I use the word "image" throughout, but that is somewhat misleading. The resulting ideal in R/S is not the image of a ring homomorphism applied to H. Rather, we are slapping denominators onto H.
We know that R/S is a ring, and it is also an R/S left module. If H is a left ideal in R, let H/1 generate an R/S submodule inside R/S. To accomplish this, apply all the denominators of S to all the elements of H. The result is indeed a left ideal in R/S. The only tricky part is showing closure under addition. Given h1/s1 + h2/s2, apply the definition of addition and write (h1s2 + h2s1) / s1s2. The numerator is in the ideal H, and the common denominator is in the set S, so the result is in fact some element of H with a denominator from S.
If S does not contain 1, there may not be an obvious way to map H into its image H/S. When I talk about H/1 generating a submodule inside R/S, I'm speaking figuratively. The elements H/1 may not live in R/S at all. It is perhaps more accurate to describe H/S as a set, elements of H cross elements of S, then prove it is a left ideal in R/S, and hence it is a left R/S module. In practice, S usually contains 1, so the distinction is rarely significant.
Let the ideal in R/S include x/s1, and multiply by s1/s2 to find another element in the ideal. This is equivalent to x/s2, hence all denominators are represented.
Select any denominator w and let H be the set of numerators associated with w. As shown above, x/w brings in all of x/S. Add x/w and y/w and get w(xy)/w2, which is equivalent to (x+y)/w. Similarly, multiply x/w by z/w for any z in R and find zx/w2. This implies zx/w, hence zx belongs to H, and H is an ideal in R.
Put the denominators back again to show that the image of the preimage gives the same ideal in R/S. However, the converse does not hold.
Consider the integers, with the rationals as fraction field. The even numbers form an ideal in Z. Apply all denominators and get every rational number, including 2/14, which is equivalent to 1/7. Now the numerators associated with the denominator 7 include 1, and all the other integers. The preimage of the image of 2Z is Z.
If x is in H1 and H2, x/S is in both images, and in their intersection. Conversely, assume a1/s1 and a2/s2 represent the same fraction, and are in the intersection of the image ideals. Now a1s2 and a2s1 are equal. (This is where we need no zero divisors in S.) This common value, call it x, is present in H1 and H2. Note that x/(s1s2) is in the image of H1∩H2. Also, x/(s1s2) is the same fraction as a1/s1 and a2/s2, hence the intersection of the images comes from H1∩H2.
This works, by induction, for finitely many ideals, but I'm not sure if it generalizes to an arbitrary intersection. Certainly the intersection of the images contains the image of the intersection.
As for the sum, start with a1+a2 in H1+H2, and its fraction, (a1+a2)/w, is equivalent to a1/w+a2/w, which is in the sum of the images. Conversely, add a1/s1 and a2/s2, giving something in H1 + something in H2 over s1s2. Therefore the sum of the images is the image of the sum.
The image of the product ideal is generated by a1a2/w2, for any fixed w in S, and every a1 in H1 and a2 in H2. Fixing the denominator is not an issue, since all the other denominators are brought in. Now each of these generators is produced by a1/w times a2/w, hence the image of the product lies inside the product of the images.
Conversely, the image of H1 is generated by H1/w for a fixed w, and similarly for H2, hence their product is generated by a1a2/w2. Each of these generators lives in the image of the product, hence the image of the product ideal is the product of the images.
In the above, H1 and H2 could be arbitrary R modules, or in the case of the product, one could be an R module while the other is an ideal acting on said module.
Let a1/s1 * a2/s2 be a fraction in the ideal H/S. In other words, there is some c/d from H/S that is equivalent to a1a2/s1s2. Write ucs1s2 = uda1a2. Either u, d, a1, or a2 is in H. We said H and S are disjoint, hence H contains a1 or a2, H/S contains a1/s1 or a2/s2, and H/S is prime.
Now let H1 and H2 be distinct prime ideals, with a1 in H1, but not in H2. Suppose a1/S is in the image of H2. That is, a1/s1 = a2/s2, or ua1s2 = ua2s1. Now ua1s2 is in H2, and if H1 and H2 are disjoint from S, a1 must belong to H2, which is a contradiction. The prime ideals in R that are disjoint from S map 1-1 into prime ideals in R/S.
Now let's complete the correspondence. Let J be any prime ideal in R/S. Build H as before, taking the numerators from any given denominator. Since J is proper, H and S are disjoint. Let a1a2 lie in H, hence a1a2/w2 lies in J. Either a1/w or a2/w is in J, so either a1 or a2 is in H, and H is prime.
In summary, the prime ideals disjoint from S correspond 1-1 with the prime ideals in R/S.
As a corollary, the maximal ideals in R/S corespond to the maximal prime ideals in R that are disjoint from S. (Remember that R is commutative here.) There is always at least one maximal ideal in any ring, hence there is a maximal ideal in R/S, and a maximal prime ideal disjoint from S.
Beyond this, we can start with an ideal H disjoint from S, take its image in R/S, raise this to a maximal ideal in R/S, and pull back to a maximal prime ideal in R that contains H and misses S. We only need show H/S is not all of R/S. This is assured when S has no zero divisors, or when H is prime. The proper ideal H becomes proper in R/S, as shown earlier on this page.
Since 0 is prime in an integral domain, and maps to 0 in the fraction ring, S inverse of an integral domain is an integral domain.