nLab cancellative element of a commutative ring

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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).

Contents

Definition

Given a commutative ring RR, an element eRe \in R is left cancellative or left regular if ea=ebe \cdot a = e \cdot b for all aRa \in R and bRb \in R, then a=ba = b.

isLeftCancellative(e)a,bR.(ea=eb)(a=b)\mathrm{isLeftCancellative}(e) \coloneqq \forall a,b \in R.(e \cdot a = e \cdot b) \implies (a = b)

An element eRe \in R is right cancellative or right regular if ae=bea \cdot e = b \cdot e for all aRa \in R and bRb \in R, then a=ba = b.

isRightCancellative(e)a,bM.(ae=be)(a=b)\mathrm{isRightCancellative}(e) \coloneqq \forall a,b \in M.(a \cdot e = b \cdot e) \implies (a = b)

An element eRe \in R is cancellative or regular if it is both left cancellative and right cancellative.

isCancellative(e)isLeftCancellative(e)isRightCancellative(e)\mathrm{isCancellative}(e) \coloneqq \mathrm{isLeftCancellative}(e) \wedge \mathrm{isRightCancellative}(e)

Multiplicative subset of cancellative elements

The multiplicative subset of cancellative elements in RR is the multiplicative subset of all cancellative elements in RR

Can(R){eM|isCancellative(e)}\mathrm{Can}(R) \coloneqq \{e \in M \vert \mathrm{isCancellative}(e)\}

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 Reg(R)\mathrm{Reg}(R).

This is also called the filter of regular elements in RR, as in Lombardi & Quitté 2010. Similarly as above, ‘regular element’ could be replaced with ‘cancellative element’.

Since the multiplicative identity element 11 is always cancellative, the multiplicative subset of all cancellative elements forms a cancellative monoid.

Properties

Proposition

For a commutative ring RR, given elements a,bRa, b \in R, if the product aba \cdot b is cancellative, then both aa and bb are cancellative.

By this proposition, a commutative ring RR is a strict approximate integral domain if the following condition holds:

  • For all elements a,bRa, b \in R, if the a+ba + b is cancellative, then either aa is cancellative or bb is cancellative.

Theorem

For a strict approximate integral domain RR, the addition and multiplication operations on RR are strongly extensional with respect to the canonical apartness relation #\# defined by x#yx \# y iff xyx - y is cancellative.

In this way RR becomes an internal commutative ring object in the category Apart\mathrm{Apart}, consisting of sets with apartness relations and maps (strongly extensional functions) between them.

Proof

Recall that products X×YX \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)(x, y) \# (x', y') iff x#xx \# x' in XX or y#yy \# y' in YY. Recall also that a function f:XYf: X \to Y between sets with apartness relations is strongly extensional if f(x)#f(y)f(x) \# f(y) implies x#yx \# y.

For addition, for all elements aRa \in R, bRb \in R, aRa' \in R, bRb' \in R, if (a+b)#(a+b)(a + b) \# (a' + b'), then a+b(a+b)=(ab)+(ab)a + b - (a' + b') = (a - b') + (a - b') is cancellative, so aaa - a' or bbb - b' is cancellative since for all elements aRa \in R and bRb \in R, if the sum a+ba + b is cancellative, then either aa is cancellative or bb is cancellative, whence (a,b)#(a,b)(a, b) # (a', b'). Thus addition is strongly extensional.

For multiplication, for all elements aRa \in R, bRb \in R, aRa' \in R, bRb' \in R, if ab#aba \cdot b # a' \cdot b', then ababa \cdot b - a' \cdot b' is cancellative. Write abab=(aa)b+a(bb)a \cdot b - a' \cdot b' = (a - a') \cdot b + a' \cdot (b - b'). Since for all elements aRa \in R and bRb \in R, if the sum a+ba + b is cancellative, then either aa is cancellative or bb is cancellative, either (aa)b(a - a') \cdot b is cancellative or a(bb)a' \cdot (b - b') is cancellative. Since for all elements aRa \in R and bRb \in R, if aba \cdot b is cancellative, then aa is cancellative and bb is cancellative, either aaa - a' is cancellative or bbb - b' is, whence (a,b)#(a,b)(a, b) # (a', b'). So multiplication is also strongly extensional.

