# 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. Nothing more in A can be given an inverse, if one wants the homomorphism from A to the new ring to be injective.

## 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.

## Notes

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

## References

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