Zero divisor
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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, or equivalently if the map from R to R that sends x to ax is not injective. 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. 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, 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}},}
![{\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}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1a41fc370c9b2b4e7d8b88f0c4239fc4da741e4)
(
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
{\displaystyle K}
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
{\displaystyle K}
.
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
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. The set of M-regular elements is a multiplicative set in R.
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
- ^ Since the map is not injective, we have ax = ay, in which x differs from y, and thus a(x − y) = 0.
References
- ^ N. Bourbaki (1989), Algebra I, Chapters 1–3, Springer-Verlag, p. 98
- ^ Charles Lanski (2005), Concepts in Abstract Algebra, American Mathematical Soc., p. 342
- ^ Nicolas Bourbaki (1998). Algebra I. Springer Science+Business Media. p. 15.
- ^ a b Hideyuki Matsumura (1980), Commutative algebra, 2nd edition, The Benjamin/Cummings Publishing Company, Inc., p. 12
Further reading