Total ring of fractions

Definition

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-{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.

Examples

• 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.

• 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^{\times}, and so Q(R) = (R^{\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.

The total ring of fractions of a reduced ring

:Q(A) \simeq \prod_{i=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 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 \mathfrak{p}_i Q(A) since 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 maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A), :Q(A) \simeq \prod_i Q(A)/\mathfrak{p}_i Q(A). Let S be the multiplicatively closed set of non-zero-divisors of A. By 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

Generalization

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 \in S, then S^{-1}R is the trivial ring.

Citations

References

de:Lokalisierung (Algebra)#Totalquotientenring