If you don't know what a projective module is, don't worry about it. This theorem isn't critical for an understanding of dedekind domains; it's more of an interesting side note. So you can skip ahead to the next section if you like.
Let H be an invertible ideal, even an invertible fractional ideal, hence it is finitely generated.
Let Y be a free R module whose rank is equal to the number of generators in H. Now you can see what we're going to do. If the generators of H are a1 through an, then a module homomorphism maps the basis e1 through en of Y onto these generators. The first coordinate e1 maps to a1, the second coordinate e2 maps to a2, and so on.
Let b1 through bn be elements of H inverse (some of them may be 0), so that the sum of aibi equals 1.
Build a map from H into Y as follows. For c in H, map c to the sum of cbiei. Note that cbi is in R, hence cbiei is a valid element of Y. Verify that this map is indeed a module homomorphism.
Compose the two homomorphisms; from H into Y, and back onto H. The result is the sum of caibi, which is c. The sequence is split exact, H is a summand of Y, and H is projective.
Now for the converse. Let Y be a free module with a (possibley infinite) basis e1 e2 …, and let Y map onto a projective module H, that happens to be a fractional ideal in the ring R. We'll call this a projective fractional ideal.
As before, Y has a basis whose rank is equal to the cardinality of the generators of H. (Assume all these generators are nonzero.) Map e1 onto a1, e2 onto a2, and so on.
The sequence is split exact, so a reverse homomorphism carries H back into Y. If the rank of Y is infinite, this reverse homomorphism maps H into the direct sum, rather than the direct product. That's the definition of a free R module.
Compose the reverse homomorphism with the projection from Y onto ej to give the homomorphism gj(H) into R.
Let x be any one of the generators ai in H.
Let wj = gj(x).
Let c be any element of H.
By the symmetry of module homomorphisms on H, cwj = xgj(c).
In the fraction field F, c times wj/x = gj(c), which is an element of R. This holds for every c in H. Therefore wj/x drives H into R, and lies in H inverse.
With luck, the set wj/x will generate H inverse. We only need show these generators span 1.
Select any nonzero c in H and let S be the finite set of indices for which gj(c) is nonzero. Thus the reverse image of c in Y is the finite sum of gj(c)*ej, for all j ∈ S.
Remember that gj(c) is cwj/x. Make this substitution and get the following sum.
c = sum(j ∈ S) cejwj/x
Apply the ring homomorphism forward from Y into H. This is where we use split exact. The image of the sum, in H, is still c. So we have the following.
c = sum(j ∈ S) cajwj/x
Cancel c on both sides, and a linear combination of the generators wj/x is equal to 1. The submodule spanned by these generators is indeed the inverse of H.
In summary, a fractional ideal is invertible iff it is projective.