ring of the additive group S {\displaystyle S} .) Three examples of elements of this ring are the right shift R ( a 1 , a 2 , a 3 , . . . ) = ( 0 , a 1 , a 2 , . . . ) {\displaystyle R(a_{1},a_{2},a_{3},...)=(0,a_{1},a_{2},...)} , the left shift L ( a 1 , a 2 , a 3 , . . . ) = ( a 2 , a 3 , a 4 , . . . ) {\displaystyle L(a_{1},a_{2},a_{3},...)=(a_{2},a_{3},a_{4},...)} , and the projection map onto the first factor P ( a 1 , a 2 , a 3 , . . . ) = ( a 1 , 0 , 0 , . . . ) {\displaystyle P(a_{1},a_{2},a_{3},...)=(a_{1},0,0,...)} . All three of these additive maps are not zero, and the composites L P {\displaystyle LP} and P R {\displaystyle PR} are both zero, so L {\displaystyle L} is a left zero divisor and R {\displaystyle R} is a right zero divisor in the ring of additive maps from S {\displaystyle S} to S {\displaystyle S} . However, L {\displaystyle L} is not a right zero divisor and R {\displaystyle R} is not a left zero divisor: the composite L R {\displaystyle LR} is the identity. R L {\displaystyle RL} is a two-sided zero-divisor since R L P = 0 = P R L {\displaystyle RLP=0=PRL} , while L R = 1 {\displaystyle LR=1} is not in any direction.

Non-examples

  • The ring of integers modulo a prime number has no nonzero zero divisors. Since every nonzero element is a unit, this ring is a finite field.
  • More generally, a division ring has no nonzero zero divisors.
  • A non-zero commutative ring whose only zero divisor is 0 is called an integral domain.

Properties

  • In the ring of n × n matrices over a field, the left and right zero divisors coincide; they are precisely the singular matrices. In the ring of n × 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 = x 0.
  • 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 M a M {\displaystyle M\,{\stackrel {a}{\to }}\,M} 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

Notes

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

References

  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. ^ a b Hideyuki Matsumura (1980), Commutative algebra, 2nd edition, The Benjamin/Cummings Publishing Company, Inc., p. 12

Further reading