Let H be an ideal in R that is invertible, with H′ as inverse. Recall that H embeds in, and generates, the ideal H/S inside the ring R/S. Simply apply all the denominators in S to H and to R. The embedding is implemented via H → H/1.
In the last section we showed H′ is finitely generated. Assign it the generators x1 x2 x3 … xn. Remember, these generators live in F, and might not be contained in R.
These generators span a finitely generated R/S module in F, which is a fractional ideal in the ring R/S.
Let a/b be any element of H/S and multiply by one of the generators, say x1. Since ax1 is in R, the result lies in the ring R/S. This holds for all generators times all elements in H/S, and all sums thereof, hence H/S times H′/S is contained in R/S.
Since R is an integral domain, the map from R into R/S is 1-1. In other words, R embeds in R/S, and 1 in R becomes 1 in R/S.
We know that H times H′ spans 1, so write out the equation as sum aibi = 1, where ai comes from H and bi comes from H′. These elements embed in R and F respectively, and F is still the fraction field of R/S, and multiplication and addition behave just as they did before. Therefore H/S times H′/S spans 1, and H/S is invertible.
An ideal that is invertible remains invertible in the fraction ring.