Total ring of fractions

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.

• 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 the definition of a Cartier divisor.

The total ring of fractions of a reduced ring

Proposition—Let A be a reduced ring that has only finitely many minimal prime ideals, ${\displaystyle {\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r}$ (e.g., a Noetherian reduced ring). Then

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

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

Proof: Every element of Q(A) is either a unit or a zero divisor. Thus, any proper ideal I of Q(A) is contained in the set of zero divisors of Q(A); that set equals the union of the minimal prime ideals ${\displaystyle {\mathfrak{p}}_{i}Q(A)}$ since 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 maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A),

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

Let S be the multiplicatively closed set of non-zero-divisors of A. By 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 ${\displaystyle Q(A/{\mathfrak{p}}_i)}$. ${\displaystyle ◻}$

Generalization

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.