Total ring of fractions

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 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. If the homomorphism from R to the new ring is to be injective, no further elements can be given an inverse.

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
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.