Total ring of fractions

(Redirected from Total quotient ring)

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

edit

Let   be a commutative ring and let   be the set of elements that are not zero divisors in  ; then   is a multiplicatively closed set. Hence we may localize the ring   at the set   to obtain the total quotient ring  .

If   is a domain, then   and the total quotient ring is the same as the field of fractions. This justifies the notation  , which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain.

Since   in the construction contains no zero divisors, the natural map   is injective, so the total quotient ring is an extension of  .

Examples

edit
  • 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.
  • 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,  , and so  . But since all these elements already have inverses,  .
  • 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,  .

The total ring of fractions of a reduced ring

edit

Proposition — Let A be a reduced ring that has only finitely many minimal prime ideals,   (e.g., a Noetherian reduced ring). Then

 

Geometrically,   is the Artinian scheme consisting (as a finite set) of the generic points of the irreducible components of  .

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   since Q(A) is reduced. By prime avoidance, I must be contained in some  . Hence, the ideals   are maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A),

 .

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

 ,

which is already a field and so must be  .  

Generalization

edit

If   is a commutative ring and   is any multiplicatively closed set in  , the localization   can still be constructed, but the ring homomorphism from   to   might fail to be injective. For example, if  , then   is the trivial ring.

Citations

edit

References

edit
  • 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