Begin with a local ring R, contained in an integral extension S, where S is a free R module. Thus S has a discriminant d, and S is finite atol iff d is a unit in R.
Let P be the maximal ideal of R. Let H be the extension of P in S. In other words, H is the action of P on S, as an R module, or H = P*S.
There are various prime ideals Qi in S, lying over P, and these primes are maximal in S. Assume we know that H is a certain product of these primes Qi raised to the ei. For example, S could be dedekind, whence P factors uniquely into prime ideals that contain P, i.e. prime ideals lying over P. Do we know S is dedekind? Often we do. If R is dedekind, and S is an integral domain with a group of automorphisms fixing R, and d is a unit, then S is integrally closed, and a dedekind domain. So this is not a stretch at all.
Assume S is finite atol over R, and tensor with the new ring R/P. The result remains finite atol. The base ring R/P has become a field, and we have already characterized finite atol extensions over a field.
Review the formula for tensoring with a quotient ring. The new ring becomes S/H, which is indeed an R/P vector space. It has the same rank as S/R, using the same basis. (Tensor each copy of R with R/P and it works out.)
Consider two maximal ideals Qi and Qj, raised to the ei and ej respectively, and suppose a maximal ideal contains both these ideals. A maximal ideal is prime, hence it contains both Qi and Qj. Since Qi and Qj are maximal, this is a contradiction. Therefore Qi to the ei and Qj to the ej span the ring. This allows us to invoke the chinese remainder theorem. Now S/H can be rewritten as a product of quotient rings. Each summand in the product looks like S mod Qi to the ei, according to the factorization of H into prime ideals.
1 in S/H becomes 1 in each component, and the action of R follows. Since P is part of H, is part of Qi to the ei, P drives each component to 0. Thus each component is an R/P module, or a vector space. This is a field extension iff the component is a field, iff the maximal ideal Qi becomes trivial, iff ei = 1, iff P is unramified at Qi. Each Qi that participates in the aforementioned factorization of H appears only once.
Most of the time we know that all primes participate. For instance, when S is dedekind, to contain is to divide, so each Qi is a factor of H. And if S has a unit discriminant, each Qi appears only once.
Let d be the discriminant of S. Thus the image of d in R/P becomes the discriminant of S/H over R/P. Units map to units in the quotient ring, which is why the forward direction is easy. However, in this case P is the maximal ideal, and the jacobson radical. This means the map can be reversed. Thus d is a unit iff d/P is a unit. That completes the proof.
This happens most of the time. If R has infinitely many primes, most of them will not divide d, whence dP becomes a unit. This assures each prime in the factorization of P appears only once.
Conversely, when P does divide d, there is some Q lying over P such that Q has a higher multiplicity in the factorization of P.