# 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 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 $R$ .

## Examples

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 $Q(A/{\mathfrak {p}}_{i})$ . $\square$ ## Generalization

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.

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