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Math 615: Lecture of April 4, 2007
Let p > 0 be a prime integer. We now know that a coefficient ring of mixed characteristic p and characteristic pn, where n ≥ 2, is determined up to isomorphism by its residue class field. Let K be a given field of characteristic p. We also know that there is a complete mixed characteristic Noetherian discrete valuation ring (V, pV, K) with residue class field K. This implies that V /pnV is a coefficient ring of characteristic pn. Hence, as asserted earlier:
Theorem. A mixed characteristic coefficient ring of characteristc pn, where p > 0 is prime, has the form V /pnV , where V is a complete Noetherian discrete valuation ring that is a coefficient field.
This completes our proof of all of the structure theorems for complete local rings. We restate the following:
Theorem. Every complete local ring is either a homomorphic image of K[[x 1 ,... , xn]], a power series ring over a field K, or of V [[x 1 ,... , xn]], a power series ring over a mixed characteristic coefficient ring (V, pV, K) that is a Noetherian discrete valuation ring.
Both K[[x 1 ,... , xn]] and V [[x 1 ,... , xn]] are Cohern-Macaulay, as is every regular local ring, since a minimal system of generators for the maximal ideal is a regular sequence. But Cohen-Macaulay rings are universally catemary. Hence:
Corollary. Every complete local ring is universally catenary.
Complete regular local rings are formal power series rings in equal chacteristic, and also in mixed characteristic if unramified. The following is an important tool in working with formal power series rings.
Theorem (Weierstrass preparation theorem). Let (A, m, K) be a complete local ring and let x be a formal indeterminate over A. Let f =
n=0 anx
n (^) ∈ A[[x]], where
ah ∈ A − m is a unit and an ∈ m for n < h. (Such an element f is said to be regular in x of order h.) Then the images of 1 , x,... , xh−^1 are a free basis over A for the ring A[[x]]/f A[[x]], and every element g ∈ A[[x]] can be written uniquely in the form qf + r where q ∈ A[[x]], and r ∈ A[x] is a polynomial of degree ≤ h − 1.
Proof. Let M = A[[x]]/(f ), which is a finitely generated A[[x]]-module, and so will be sep- arated in the M-adic topology, where M = (m, x)A[[x]]. Hence, it is certainly separated in the m-adic topology. Then M/mM ∼= K[[x]]/(f ), where f is the image of f under the map A[[x]] K[[x]] induced by A K: it is the result of reducing coefficients of f mod 1
m. It follows that the lowest nonzero term of f has the form cxh, where c ∈ K, and so f = xhγ where γ is a unit in K[[x]]. Thus,
M/mM ∼= K[[x]]/(f ) = K[[x]]/(xh),
which is a K-vector space for which the images of 1, x,... , xh−^1 form a K-basis. By the Proposition on p. 2 of the Lecture Notes of March 23, the elements 1, x,... , xh−^1 span A[[x]]/(f ) as an A-module. This means precisely that every g ∈ A[[x]] can be written g = qf + r where r ∈ A[x] has degree at most h − 1.
Suppose that g′f + r′^ is another such representation. Then r′^ − r = (q − q′)f. Thus, it will suffice to show if r = qf is a polynomial in x of degree at most h − 1, then q = 0 (and r = 0 follows). Suppose otherwise. Since some coefficient of q is not 0, we can choose t such that q is not 0 when considered mod mtA[[x]]. Choose such a t as small as possible, and let d be the least degree such that the coefficient of xd^ is not in mt. Pass to R/mt. Then q has lowest degree term axd, and both a and all higher coefficients are in mt−^1 , or we could have chosen a smaller value of t. When we multiply by f (still thinking mod mt), note that all terms of f of degree smaller than h kill q, because their coefficients are in m. There is at most one nonzero term of degree h + d, and its coefficient is not zero, because the coefficient of xh^ in f is a unit. Thus, qf has a nonzero term of degree ≥ h + d > h − 1, a contradiction. This completes the proof of the existence and uniqueness of q and r.
Corollary. Let A[[x]] and f be as in the statement of the Weierstrass Preparation The- orem, with f regular of order h in x. Then f has a unique multiple f q which is a monic polynomial in A[x] of degree h. The multiplier q is a unit, and qf has all non-leading coefficients in m. The polynomial qf called the unique monic associate of f.
Proof. Apply the Weierstrass Preparation Theorem to g = xh. Then xh^ = qf + r, which says that xh^ − r = qf. By the uniqueness part of the theorem, these are the only choices of q, r that satisfy the equation, and so the uniqueness statement follows. It remains only to see that q is a unit, and that r has coefficients in m. To this end, we may work mod mA[[x]]. We use u for the class of u ∈ A[[x]] mod mA[[x]], and think of u as an element of K[[x]].
Then xh^ − r = qf. Since f is a unit γ times xh, we must have r = 0. It follows that xh^ = xhqγ. We may cancel xh, and so q is a unit of of K[[x]]. It follows that q is a unit of A[[x]], as asserted.
