Let E/F have dimension n, and find a basis for E as an F vector space. Each basis element is algebraic over R, and some multiple becomes integral over R. Scale the basis elements accordingly. The result is still a basis for E over F, but the basis is now integral over R. This basis is linearly independent in F, and in R, and spans a free R module of rank n. Linear combinations of integral elements remain integral, so the entire span is integral. Therefore S contains a free R module of rank n.
Let b1 through bn form a basis for E over F, with each bi contained in S. Assume E is galois over F. Let y lie in S, and represent y as a sum over aibi. Note that ybj lies in S for any basis element bj. The trace of ybj lies in F. Call this trace tj.
At this point we need ybj to be integral, to show tj lies in R. In the last section we showed that the minimum polynomial of an integral element is monic and lies in R. (This is where we need R to be integrally closed.) The second coefficient of this polynomial lies in R, and is minus the trace of ybj, in a (possibly intermediate) field extension. Multiply this by an integer to get the trace relative to E/F. Thus each tj lies in R.
The trace of the sum is the sum of the trace, so rewrite tj as the sum of the trace of aibibj. Since ai lies in F it is a constant, and can be pulled out of the trace.
∑ ai * trace(bibj) = tj
There are n such equations, as j runs from 1 to n. Put them together as a system of simultaneous equations, like this.
M*a = t
Here a is a column vector a1 through an, the coefficients of y using the basis b1 through bn. The right side t is another column vector t1 through tn, where tj is the trace of ybj. The symmetric matrix M holds the trace of bibj. the entries of M and t lie in R, and the column vector a lies in F.
Build a new matrix W, whose first row is the basis b1 through bn. This is the trivial automorphism applied to the basis. For the second row of W, take another automorphism of E/F and apply it to the basis. The third row of W is the image of b1 through bn under the third automorphism, and so on. Since E/F is galois there are n automorphisms, and W is a square matrix.
Suppose the rows of W are linearly dependent. A linear combination of automorphisms takes the basis to 0, and since F commutes past these automorphisms, the same linear combination of automorphisms maps all of E to 0. However, distinct automorphisms are linearly independent. Therefore W is a nonsingular matrix, with a nonzero determinant in R.
Multiply the transpose of W by W, in that order, to get M. Thus det(M) = det(W)2, in an integral domain. With det(M) nonzero, M is nonsingular.
Let d = det(M). Apply cramer's rules, and the inverse of M lies in R, with a common denominator of d. Multiply by M inverse on the left, and the coefficients a1 through an are elements of R with denominators of d. This holds for every integral y in S.
Divide the basis elements by d and find another basis of E/F, that spans a free R module of rank n, that contains S. Therefore S is trapped between two free R modules of rank n.
Note that S need not be integrally closed - as long as it contains a basis for E/F. The integral closure of R, and of S, is trapped between the free R modules described above. In fact, det(M) is a handy way to bound the integral ring inside E.
What if E/F is not galois? Let K be the normal closure of E/F. (This is where we need E/F to be separable.) Let T be the integral ring of K/F/R. Note that T contains S.
Now b1 through bn is part of a basis for K/F. Let y lie in T, and write it as a linear combination of basis elements. Use the above to find a matrix M and a determinant d, and the coefficients on y lie in R/d. Divide the basis through by d, and all of T lies in the span of this basis. Restrict attention to b1 through bn. with coefficients from F, the span is E. Intersect T and E to get S. Therefore S is spanned by b1 through bn alone, and since S lies in T, we only need use coefficients from R. Once again S is contained in a free module of rank n.
Recall that S is trapped between two free R modules of rank n. Now S is free, and its rank cannot exceed n. Neither can its rank be smaller than n. Therefore S is a free R module of rank n.
There is another way to prove S is free, though it doesn't come up very often. Let R be noetherian and integrally closed, and trap S between two free R modules of rank n. Find n generators that span S as an R module, and apply this proof.
When R is a pid, it is also dedekind, hence S is a dedekind domain that happens to be a free R module of rank n.
Let S be an integral domain containing R, such that S is a finitely generated R module. Thus S is integral over R. Further assume S is separable, and integrally closed in its fraction field, which I will call E. If S is a free R module, or if S contains a free R module with proper spillover, then S×F = E, and S is the integral ring of E.
The correspondence is perfect when R is a pid. The integral ring of E/F becomes a free R module, while the fraction field of a free R module of finite rank, that is also an integral domain, reproduces the field extension.
Remember that field and ring extensions must be separable. This is assured when R has characteristic 0.