Total ring of fractions

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 ${\displaystyle R}$.

• 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, ${\displaystyle R^{\times }}$, and so ${\displaystyle Q(R)=(R^{\times })^{-1}R}$. But since all these elements already have inverses, ${\displaystyle Q(R)=R}$.

• 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, ${\displaystyle Q(R)=R}$.

• In algebraic geometry one considers a sheaf of total quotient rings on a scheme, and this may be used to give one possible definition of a Cartier divisor.

There is an important fact:

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

If ${\displaystyle R}$ is a commutative ring and ${\displaystyle S}$ is any multiplicatively closed set 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.

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