In abstract algebra, the total quotient ring, or total ring of fractions, is a construction that generalizes the notion of the field of fractions of an integral domain to commutative ringsR that may have zero divisors. The construction embeds R in a larger 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.
Let be a commutative ring and let be the set of elements which are not zero divisors in ; then is a multiplicatively closed set. Hence we may localize the ring at the set to obtain the total quotient ring .
If is a domain, then and the total quotient ring is the same as the field of fractions. This justifies the notation , which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain.
Since in the construction contains no zero divisors, the natural map is injective, so the total quotient ring is an extension of .
For a product ring A × B, the total quotient ring Q(A × B) is the product of total quotient rings Q(A) × Q(B). In particular, if A and B are integral domains, it is the product of quotient fields.
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, , and so . But since all these elements already have inverses, .
In a commutative von Neumann regular ringR, 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, .
Proposition — Let A be a reduced ring that has only finitely minimal prime ideals, . Then
Geometrically, is the Artinian scheme consisting (as a finite set) of the generic points of the irreducible components of .
Proof: Every element of Q(A) is either a unit or a zerodivisor. Thus, any proper ideal I of Q(A) is contained in the set of zerodivisors of Q(A); that set equals the union of the minimal prime ideals since Q(A) is reduced. By prime avoidance, I must be contained in some . Hence, the ideals are maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A),
Let S be the multiplicatively closed set of non-zerodivisors of A. By exactness of localization,
which is already a field and so must be .
If is a commutative ring and is any multiplicatively closed set in , the localization can still be constructed, but the ring homomorphism from to might fail to be injective. For example, if , then is the trivial ring.