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## Summary

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

• 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, $R^{\times }$ , 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, $Q(R)=R$ .

## The total ring of fractions of a reduced ring

Proposition — Let A be a reduced ring that has only finitely minimal prime ideals, ${\mathfrak {p}}_{1},\dots ,{\mathfrak {p}}_{r}$ . Then

$Q(A)\simeq \prod _{i=1}^{r}Q(A/{\mathfrak {p}}_{i}).$

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 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 ${\mathfrak {p}}_{i}Q(A)$  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-zerodivisors 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.