Localization, Same Quotient Ring

Same Quotient Ring

Let S be a multiplicatively closed set in a commutative ring R, so that the fraction ring R/S makes sense. Let H be a saturated ideal in R. This means that H, pushed forward into R/S, and pulled back into R, yields H. The ideal H/S doesn't bring in any equivalent fractions with new numerators that we haven't seen before. A typical example is any prime ideal missing S. Since primes correspond 1-1, this becomes prime in R/S, and pulls back to the same prime in R.

Since H is an ideal it defines a quotient ring Q. Since H/S is an ideal in R/S it defines a quotient ring Q′. We will show that Q embeds in Q′, and in some cases they are isomorphic.

If u represents a coset of H in R, let v = u/1 represent a coset of H/S in R/S. Since H maps into H/S, cosets of H map into cosets of H/S, and the function is well defined on Q. Clearly this map respects addition and multiplication, and is a ring homomorphism from Q into Q′.

To show injective, suppose u maps into the kernel H/S. Thus u is a numerator of some fraction in H/S. Pull this back to u in R, and since H is saturated, u lies in H. Thus Q embeds in Q′.

When H is a maximal ideal and S misses H, the map is surjective. In other words, Q and Q′ are isomorphic. (These rings are sometimes called residue fields.)

Let a/b represent a coset of H/S. Since b lies outside of H, b and H generate 1. For v in H, let v plus a multiple of b equal 1. Mod out by the principal ideal generated by b, and v becomes 1. Now subtract av/b from a/b, giving another fraction in the same coset of a/b. Reduce the numerator mod {b} and get 0, whence the numerator lives in the principal ideal generated by b, whence the numerator is xb for some x in R. The fraction is equivalent to x/1. This is the image of x, hence the map is surjective.

All we need for this to work is that every b in S generates a principal ideal that is coprime to H.