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

## Definition

Let ${\displaystyle R}$ be a commutative ring and let ${\displaystyle S}$ be the set of elements which are not zero divisors in ${\displaystyle R}$; then ${\displaystyle S}$ is a multiplicatively closed set. Hence we may localize the ring ${\displaystyle R}$ at the set ${\displaystyle S}$ to obtain the total quotient ring ${\displaystyle S^{-1}R=Q(R)}$.

If ${\displaystyle R}$ is a domain, then ${\displaystyle S=R-\{0\}}$ and the total quotient ring is the same as the field of fractions. This justifies the notation ${\displaystyle Q(R)}$, which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain.

Since ${\displaystyle S}$ in the construction contains no zero divisors, the natural map ${\displaystyle R\to Q(R)}$ is injective, so the total quotient ring is an extension of ${\displaystyle R}$.

## Examples

The total quotient ring ${\displaystyle Q(A\times B)}$ of a product ring is the product of total quotient rings ${\displaystyle Q(A)\times Q(B)}$. 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, ${\displaystyle R^{\times }}$, and so ${\displaystyle Q(R)=(R^{\times })^{-1}R}$. But since all these elements already have inverses, ${\displaystyle Q(R)=R}$.

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, ${\displaystyle Q(R)=R}$.

## 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 ${\displaystyle {\mathfrak {p}}_{1},\dots ,{\mathfrak {p}}_{r}}$. Then

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

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

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 ${\displaystyle {\mathfrak {p}}_{i}Q(A)}$ as Q(A) is reduced, by prime avoidance, I must be contained in some ${\displaystyle {\mathfrak {p}}_{i}Q(A)}$. Hence, the ideals ${\displaystyle {\mathfrak {p}}_{i}Q(A)}$ are the maximal ideals of Q(A), whose intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A), we have:

${\displaystyle Q(A)\simeq \prod _{i}Q(A)/{\mathfrak {p}}_{i}Q(A)}$.

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

${\displaystyle Q(A)/{\mathfrak {p}}_{i}Q(A)=A[S^{-1}]/{\mathfrak {p}}_{i}A[S^{-1}]=(A/{\mathfrak {p}}_{i})[S^{-1}]}$,

which is already a field and so must be ${\displaystyle Q(A/{\mathfrak {p}}_{i})}$. ${\displaystyle \square }$

## Generalization

If ${\displaystyle R}$ is a commutative ring and ${\displaystyle S}$ is any multiplicative subset in ${\displaystyle R}$, the localization ${\displaystyle S^{-1}R}$ can still be constructed, but the ring homomorphism from ${\displaystyle R}$ to ${\displaystyle S^{-1}R}$ might fail to be injective. For example, if ${\displaystyle 0\in S}$, then ${\displaystyle S^{-1}R}$ is the trivial ring.

## Notes

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

## References

• Hideyuki Matsumura, Commutative algebra, 1980
• Hideyuki Matsumura, Commutative ring theory, 1989