Inabstract algebra, thetotal quotient ring^[1] ortotal ring of fractions^[2] is a construction that generalizes the notion of thefield of fractions of anintegral domain tocommutative ringsR that may havezero divisors. The constructionembedsR in a largerring, giving every non-zero-divisor ofR an inverse in the larger ring. If thehomomorphism fromR to the new ring is to beinjective, no further elements can be given an inverse.

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

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

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

• For aproduct ringA ×B, the total quotient ringQ(A ×B) is the product of total quotient ringsQ(A) ×Q(B). In particular, ifA andB are integral domains, it is the product of quotient fields.

• For the ring ofholomorphic functions on anopen setD ofcomplex numbers, the total quotient ring is the ring ofmeromorphic functions onD, even ifD is notconnected.

• In anArtinian ring, all elements areunits or zero divisors. Hence the set of non-zero-divisors is thegroup of units of the ring,${\displaystyle R^{\times }}$, and so${\displaystyle Q(R)=(R^{\times }{)}^{-1}R}$. But since all these elements already have inverses,${\displaystyle Q(R)=R}$.

• In a commutativevon Neumann regular ringR, the same thing happens. Supposea inR is not a zero divisor. Then in a von Neumann regular ringa = axa for somex inR, giving the equationa(xa − 1) = 0. Sincea is not a zero divisor,xa = 1, showinga is a unit. Here again,${\displaystyle Q(R)=R}$.

• Inalgebraic geometry one considers asheaf of total quotient rings on ascheme, and this may be used to give the definition of aCartier divisor.

Proof: Every element ofQ(A) is either a unit or a zero divisor. Thus, anyproperidealI ofQ(A) is contained in the set of zero divisors ofQ(A); that set equals theunion of the minimal prime ideals${\displaystyle {\mathfrak{p}}_{i}Q(A)}$ sinceQ(A) isreduced. Byprime avoidance,I must be contained in some${\displaystyle {\mathfrak{p}}_{i}Q(A)}$. Hence, the ideals${\displaystyle {\mathfrak{p}}_{i}Q(A)}$ aremaximal ideals ofQ(A). Also, theirintersection iszero. Thus, by theChinese remainder theorem applied toQ(A),

${\displaystyle Q(A)\simeq \prod _{i}Q(A)/{\mathfrak{p}}_{i}Q(A)}$.

LetS be themultiplicatively closed set of non-zero-divisors ofA. Byexactness of localization,

${\displaystyle Q(A)/{\mathfrak{p}}_{i}Q(A)=A[S^{-1}]/{\mathfrak{p}}_{i}A[S^{-1}]=(A/{\mathfrak{p}}_i)[S^{-1}]}$,

which is already afield and so must be${\displaystyle Q(A/{\mathfrak{p}}_i)}$.${\displaystyle ◻}$

If${\displaystyle R}$ is a commutative ring and${\displaystyle S}$ is anymultiplicatively closed set in${\displaystyle R}$, thelocalization${\displaystyle S^{-1}R}$ can still be constructed, but thering homomorphism from${\displaystyle R}$ to${\displaystyle S^{-1}R}$ might fail to be injective. For example, if${\displaystyle 0\in S}$, then${\displaystyle S^{-1}R}$ is thetrivial ring.

• Commutative algebra
• Ring theory