If the chain in R starts with P1, find Q1 in S lying over P1. This starts the inductive process. From here we will go up, or down, for ascending or descending chains respectively.
If the chain goes beyond infinity, use transfinite induction. Let U be the union of an ascending chain of prime ideals in R, and let V be the union of the overlying prime ideals in S. Of course V need not be prime; but we assume U is, or U is contained in a larger prime. Since V does not contain 1 it is a proper ideal. Again, employing the previous theorem, there is a prime ideal in S, lying over U, and containing V. Therefore, arbitrary chains in R lift to arbitrary chains in S.
Descending chains do not lift so easily. We need S to be an integral domain, with R integrally closed in its fraction field K. First we need a couple of lemmas.
We don't usually talk about the integral closure of an ideal; this is a special case, just for this theorem.
First assume x is in the integral closure of H. This means there is a polynomial p with p(x) = 0, and lead coefficient 1, and all other coeffficients in H. Since x is in S, and is integral over R, it lies in C.
Move xn to one side, and what remains is an expression in x and various elements of H, which becomes an element of H′. Thus xn is in H′, and x is in rad(H′).
Conversely let xn be a finite sum of pairwise products from H and C, c1h1 + c2h2 + c3h3 etc. Let the elements c1 c2 c3 etc act as generators for a new ring over R, somewhere between R and C. Call this ring V.
Since V is R adjoin finitely many integral elements, V is a finitely generated R module.
Let w = xn. Remember that w is a sum of products drawn from H cross C. This is still the case when w is multiplied by anything in V. thus w drives V into HV.
Since w is an R module endomorphism from V into V, and since V is a finitely generated R module, the action of w can be represented by a matrix M. The representation may not be unique, but there is a matrix M, which designates the image of each generator. The ith row determines w times the ith generator of V as an R module.
Let V be generated by g1 g2 g3 etc, as an R module. Everything in V is a linear combination of these generators, using coefficients from R. If instead we use coefficients from H, the result lies in HV. Conversely, consider any element in HV. This is a sum of products xy for x in H and y in V. Represent y as a linear combination of our generators with coefficients in R. Multiply by x and the coefficients lie in H. Add this up over all pairs xy and the coefficients still lie in H. In other words, the R module HV is the span of the generators of V, with coefficients from H.
Since w maps V into HV, the matrix M consists entirely of elements of H.
Let p be the characteristic polynomial associated with M, i.e. the polynomial whose roots are the eigen values. By cayley hamilton, p(M) = 0. Since multiplication by M implements the action of w, p(w) maps the entire ring to 0. However, V contains 1. Therefore p(w) = 0. Replace w with xn and find p(x) = 0. The characteristic polynomial of any matrix is monic, hence p is monic. Since M comes from H, all remaining coefficients lie in H. Therefore x is integral over H.
As a corollary, the integral closure of H in S is an ideal in C.
What happens if H = R? The integral closure of R is C, and the extension of R into C is all of C, and rad(C) = C, so C = C, and the lemma holds true.
Since x is algebraic over K, it satisfies an irreducible polynomial f(x), and f(x) is a factor of e(x). Let the field extension L/K split f(x).
Clearly L is a ring extension of R. This is going to play the role of S/R in the previous lemma. Let C be the integral closure of R in L, which contains x. The first lemma tells us the integral closure of H forms an ideal in C. We'll need this in just a moment.
Each root of f(x) is a root of e(x), and is integral over H. Thus C contains all the conjugates of x. Show that the coefficients of f(x) are integral over H. The constant term, for instance, is ± the product of the roots of f, and since the integral closure of H forms an ideal in C, the constant term is integral over H. The other coefficients of f are expressions in the roots of f, which are integral over H, thus all the coefficients of f (save the lead coefficient) are integral over H. But remember, these coefficients also lie in K. Let's see how this helps.
Invoke the first lemma again, this time with K/R as the ring extension. The integral closure of R is now R, and the extension of H into R is simply H. So the integral closure of H equals rad(H). Therefore the coefficients of f(x) lie in rad(H), inside R. If H happens to be prime, its own radical ideal, the coefficients of f lie in H.
If H = R, this lemma remains valid; the coefficients of f lie in R.
Since f is monic, with its nonlead coefficients in H or in R, f can be used to prove x is integral over H or over R. We don't need e(x) any more. We can ratchet e down to f.
Let Q1 lie over P1 with P2 properly contained in P1.
Remember that P2 may not be an ideal in S. Let J be the extension of P2 into S. Since Q1 contains the extension of P1, which is larger than J, Q1 contains J.
When P2 extends into S, producing J, it may bring in additional elements of R, between P2 and P1. However, when R and S meet our criteria, this cannot happen.
The localization of J about Q1 is the same as the extension of the fractions of P2, within the local ring. In other words, extension and localization commute.
Let y/z be in the localization of J, where y is in the extension of P2 into S, and z is in S-Q1. It's time to apply the first lemma again. This time we're talking about S over R, and the integral closure of R in S is all of S. Thus the integral closure of P2 is equal to rad(J). Since y is in J it is in rad(J), and y is integral over P2.
Let y satisfy a monic irreducible polynomial f(y) over K. The second lemma tells us the coefficients of f (other than the lead coefficient) are in rad(P2), which is equal to P2.
Suppose y/z = x/1 for some x in R-P2. Thus we may write z = y/x in the fraction field of S.
Start with f(y) and leave the lead coefficient alone. Divide the next coefficient by x, the next one by x2, and so on to the constant, which is divided by xn. This new polynomial f′ has y/x as a root. Thus z is the root of a monic polynomial f′ over K.
Adjoining z is the same as adjoining y; they both give the same field extension over K. Thus f′ is irreducible.
Now z is still a member of S, and is integral over R. Apply the second lemma, and z is the root of an irreducible monic polynomial dividing f′, (which must equal f′), whose coefficients lie in R. So the coefficients of f′ lie in R.
Each coefficient of f is now some power of x times its counterpart in f′, and the coefficients of f lie in P2. Since x does not lie in P2, the coefficients of f′ must. This means zn lies in J, and z lies in rad(J). This implies z is in Q1, yet z is drawn from S-Q1. Therefore y/z cannot equal x/1, and J intersects R in precisely P2.
Embed S, and J, into the ring of fractions with denominators in S-Q1. This is traditional localization about Q1. Then bring in the denominators in the multiplicatively closed set R-P2. Since J misses these denominators, it remains a proper ideal. Drive J up to a maximal ideal, which is prime. This pulls back to a prime ideal Q2 in S. Since Q2 includes J, and misses R-P2, it intersects R in precisely P2. In other words, Q2 lies over P2. Since Q2 misses S-Q1, it is contained in Q1. We have lifted P2, and by induction we can lift an infinite descending chain.
What if the chain descends beyond infinity? Let the intersection of a descending chain of prime ideals in R be prime, or contain a prime. Call this prime P, and extend it into an ideal J in S. The intersection of J and R is still P. Let T be the union of S-Qi for all the primes that have been lifted heretofore. This is a multiplicatively closed set that misses J. Embed S in the ring of fractions with denominators from T, then bring in the denominators from R-P. Push J up to a prime ideal and find Q in S, lying over P, and inside all the primes of S that have come before. Therefore, transfinite induction can lift any chain from R up to S.