Total ring of fractions

In abstract algebra, the total quotient ring,[1] or total ring of fractions,[2] is a construction that generalizes the notion of the field of fractions of an integral domain to commutative rings R 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. Nothing more in A can be given an inverse, if one wants the homomorphism from A to the new ring to be injective.


Let R be a commutative ring and let S be the set of elements which 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-\{0\} 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 \to Q(R) is injective, so the total quotient ring is an extension of R.


The total quotient ring Q(A \times B) of a product ring is the product of total quotient rings Q(A) \times Q(B). In particular, if A and B are integral domains, it is the product of quotient fields.

The total quotient ring of the ring of holomorphic functions on an open set D of complex numbers 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^{\times}, and so Q(R) = (R^{\times})^{-1}R. But since all these elements already have inverses, Q(R) = R.

The same thing happens in a commutative von Neumann regular ring R. 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.

The total ring of fractions of a reduced ring

There is an important fact:

Proposition  Let A be a Noetherian reduced ring with the minimal prime ideals \mathfrak{p}_1, \dots, \mathfrak{p}_r. Then

Q(A) \simeq \prod_1^r Q(A/\mathfrak{p}_i).

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

Proof: Every element of Q(A) is either a unit or a zerodivisor. Thus, any proper ideal I of Q(A) must consist of zerodivisors. Since the set of zerodivisors of Q(A) is the union of the minimal prime ideals \mathfrak{p}_i Q(A) as Q(A) is reduced, by prime avoidance, I must be contained in some \mathfrak{p}_i Q(A). Hence, the ideals \mathfrak{p}_i Q(A) are the maximal ideals of Q(A), whose intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A), we have:

Q(A) \simeq \prod_i Q(A)/\mathfrak{p}_i Q(A).

Finally, Q(A)/\mathfrak{p}_i Q(A) is the residue field of \mathfrak{p}_i. Indeed, writing S for the multiplicatively closed set of non-zerodivisors, by the exactness of localization,

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 a field and so must be Q(A/\mathfrak{p}_i). \square


If R is a commutative ring and S is any multiplicative subset 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 \in S, then S^{-1}R is the trivial ring.


  1. Matsumura (1980), p. 12
  2. Matsumura (1989), p. 21


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