Definition:Zero Divisor/Ring

From ProofWiki
Jump to navigation Jump to search

Definition

Let $\struct {R, +, \circ}$ be a ring.


A zero divisor (in $R$) is an element $x \in R$ such that either:

$\exists y \in R^*: x \circ y = 0_R$

or:

$\exists y \in R^*: y \circ x = 0_R$

where $R^*$ is defined as $R \setminus \set {0_R}$.


That is, such that $x$ is either a left zero divisor or a right zero divisor.


The expression:

$x$ is a zero divisor

can be written:

$x \divides 0_R$


Left Zero Divisor

A left zero divisor (in $R$) is an element $x \in R$ such that:

$\exists y \in R^*: x \circ y = 0_R$


Right Zero Divisor

A right zero divisor (in $R$) is an element $x \in R$ such that:

$\exists y \in R^*: y \circ x = 0_R$


Also defined as

Some sources define a zero divisor as an element $x \in R_{\ne 0_R}$ such that:

$\exists y \in R_{\ne 0_R}: x \circ y = 0_R$

where $R_{\ne 0_R}$ is defined as $R \setminus \set {0_R}$.

That is, the element $0_R$ itself is not classified as a zero divisor.

This definition is the same as the one given on this website as a proper zero divisor.


Also known as

Some sources hyphenate, as: zero-divisor.

Some sources run the words together: zerodivisor.

Some use the more explicit and pedantic divisor of zero.


Warning

Beware the terminology divisor of zero.

It is easy to confuse this with the fact that, for every element $a$ of a ring $R$, from Ring Product with Zero we have that $a \circ 0_R = 0_R$.

Hence one may say that every such element is a divisor of zero.

However, the concept of a zero divisor specifically requires that the $b$ in $a \circ b = 0_R$ is not zero.


Notation

The conventional notation for $x$ is a divisor of $y$ is "$x \mid y$", but there is a growing trend to follow the notation "$x \divides y$", as espoused by Knuth etc.

From Ronald L. Graham, Donald E. Knuth and Oren Patashnik: Concrete Mathematics: A Foundation for Computer Science (2nd ed.):

The notation '$m \mid n$' is actually much more common than '$m \divides n$' in current mathematics literature. But vertical lines are overused -- for absolute values, set delimiters, conditional probabilities, etc. -- and backward slashes are underused. Moreover, '$m \divides n$' gives an impression that $m$ is the denominator of an implied ratio. So we shall boldly let our divisibility symbol lean leftward.


An unfortunate unwelcome side-effect of this notational convention is that to indicate non-divisibility, the conventional technique of implementing $/$ through the notation looks awkward with $\divides$, so $\not \! \backslash$ is eschewed in favour of $\nmid$.


Some sources use $\ \vert \mkern -10mu {\raise 3pt -} \ $ or similar to denote non-divisibility.


Examples

Order $2$ Square Matrices: Example $1$

Let $R$ be the ring square matrices of order $2$ over a field with unity $1$ and zero $0$.

Let:

\(\ds \mathbf A\) \(=\) \(\ds \begin {bmatrix} 1 & 0 \\ 0 & 0 \end {bmatrix}\)
\(\ds \mathbf B\) \(=\) \(\ds \begin {bmatrix} 0 & 0 \\ 0 & 1 \end {bmatrix}\)

Then:

$\mathbf A \mathbf B = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix} = \mathbf B \mathbf A$

Thus both $\mathbf A$ and $\mathbf B$ are zero divisors of $R$.


Order $2$ Square Matrices: Example $2$

Let $R$ be the ring square matrices of order $2$ over the real numbers.

Then:

$\begin {bmatrix} 0 & 1 \\ 0 & 0 \end {bmatrix} \begin {bmatrix} 1 & 0 \\ 0 & 0 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$

demonstrating that $\begin {bmatrix} 0 & 1 \\ 0 & 0 \end {bmatrix}$ and $\begin {bmatrix} 1 & 0 \\ 0 & 0 \end {bmatrix}$ are zero divisors of $R$.


Order $2$ Square Matrices: Example $3$

Let $R$ be the ring square matrices of order $2$ over the real numbers.

Then:

$\begin {bmatrix} 0 & 0 \\ 1 & 1 \end {bmatrix} \begin {bmatrix} 0 & 1 \\ 0 & -1 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$

demonstrating that $\begin {bmatrix} 0 & 0 \\ 1 & 1 \end {bmatrix}$ and $\begin {bmatrix} 0 & 1 \\ 0 & -1 \end {bmatrix}$ are zero divisors of $R$.


Order $2$ Square Matrices: Example $4$

Let $R$ be the ring of square matrices of order $2$ over the real numbers.

Then:

$\begin {bmatrix} 1 & 1 \\ 1 & 1 \end {bmatrix} \begin {bmatrix} -1 & 1 \\ 1 & -1 \end {bmatrix} = \begin {bmatrix} 0 & 0 \\ 0 & 0 \end {bmatrix}$

demonstrating that $\begin {bmatrix} 1 & 1 \\ 1 & 1 \end {bmatrix}$ and $\begin {bmatrix} -1 & 1 \\ 1 & -1 \end {bmatrix}$ are zero divisors of $R$.


Also see

  • Results about zero divisors can be found here.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Definition:Zero_Divisor/Ring&oldid=727665"