Let R/T map into R/S in the obvious way. Is this a bijection? Are the rings essentially the same?
Assume the map is a bijection. Let w be a fixed element in T, and let c be an arbitrary element of S. Thus cw/w belongs to R/S and R/T. The inverse, w/cw, has to have an equivalent fraction in R/T. Write a/b = w/cw, or uacw = ubw. Something times c produces an element in T. If H is an ideal containing c, it intersects T. This holds for every c in S, so any ideal that intersects S also intersects T. This certainly holds for prime ideals.
Conversely, assume the prime ideals that intersect S always intersect T. You may recall from an earlier section that the saturation of T is the complement of the union of the prime ideals missing T, hence S is contained in the saturation of T.
Let H be an ideal that misses T. Let a/b be a fraction in H/T. This is equivalent to 1 only if ua = ub for some u in T. This causes H and T to intersect, which is a contradiction. Thus a/b cannot equal 1, and H/T is a proper ideal in R/T. This lets us drive H up to a prime ideal P that misses T. (This is where we need R to be commutative.) Since P misses S, H also misses S. Every ideal that misses T misses S.
Return to the ring homomorphism from R/T into R/S. Let x/y lie in the kernel, hence x/y = 0/w, or uwx = 0. The elements of R that kill x form an ideal H, and H intersects S in uw, hence H intersects T. Something in T kills x, and x/y is equivalent to 0 in R/T. Only 0 maps to 0, and the map is an injection.
To demonstrate a bijection, we need a/b in R/T that is equivalent to c/d in R/S. Let d generate an ideal H, which certainly intersects S. Since H also intersects T, Let d*a = b, for some b in T. Now ca/b is equivalent to c/d, and the map is onto.
The rings R/T and R/S are isomorphic iff all the prime ideals that intersect S also intersect T.