Since S is the inverse limit, ring homomorphisms map S into each R/Hn. These reduce f(x) mod Hn, which is implemented by retaining the first n digits of the coefficients of f, and any x that you care to evaluate. If u is a root of f, it is still a root when everything is reduced mod Hn. Thus each root in S maps to a consistent set of roots, or a "root system", in the descending chain of quotient rings.
If u1 and u2 are different roots, they differ in their nth digits, for some n. Thus they map to distinct roots in R/Hn. The roots upstairs map injectively into root systems downstairs.
Given a root system downstairs, collect the successive cosets of the roots and build u in S. Evaluate f(u), and the result is 0 in each R/Hn, hence the result is 0. This makes u a root of f, and that completes the correspondence. We only need tie the root system back to R/H. More accurately, a root in R/H should lift, uniquely, to a root system. There are certain situations, not at all uncommon, where this is assured. In these cases, the roots of f, in S, correspond 1-1 with the roots in R/H.
Consider a quotient ring, where the kernel squared is 0. If a root in the quotient lifts to a unique root in the original ring, this can be pushed all the way up to a root system by induction. (This was described in the previous section.) So the problem has been reduced to a quotient ring, one link in the chain.
Let H be an ideal of R, such that H2 = 0. Let f be a polynomial over R, with q = f/H, the image of the polynomial in the quotient ring. Let q(v) = 0, with q′(v) a unit in R/H. We want to lift v up to u, such that f(u) = 0. Also, f′(u) has to be a unit in R, to keep the chain going.
Take the second criteria first; it's not hard. If the elements of H are nilpotent, and u in R maps to a unit v in R/H, we have u*w = 1+z, for some z in H. Yet 1 + a nilpotent is a unit, so u is a unit. If q′(v) is a unit in R/H, then f′(u) will be a unit in R. No trouble there.
Let y be anything in the preimage of v. By the above, f′(y) is a unit in R. We wish to solve for b, such that f(y+b) = 0, and b lies in H.
Use the taylor expansion to write f(y+b) as f(y) + b*f′(y) +b2 times some other stuff. Of course b2 = 0, hence b = -f(y)/f′(y). Since the denominator is a unit, there is one and only one solution. Set u = y+b for the unique lift of v. That completes the proof.
The same thing occurs when R/H is a pid, and f and f′ are relatively prime. We know f and f′ generate 1, and if f drops to 0 at v, f′(v) must be a unit. It is equivalent to say f has no repeated irreducible polynomials in its unique factorization.
Since each sequence in S begines with something from the field K, S is an integral domain. Assume S is integrally closed. This happens when R is a dvr, and its completion is a dvr, which is integrally closed. Since f is monic, a root of f in the fraction field of S is integral, and lies in S. The roots in K correspond 1-1 with the roots in S, or the fraction field thereof.
Let R be Z localized about p, which is a dvr. Let K be the residue field Z mod p. Let S be the completion of R, which is also a dvr. (You know this better as the p-adic integers.) Let f be a monic polynomial in Z or in R, and push this down to a monic polynomial in K. Assume f does not have a multiple root in K[x]. Let f have a root v, hence f = (x-v)*g(x). The derivative, evaluated at v, is simply g(v). Since v is not a root of g, this is nonzero. The roots of f in K correspond 1-1 with the roots of f in the p-adic integers, or the p-adic numbers.
Let f = xm-c. If p does not divide m, mth roots of c mod p correspond 1-1 with the mth roots of c in the p-adic numbers.
Some of these results were proved elsewhere, in a brute force fashion.