 Relation with zero-divisors

Theorem

Given a commutative ring RR, if an element eRe \in R is a cancellative element, then it is a non-zero-divisor, where non-zero-divisor is defined as an element aRa \in R such that for every element b:Rb:R, ba=0b \cdot a = 0 or ab=0a \cdot b = 0 implies that b=0b = 0.

Proof

Suppose that eRe \in R is cancellative. This means that for all elements aRa \in R and bRb \in R, ae=bea \cdot e = b \cdot e implies that a=ba = b and ea=ebe \cdot a = e \cdot b implies that a=ba = b. For the first equation, subtracting beb \cdot e from both sides of the equation leads to (ab)e=0(a - b) \cdot e = 0, and for the second equation, subtraction ebe \cdot b from both sides of the equation leads to e(ab)=0e \cdot (a - b) = 0. Subtracting bb from both sides of a=ba = b leads to ab=0a - b = 0. Defining the element c=abc = a - b results in the condition that for every element cRc \in R, ce=0c \cdot e = 0 implies that c=0c = 0 and ec=0e \cdot c = 0 implies that c=0c = 0, which implies that ee is a non-zero-divisor.

Theorem

Given a commutative ring RR, if an element eRe \in R is a non-zero-divisor, then it is cancellative.

Proof

Suppose that eRe \in R is a non-zero-divisor. This means that for every element cRc \in R, ce=0c \cdot e = 0 or ec=0e \cdot c = 0 implies that c=0c = 0. However, since RR is a commutative ring, if ce=0c \cdot e = 0, then ec=0e \cdot c = 0, so the statement ce=0c \cdot e = 0 or ec=0e \cdot c = 0 implies that c=0c = 0 implies the statement that ce=0c \cdot e = 0 implies that c=0c = 0 and ec=0e \cdot c = 0 implies that c=0c = 0. Since RR is an abelian group, by definition of an abelian group, the image of the binary subtraction function aba - b is RR itself, and thus, one could replace cc with aba - b for elements aRa \in R and bRb \in R, resulting in the statement that (ab)e=0(a - b) \cdot e = 0 implies that (ab)=0(a - b) = 0 and e(ab)=0e \cdot (a - b) = 0 implies that (ab)=0(a - b) = 0. Adding beb \cdot e to each side of the first equation, ebe \cdot b to each side of the second equation, and bb to each side fo the third equation leads to the statement that ae=bea \cdot e = b \cdot e implies that a=ba = b and ea=ebe \cdot a = e \cdot b implies that a=ba = 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,

Theorem

In a commutative ring RR, an element eRe \in R is a zero-divisor if and only if it is non-cancellative

isZeroDivisor(e)¬isCancellative(e)\mathrm{isZeroDivisor}(e) \iff \neg \mathrm{isCancellative}(e)

Integral domains

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:

Definition

A commutative ring RR is a integral domain if an element is non-cancellative (or equivalently, a zero-divisor) iff it is zero. In addition to 010\neq 1, this condition means that every non-cancellative element (or equivalenty, zero-divisor) is zero.

Definition

A commutative ring RR is a Heyting integral domain if it is an integral domain and additionally, for all elements aRa \in R and bRb \in R, if the sum a+ba + b is cancellative, then either aa is cancellative or bb is cancellative.

Theorem

In addition to 0#10 \# 1, the above condition in a Heyting integral domain then means that every element apart from 00 is cancellative.

Definition

A commutative ring RR is a discrete integral domain if all elements eRe \in R are cancellative xor equal to zero. This condition means that every element is either 00 or cancellative, and it also implies that 010 \neq 1.

Fields

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.

Ring of fractions

For every commutative ring RR, the ring of fractions is defined in Quinn2009 to be the localization of RR at the multiplicative subset of cancellative elements, R[Can(R) 1]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[Can(R) 1]R[\mathrm{Can}(R)^{-1}] is the group of units R[Can(R) 1] ×R[\mathrm{Can}(R)^{-1}]^\times in R[Can(R) 1]R[\mathrm{Can}(R)^{-1}].

If RR is an integral domain, then R[Can(R) 1]R[\mathrm{Can}(R)^{-1}] is the field of fractions of RR.

Unique factorization domains

In Lombardi & Quitté 2010, a unique factorization domain is defined as a GCD domain RR for which the quotient monoid of the multiplicative subset of cancellative elements by the group of units Can(R)/R ×\mathrm{Can}(R)/R^\times admits a complete factorization.

See also

References

Last revised on August 19, 2024 at 15:07:48. See the history of this page for a list of all contributions to it.