This entry is about regular elements in ring theory and commutative algebra. For regular elements in formal logic and topology, see regular element. For regular elements in physics/quantum field theory see at regularization (physics).
symmetric monoidal (∞,1)-category of spectra
Given a commutative ring $R$, an element $e \in R$ is left cancellative or left regular if $e \cdot a = e \cdot b$ for all $a \in R$ and $b \in R$, then $a = b$.
An element $e \in R$ is right cancellative or right regular if $a \cdot e = b \cdot e$ for all $a \in R$ and $b \in R$, then $a = b$.
An element $e \in R$ is cancellative or regular if it is both left cancellative and right cancellative.
The multiplicative subset of cancellative elements in $R$ is the multiplicative subset of all cancellative elements in $R$
The term ‘cancellative element’ could be replaced with the synonym ‘regular element’. If the synonym ‘regular element’ is used in place of ‘cancellative element’, such as in Lombardi & Quitté 2010, then this multiplicative subset is typically written as $\mathrm{Reg}(R)$.
This is also called the filter of regular elements in $R$, as in Lombardi & Quitté 2010. Similarly as above, ‘regular element’ could be replaced with ‘cancellative element’.
Since the multiplicative identity element $1$ is always cancellative, the multiplicative subset of all cancellative elements forms a cancellative monoid.
Given elements $a \in R$ and $b \in R$, if the product $a \cdot b$ is cancellative, then both $a$ and $b$ is cancellative.
Given an approximate integral domain $R$, if for all elements $a \in R$ and $b \in R$, if the sum $a + b$ is cancellative, then either $a$ is cancellative or $b$ is cancellative, then the addition and multiplication operations on $R$ are strongly extensional with respect to the canonical apartness relation $\#$ defined by $x \# y$ iff $x - y$ is cancellative.
In this way $R$ becomes an internal commutative ring object in the category $\mathrm{Apart}$, consisting of sets with apartness relations and maps (strongly extensional functions) between them.
Recall that products $X \times Y$ in the category of sets with apartness relations is the cartesian product of the underlying sets equipped with the apartness relation defined by $(x, y) \# (x', y')$ iff $x \# x'$ in $X$ or $y \# y'$ in $Y$. Recall also that a function $f: X \to Y$ between sets with apartness relations is strongly extensional if $f(x) \# f(y)$ implies $x \# y$.
For addition, for all elements $a \in R$, $b \in R$, $a' \in R$, $b' \in R$, if $(a + b) \# (a' + b')$, then $a + b - (a' + b') = (a - b') + (a - b')$ is cancellative, so $a - a'$ or $b - b'$ is cancellative since for all elements $a \in R$ and $b \in R$, if the sum $a + b$ is cancellative, then either $a$ is cancellative or $b$ is cancellative, whence $(a, b) # (a', b')$. Thus addition is strongly extensional.
For multiplication, for all elements $a \in R$, $b \in R$, $a' \in R$, $b' \in R$, if $a \cdot b # a' \cdot b'$, then $a \cdot b - a' \cdot b'$ is cancellative. Write $a \cdot b - a' \cdot b' = (a - a') \cdot b + a' \cdot (b - b')$. Since for all elements $a \in R$ and $b \in R$, if the sum $a + b$ is cancellative, then either $a$ is cancellative or $b$ is cancellative, either $(a - a') \cdot b$ is cancellative or $a' \cdot (b - b')$ is cancellative. Since for all elements $a \in R$ and $b \in R$, if $a \cdot b$ is cancellative, then $a$ is cancellative and $b$ is cancellative, either $a - a'$ is cancellative or $b - b'$ is, whence $(a, b) # (a', b')$. So multiplication is also strongly extensional.
Given a commutative ring $R$, if an element $e \in R$ is a cancellative element, then it is a non-zero-divisor, where non-zero-divisor is defined as an element $a \in R$ such that for every element $b:R$, $b \cdot a = 0$ or $a \cdot b = 0$ implies that $b = 0$.
Suppose that $e \in R$ is cancellative. This means that for all elements $a \in R$ and $b \in R$, $a \cdot e = b \cdot e$ implies that $a = b$ and $e \cdot a = e \cdot b$ implies that $a = b$. For the first equation, subtracting $b \cdot e$ from both sides of the equation leads to $(a - b) \cdot e = 0$, and for the second equation, subtraction $e \cdot b$ from both sides of the equation leads to $e \cdot (a - b) = 0$. Subtracting $b$ from both sides of $a = b$ leads to $a - b = 0$. Defining the element $c = a - b$ results in the condition that for every element $c \in R$, $c \cdot e = 0$ implies that $c = 0$ and $e \cdot c = 0$ implies that $c = 0$, which implies that $e$ is a non-zero-divisor.
