Total ring of fractions

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

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

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

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

• 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, [math]\displaystyle{ Q(R) = R }[/math].

• 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, [math]\displaystyle{ \mathfrak{p}_1, \dots, \mathfrak{p}_r }[/math] (e.g., a Noetherian reduced ring). Then

[math]\displaystyle{ Q(A) \simeq \prod_{i=1}^r Q(A/\mathfrak{p}_i). }[/math]

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

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 [math]\displaystyle{ \mathfrak{p}_i Q(A) }[/math] since Q(A) is reduced. By prime avoidance, I must be contained in some [math]\displaystyle{ \mathfrak{p}_i Q(A) }[/math]. Hence, the ideals [math]\displaystyle{ \mathfrak{p}_i Q(A) }[/math] are maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A),

[math]\displaystyle{ Q(A) \simeq \prod_i Q(A)/\mathfrak{p}_i Q(A) }[/math].

Let S be the multiplicatively closed set of non-zero-divisors of A. By exactness of localization,

[math]\displaystyle{ Q(A)/\mathfrak{p}_i Q(A) = A[S^{-1}] / \mathfrak{p}_i A[S^{-1}] = (A / \mathfrak{p}_i)[S^{-1}] }[/math],

which is already a field and so must be [math]\displaystyle{ Q(A/\mathfrak{p}_i) }[/math]. [math]\displaystyle{ \square }[/math]

Generalization

If [math]\displaystyle{ R }[/math] is a commutative ring and [math]\displaystyle{ S }[/math] is any multiplicatively closed set in [math]\displaystyle{ R }[/math], the localization [math]\displaystyle{ S^{-1}R }[/math] can still be constructed, but the ring homomorphism from [math]\displaystyle{ R }[/math] to [math]\displaystyle{ S^{-1}R }[/math] might fail to be injective. For example, if [math]\displaystyle{ 0 \in S }[/math], then [math]\displaystyle{ S^{-1}R }[/math] is the trivial ring.

References

• Matsumura, Hideyuki (1980), Commutative algebra (2nd ed.), Benjamin/Cummings, ISBN 978-0-8053-7026-3, OCLC 988482880 
• Matsumura, Hideyuki (1989), Commutative ring theory, Cambridge University Press, ISBN 978-0-521-36764-6, OCLC 23133540 

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