As usual, we have to show this is well defined, i.e. basis invariant. We're going to use field theory to do this.
Let F be the fraction field of R and let E be the fraction field of S. Recall that E is S tensor F, or the linear combinations of b1 through bn with coefficients in F. Thus E/F is an n dimensional field extension.
If x is an element of S, multiplication by x builds the same matrix, whether we are working over R or over F. After all, x*bi has not changed. Therefore the trace of x is the same in S/R and in E/F. Furthermore, each basis of S/R is also a basis for E/F. Therefore it is enough to show the discriminant is basis invariant for field extensions.
Build a matrix W that runs the basis elements through all the embeddings of E into the algebraic closure of F. Thus Wi,j is the image of bj under the ith embedding of E into the algebraic closure of F. If E/F includes a purely inseparable subfield, repeat the rows of W over and over, until W becomes a square matrix.
Consider WT*W, in that order. The result is M, which holds the trace of bibj, as described above. When there is an inseparable subfield, the replicated rows produce just the right fudge factor. Take a moment to write this all out; it actually works. Therefore the discriminant is det(W)2.
Now imagine a new basis for R, consisting of c1 through cn. Let Q be a matrix in R that performs the linear transformation. Thus the row vector b1 through bn times Q gives the row vector c1 through cn.
If h is a field homomorphism, i.e. one of the embeddings of E into the algebraic closure of F, apply h to b1 through bn, then multiply by Q. Since R is fixed by h, we can move the entries of Q into the homomorphism. Thus the result is h applied to c1 through cn. Put this all together and W*Q is the matrix that corresponds to the new basis.
Take the determinant of WTQTWQ and get det(W)2 times the square of a unit in R. Therefore the discriminant is well defined, up to the square of a unit in R. If R = Z, the units are ±1, and the discriminant is a fixed integer.
Take a step back and look at x*M. This gives a row vector u, such that u.y = trace(xy). thus u determines the function we know as φ(x). The map φ is not injective iff a nonzero x produces the zero vector u, iff M has a zero determinant, iff the discriminant is 0.
At the same time, every R homomorphism from S into R is uniquely determined by a certain vector u. The map φ is onto iff every possible vector u is x*M for some x. Work within the field F and find the inverse of M. Apply it to u to find x. The values of x, for all vectors u, lie in R, iff the inverse of M lies in R, iff the determinant of M is a unit.
In summary, a ring extension that is an integral domain, and a free R module, is finite atol iff its discriminant is a unit in R. Since a change of basis alters the discriminant by the square of a unit, this criterion is well defined.
Let a group of automorphisms G act on S, such that G fixes R. Pass to the fraction fields E and F, and G fixes F. In fact E/F is galois, with galois group G.
Let the discriminant of S = u, where u is a unit in R. Trap the integral closure of S between two free R modules, generated by b and b/u. Since u is a unit, these free modules are the same. Therefore S equals its integral closure, and S is integrally closed.
Suppose u satisfies a smaller polynomial over F. Since F[u] is a pid, Q divides p. Thus p is the product of two polynomials with coefficients in F. Multiply both factors by a common denominator and two polynomials over R have a product that is a scale multiple of p. Of course p(u) becomes 0 in S, hence two nonzero elements have a product of 0. This contradicts S being an integral domain, hence p is the minimum polynomial for u. In other words, E = F(u).
Let the powers of u form a basis for S. Let this be the first row of a matrix W. The second row of W is the powers of u run through another embedding of E into the algebraic closure of F. These are the powers of one of the conjugates of u. This continues all the way down the rows of W. Each row is the powers of a different conjugate of u.
Notice that W is a vandermonde matrix. Its determinant is the product of the pairwise differences of the conjugates of u. Since E/F is separable, the roots are all distinct, each difference is nonzero, and the product is nonzero. The discriminant is the square of this product, which is also nonzero. If R = F, i.e. R is already a field, the discriminant is a unit, and E/F is finite atol. This confirms something we already knew; a finite separable field extension is finite atol.
In summary, an integral domain S that is a simple ring extension R[u], via a monic polynomial p(x), has, as its discriminant, the square of the product of the pairwise differences of the roots of p(x). The extension is finite atol iff this product is a unit in R.
Oddly enough, the same result holds if u is purely inseparable. The trace of anything in such an extension is 0, the matrix is 0, and the discriminant is 0. At the same time, the roots of p(x) are equal, their differences are 0, and the discriminant is 0.
With the basis set, the proof, shown above, that equates a unit discriminant with finite atol, remains valid. Therefore, for any basis, d is a unit iff S/R is finite atol.
Let b be a basis for S/R. Tensor this free module with T and b becomes a basis for U/T. Evaluate the discriminant of U/T over b. Thus M is a matrix with Mi,j = trace(bibj). Remember that trace commutes with tensor product. Thus each entry is h(trace(bibj)), where h is the ring homomorphism that maps R into T. Since h respectes multiplication and addition, apply h to the entire determinant. Therefore the discriminant of U is h applied to the discriminant of S.