Given a commutative ring $R$, if an element $e \in R$ is a non-zero-divisor, then it is cancellative.
Suppose that $e \in R$ is a non-zero-divisor. This means that for every element $c \in R$, $c \cdot e = 0$ or $e \cdot c = 0$ implies that $c = 0$. However, since $R$ is a commutative ring, if $c \cdot e = 0$, then $e \cdot c = 0$, so the statement $c \cdot e = 0$ or $e \cdot c = 0$ implies that $c = 0$ implies the statement that $c \cdot e = 0$ implies that $c = 0$ and $e \cdot c = 0$ implies that $c = 0$. Since $R$ is an abelian group, by definition of an abelian group, the image of the binary subtraction function $a - b$ is $R$ itself, and thus, one could replace $c$ with $a - b$ for elements $a \in R$ and $b \in R$, resulting in the statement that $(a - b) \cdot e = 0$ implies that $(a - b) = 0$ and $e \cdot (a - b) = 0$ implies that $(a - b) = 0$. Adding $b \cdot e$ to each side of the first equation, $e \cdot b$ to each side of the second equation, and $b$ to each side fo the third equation leads to the statement that $a \cdot e = b \cdot e$ implies that $a = b$ and $e \cdot a = e \cdot b$ implies that $a = b$, which is precisely the definition of cancellative element. Thus, every non-zero-divisor is a cancellative element.
Thus, we have established that cancellative elements and non-zero-divisors are the same thing in a commutative ring. However, the proof relies on the abelian group structure of commutative rings, and this property does not necessarily hold in other algebraic structures where the concepts of cancellative element and non-zero-divisor make sense, such as in rigs or absorption monoids.
Since a zero-divisor is defined in the nLab as not being a non-zero-divisor,
In a commutativ ring $R$, an elmement $e \in R$ is a zero-divisor if and only if it is non-cancellative
The theorems relating cancellative elements to zero-divisors provide alternative definition of the various (commutative) integral domains in constructive mathematics in terms of cancellative elements, in analogy with the definition of fields in terms of invertible elements:
A commutative ring $R$ is a integral domain if an element is non-cancellative (or equivalently, a zero-divisor) iff it is zero. In addition to $0\neq 1$, this condition means that every non-cancellative element (or equivalenty, zero-divisor) is zero.
A commutative ring $R$ is a Heyting integral domain if it is an integral domain and additionally, for all elements $a \in R$ and $b \in R$, if the sum $a + b$ is cancellative, then either $a$ is cancellative or $b$ is cancellative.
In addition to $0 \# 1$, the above condition in a Heyting integral domain then means that every element apart from $0$ is cancellative.
A commutative ring $R$ is a discrete integral domain if all elements $e \in R$ are cancellative xor equal to zero. This condition means that every element is either $0$ or cancellative, and it also implies that $0 \neq 1$.
Given the above definitions of an integral domain, a field could be defined as an integral domain where every cancellative element is a unit, or equivalently, an integral domain whose multiplicative subset of cancellative elements is the group of units.
For every commutative ring $R$, the ring of fractions is defined in Quinn2009 to be the localization of $R$ at the multiplicative subset of cancellative elements, $R[\mathrm{Can}(R)^{-1}]$. This is similar to the Grothendieck group construction of a general cancellative monoid: the multiplicative subset of cancellative elements in $R[\mathrm{Can}(R)^{-1}]$ is the group of units $R[\mathrm{Can}(R)^{-1}]^\times$ in $R[\mathrm{Can}(R)^{-1}]$.
If $R$ is an integral domain, then $R[\mathrm{Can}(R)^{-1}]$ is the field of fractions of $R$.
In Lombardi & Quitté 2010, a unique factorization domain is defined as a GCD domain $R$ for which the quotient monoid of the multiplciative subset of cancellative elements by the group of units $\mathrm{Can}(R)/R^\times$ admits a complete factorization.
Henri Lombardi, Claude Quitté (2010): Commutative algebra: Constructive methods (Finite projective modules) Translated by Tania K. Roblo, Springer (2015) (doi:10.1007/978-94-017-9944-7, pdf)
Frank Quinn, Proof Projects for Teachers (2009) [pdf, pdf]
Last revised on December 9, 2022 at 00:44:08. See the history of this page for a list of all contributions to it.