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In [[abstract algebra]], the '''total quotient ring'''{{sfn|Matsumura|1980|p=12}} or '''total ring of fractions'''{{sfn|Matsumura|1989|p=[{{GBurl|yJwNrABugDEC|p=21}} 21]}} is a construction that generalizes the notion of the [[field of fractions]] of an [[integral domain]] to [[commutative ring]]s ''R'' that may have [[zero divisor]]s. The construction [[embedding|embeds]] ''R'' in a larger [[ring (mathematics)|ring]], giving every non-zero-divisor of ''R'' an inverse in the larger ring.  If the [[homomorphism]] from ''R'' to the new ring is to be [[injective]], no further elements can be given an inverse.
<!-- The idea is to formally invert as many elements of the ring as possible without making the ring smaller (i. e. without trivializing any nonzero element of the ring).-->

== Definition ==

Let <math>R</math> be a commutative ring and let <math>S</math> be the [[set (mathematics)|set]] of elements that are not zero divisors in <math>R</math>; then <math>S</math> is a [[multiplicatively closed set]]. Hence we may [[localization of a ring|localize]] the ring <math>R</math> at the set <math>S</math> to obtain the total quotient ring <math>S^{-1}R=Q(R)</math>.

If <math>R</math> is a [[integral domain|domain]], then <math>S = R-\{0\}</math> and the total quotient ring is the same as the field of fractions. This justifies the notation <math>Q(R)</math>, which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain.

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

== Examples ==

*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 function]]s on an [[open set]] ''D'' of [[complex number]]s, the total quotient ring is the ring of [[meromorphic function]]s on ''D'', even if ''D'' is not [[connected set|connected]].

*In an [[Artinian ring]], all elements are [[unit (ring theory)|units]] or zero divisors. Hence the set of non-zero-divisors is the [[group (mathematics)|group]] of units of the ring, <math>R^{\times}</math>, and so <math>Q(R) = (R^{\times})^{-1}R</math>. But since all these elements already have inverses, <math>Q(R) = R</math>. 

*In a commutative [[von Neumann regular ring|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''&nbsp;=&nbsp;''axa'' for some ''x'' in ''R'', giving the equation ''a''(''xa''&nbsp;−&nbsp;1)&nbsp;=&nbsp;0. Since ''a'' is not a zero divisor, ''xa''&nbsp;=&nbsp;1, showing ''a'' is a unit. Here again, <math>Q(R) = R</math>.

*In [[algebraic geometry]] one considers a [[sheaf (mathematics)|sheaf]] of total quotient rings on a [[scheme (mathematics)|scheme]], and this may be used to give the definition of a [[Cartier divisor]].

== The total ring of fractions of a reduced ring ==
{{math_theorem|name=Proposition|Let ''A'' be a [[reduced ring]] that has only finitely many [[minimal prime ideal]]s, <math>\mathfrak{p}_1, \dots, \mathfrak{p}_r</math> (e.g., a [[Noetherian ring|Noetherian]] reduced ring). Then
:<math>Q(A) \simeq \prod_{i=1}^r Q(A/\mathfrak{p}_i).</math>
Geometrically, <math>\operatorname{Spec}(Q(A))</math> is the [[Artinian scheme]] consisting (as a finite set) of the generic points of the irreducible components of <math>\operatorname{Spec} (A)</math>.}}

Proof: Every element of ''Q''(''A'') is either a unit or a zero divisor. Thus, any [[proper ideal|proper]] [[ideal (ring theory)|ideal]] ''I'' of ''Q''(''A'') is contained in the set of zero divisors of ''Q''(''A''); that set equals the [[union (set theory)|union]] of the minimal prime ideals <math>\mathfrak{p}_i Q(A)</math> since ''Q''(''A'') is [[reduced ring|reduced]].  By [[prime avoidance]], ''I'' must be contained in some <math>\mathfrak{p}_i Q(A)</math>. Hence, the ideals <math>\mathfrak{p}_i Q(A)</math> are [[maximal ideal]]s of ''Q''(''A'').  Also, their [[intersection (set theory)|intersection]] is [[zero ideal|zero]]. Thus, by the [[Chinese remainder theorem]] applied to ''Q''(''A''), 
:<math>Q(A) \simeq \prod_i Q(A)/\mathfrak{p}_i Q(A)</math>.
Let ''S'' be the [[multiplicatively closed set]] of non-zero-divisors of ''A''.  By [[exact functor|exactness]] of localization, 
:<math>Q(A)/\mathfrak{p}_i Q(A) = A[S^{-1}] / \mathfrak{p}_i A[S^{-1}] = (A / \mathfrak{p}_i)[S^{-1}]</math>,
which is already a [[field (mathematics)|field]] and so must be <math>Q(A/\mathfrak{p}_i)</math>. <math>\square</math>

== Generalization ==

If <math>R</math> is a commutative ring and <math>S</math> is any [[multiplicatively closed set]] in <math>R</math>, the [[Localization of a ring|localization]] <math>S^{-1}R</math> can still be constructed, but the [[ring homomorphism]] from <math>R</math> to <math>S^{-1}R</math> might fail to be injective.  For example, if <math>0 \in S</math>, then <math>S^{-1}R</math> is the [[trivial ring]].

==Citations==

<references/>

==References==
{{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]]