Discussion. This result is often applied to the formal power series ring in n variables, K[[x 1 ,... , xn]]: one may take A = K[[x 1 ,... , xn− 1 ]] and x = xn, for example, though, obviously, one might make any of the variable play the role of x. In this case, a power series f is regular in xn if it involves a term of the form cxhn with c ∈ K − { 0 }, and if one takes h as small as possible, f is regular of order h in xn. The regularity of f of order h in xn is equivalent to the assertion that under the unique continuous K[[xn]]-algebra map K[[x 1 ,... , xn]] → K[[xn]] that kills x 1 ,... , xn− 1 , the image of f is a unit times xhn. A logical notation for the image of f is f (0,... , 0 , xn). The Weierstrass preparation theorem
since dj = ej for j < i. It will be enough to show that this difference is positive. Since ei > di, the leftmost term is at least Ni. Some of the remaining terms are nonnegative, and we omit these. The terms for those j such ej < dj are negative, but what is being subtracted is bounded by dj Nj ≤ dNj. Since at most n − 1 terms are being subtracted, the sum of the quantities being subtracted is strictly bounded by nd maxj>i{dNj }. The largest of the Nj is Ni+1, which is (dn)n−(i+1). Thus, the total quantity being subtracted is strictly bounded by (dn)(dn)n−i−^1 = (dn)n−i^ = Ni. This completes the proof that
e 1 N 1 + e 2 N 2 + · · · + en− 1 Nn− 1 + en > d 1 N 1 + d 2 N 2 + · · · + dn− 1 Nn− 1 + dn,
and we see that θ(f ) is regular in xn, as required.
If the Weierstrass Preparation Theorem is proved directly for a formal or convergent power series ring R over a field K (the constructive proofs do not use a priori knowledge that the power series ring is Noetherian), the theorem can be used to prove that the ring R is Noetherian by induction on n. The cases where n = 0 or n = 1 are obvious: the ring is a field or a discrete valuation ring. Suppose the result is known for the power series ring A in n − 1 variables, and let R be the power series ring in one variable xn over A. Let I be an ideal of R. We must show that I is finitely generated over R. If I = (0) this is clear. If I 6 = 0 choose f ∈ I with f 6 = 0. Make a change of variables such that f is regular in xn over A. Then I/f R ⊆ R/f R, which is a finitely generated module over A. By the induction hypothesis, A is Noetherian, and so R/f R is Noetherian over A, and hence I/f R is a Noetherian A-module, and is finitely generated as an A-module. Lift these generators to I. The resulting elements, together with f , give a finite set of generators for I.
Although we shall later give a quite different proof valid for all regular local rings, we want to show how the Weierstrass preparation theorem can be used to prove unique factorization in a formal power series ring.
Theorem. Let K be a field and let (V, π, K) be a Noetherian discrete valuation ring. R = K[[x 1 ,... , xn]] or V [[x 1 ,... , xn]] be the formal power series ring in n variables over K or V. Then R is a unique factorization domain.
Proof. We use induction on n. If n = 0 then R is a field or a discrete valuation ring. In the latter case, R is a principal ideal domain and, hence, a unique factorization domain.
Suppose that n ≥ 1. It suffices to prove that if f ∈ m is irreducible then f is prime. If π divides f , the f is a multiple of π by a unit, since f is irreducible. We know that π is prime, since R/(π) ∼= K[[x 1 ,... , xn]], a domain. Hence, we may assume that π does not divide f. Suppose that f divides gh, where it may be assumed without loss of generality that g, h ∈ m. Then we have an equation f w = gh, and since f is irreducible, we must have that w ∈ m as well. If some power of π divides w, then π divides g or h. We may factor out π and obtain a similar equation in which a lower power of π divides w. Eventually, we obtain an equation in which π does not divide w: otherwise, w would be in every power of the maximal ideal. Then π does not divide g nor h as well. Hence, π does not divide f gh.
Therefore, by the Discussion on pp. 3 and 4, we can make a change of variables in the formal power series ring such that f gh is regular in xn modulo π. Since an element of the ring that is a unit modulo π is a unit, we have that f gh is regular in xn in R as well. Then f , g, and h are all regular in xn, and we may multiply each by a unit so as to replace it by its unique monic associate: here we view R as A[[xn]] where A = K[[x 1 ,... , xn− 1 ]] or V [[x 1 ,... , xn− 1 ]]. Thus, we may assume without loss of generality that f , g, and h are monic polynomials in A[xn] whose non-leading coefficients are in Q = (x 1 ,... , xn− 1 )A. In the process of replacing f, g, h by their products units, w is replaced by its product with a certain unit as well, so that we still have f w = gh. However, a priori, w may be a power series in xn rather than a polynomial.
It is easy, however, to see that w ∈ A[xn] as well. We can divide gh ∈ A[xn] by f , which is monic in xn, to get a unique quotient and remainder, say gh = qf +r, where the degree of r is less than the degree d of f. The Weierstrass preparation theorem guarantees a unique such representation in A[[xn]], and in the larger ring we know that r = 0. Therefore, the equation gh = qf holds in A[xn], and this means that q = w is a monic polynomial in xn as well.
By the induction hypothesis, A is a UFD, and so A[xn] is a UFD. We first note that f is still irreducible in A[xn] (this is an issue because it might factor as a polynomial with an invertible constant term in one factor: such a factorization does not contradict irreducibility in A[[xn]]). But if f factors non-trivially f = f 1 f 2 in A[xn], the factors f 1 , f 2 must be polynomials in xn of lower degree which can be taken to be monic. Mod Q, f 1 , f 2 give a factorization of xdn, and this must be into two powers of xn of lower degree. Therefore, f 1 and f 2 both have all non-leading coefficients in Q, and, in particular their constant terms are in Q. This implies that neither f 1 nor f 2 is a unit of R, and this contradicts the irreducibility of f in R. Thus, f must be irreducible in A[xn] as well. But then, in A[xn] we have that f | g or f | h, and the same obviously holds in the larger ring R, as required.
