Total ring of fractions

Let $R$ be a commutative ring and let $S$ be the set of elements that are not zero divisors in $R$; then $S$ is a multiplicatively closed set. Hence we may localize the ring $R$ at the set $S$ to obtain the total quotient ring $S^\{-1\}R=Q(R)$.

If $R$ is a domain, then $S\; =\; R-\backslash \{0\backslash \}$ and the total quotient ring is the same as the field of fractions. This justifies the notation $Q(R)$, which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain.

Since $S$ in the construction contains no zero divisors, the natural map $R\; \backslash to\; Q(R)$ is injective, so the total quotient ring is an extension of $R$.

• For a product ring {{nowrap|A × B}}, the total quotient ring {{nowrap|Q(A × B)}} is the product of total quotient rings {{nowrap|Q(A) × Q(B)}}. In particular, if A and B are integral domains, it is the product of quotient fields.

• For the ring of holomorphic functions on an open set D of complex numbers, the total quotient ring is the ring of meromorphic functions on D, even if D is not connected.

• In an Artinian ring, all elements are units or zero divisors. Hence the set of non-zero-divisors is the group of units of the ring, $R^\{\backslash times\}$, and so $Q(R)\; =\; (R^\{\backslash times\})^\{-1\}R$. But since all these elements already have inverses, $Q(R)\; =\; R$.

• In a commutative von Neumann regular ring R, the same thing happens. Suppose a in R is not a zero divisor. Then in a von Neumann regular ring a = axa for some x in R, giving the equation a(xa − 1) = 0. Since a is not a zero divisor, xa = 1, showing a is a unit. Here again, $Q(R)\; =\; R$.

• In algebraic geometry one considers a sheaf of total quotient rings on a scheme, and this may be used to give the definition of a Cartier divisor.

{{math_theorem|name=Proposition|Let A be a reduced ring that has only finitely many minimal prime ideals, $\backslash mathfrak\{p\}\_1,\; \backslash dots,\; \backslash mathfrak\{p\}\_r$ (e.g., a Noetherian reduced ring). Then

:$Q(A)\; \backslash simeq\; \backslash prod\_\{i=1\}^r\; Q(A/\backslash mathfrak\{p\}\_i).$

Geometrically, $\backslash operatorname\{Spec\}(Q(A))$ is the Artinian scheme consisting (as a finite set) of the generic points of the irreducible components of $\backslash operatorname\{Spec\}\; (A)$.}}

Proof: Every element of Q(A) is either a unit or a zero divisor. Thus, any proper ideal I of Q(A) is contained in the set of zero divisors of Q(A); that set equals the union of the minimal prime ideals $\backslash mathfrak\{p\}\_i\; Q(A)$ since Q(A) is reduced. By prime avoidance, I must be contained in some $\backslash mathfrak\{p\}\_i\; Q(A)$. Hence, the ideals $\backslash mathfrak\{p\}\_i\; Q(A)$ are maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A),

:$Q(A)\; \backslash simeq\; \backslash prod\_i\; Q(A)/\backslash mathfrak\{p\}\_i\; Q(A)$.

Let S be the multiplicatively closed set of non-zero-divisors of A. By exactness of localization,

:$Q(A)/\backslash mathfrak\{p\}\_i\; Q(A)\; =\; A[S^\{-1\}]\; /\; \backslash mathfrak\{p\}\_i\; A[S^\{-1\}]\; =\; (A\; /\; \backslash mathfrak\{p\}\_i)[S^\{-1\}]$,

which is already a field and so must be $Q(A/\backslash mathfrak\{p\}\_i)$. $\backslash square$

If $R$ is a commutative ring and $S$ is any multiplicatively closed set in $R$, the localization $S^\{-1\}R$ can still be constructed, but the ring homomorphism from $R$ to $S^\{-1\}R$ might fail to be injective. For example, if $0\; \backslash in\; S$, then $S^\{-1\}R$ is the trivial ring.

{{refbegin}}

• {{citation|last=Matsumura|first=Hideyuki|authorlink = Hideyuki Matsumura|title=Commutative algebra|year=1980 |publisher=Benjamin/Cummings |isbn=978-0-8053-7026-3 |edition=2nd |oclc=988482880}}
• {{citation|last=Matsumura|first=Hideyuki|title=Commutative ring theory|year=1989 |isbn=978-0-521-36764-6 |publisher=Cambridge University Press |oclc=23133540 |url=}}

{{refend}}

Category:Commutative algebra

Category:Ring theory

de:Lokalisierung (Algebra)#Totalquotientenring