HEITMANN’S PROOF OF THE DIRECT SUMMAND CONJECTURE IN DIMENSION 3

PAUL ROBERTS

We describe the main ideas of Ray Heitmann’s proof of the Direct Summand Conjecture in dimension 3 for a ring of mixed characteristic [1]. In the first section we describe the main methods which are used and prove the necessary lemmas. In the second section we prove the main result of Heitmann’s paper. Finally, in the third section we give a proof of the Canonical Element Conjecture using this result.

1. An outline of the methods used in the proof

Let R be a Noetherian local ring of mixed characteristic p of dimension 3. We assume that R is an integrally closed domain and is a homomorphic image of a regular local ring. In fact, the questions we consider can be reduced to the case of a complete integrally closed domain, and complete local rings are homomorphic images of regular local rings by the Cohen structure theorems.

Let (p, x, y) be a system of parameters for R. Let z be an element of R such that p^{N}z ∈ (x, y). The aim is to find a finite extension of R in which z is in the ideal generated by x and y. If this is always possible, then it is not difficult to prove the Direct Summand Conjecture (or any of several equivalent conjectures). This aim is not quite reached, but a weaker result, which is also enough to prove these conjectures, is proven instead. However, to motivate the construction in the proof, we first discuss the question of attempting to find an extension in which z ∈ (x, y). We assume throughout that R is an integrally closed domain.

The equation for which we are trying to find a solution is

z = wx + vy,

where w and v are in a finite extension of R. This means that w is integral over R, so it is a root of a polynomial

f(T) = T^{n} + a_{1}T^{n−1} + ⋯ + a_{n}

with coefficients in R. Since z, x, and y are in R, we can solve the equation z = wx + vy for v, and v will be in the field generated by w over the quotient field of R:

v = \frac{z − wx}{y}.

From this equation we can use Taylor’s formula to find a polynomial h(T) whose root is v, as we show below. If we can find a polynomial f(T) such that the corresponding polynomial h(T) with root v also has coefficients in R, we have solved the problem, since then both w and v are integral over R.

Suppose that f(T) = T^{n} +a_{1}T^{n−1} +⋯+a_{n} is a polynomial such that f(w) = 0. We show how to find the polynomial h(T) with root v. Let F (S, U) be f(T) made homogeneous; that is, if f(T) = T^{n} + a_{1}T^{n−1} + ⋯ + a_{n}, we let

F (S, U) = S^{n} + a_{1}S^{n−1}U + ⋯ + a_{n}U^{n}.

By Taylor’s Theorem, we have

F (S − T, U) = ∑_{i=0}^{n}(−1)^{i} \frac{1}{i!}F^{(i)}(S, U)T^{i},

where F^{(i)}(S, U) denotes the ith derivative of F (S, T) with respect to the first variable. We will use this notation for the derivative of a homogeneous polynomial with respect to the first variable, and we will refer to it as the derivative of the polynomial, throughout the paper. Applying this equation with S = z, T = z −wx, and U = x we obtain

F (z − (z − wx), x) = ∑_{i=0}^{n}(−1)^{i} \frac{1}{i!}F^{(i)}(z, x)(z − wx)^{i}.

On the other hand, we have

F (z − (z − wx), x) = F (wx, x) = x^{n}F (w, 1) = x^{n}f(w) = 0.

Hence if we let

b_{i} = (−1)^{n−i} \frac{1}{(n − i)!}F^{(n−i)}(z, x)

and let

g(T) = T^{n} + b_{1}T^{n−1} + ⋯ + b_{n},

then g(z − wx) = 0. Finally, if we let c_{i} = b_{i}/y^{i} and h(T) = T^{n} + c_{1}T^{n−1} + ⋯ + c_{n}, then h((z − wx)/y) = 0.

To summarize, we wish to find a polynomial f(T) with coefficients in R so that the resulting polynomial h(T) also has coefficients in R. Tracing back through the construction of h(T), we see that this means that

\frac{1}{(n − i)!}F^{(n−i)}(z, x) ∈ y^{i}R

for i = 0, …, n.

The strategy is to start with \frac{1}{(n−1)!}F^{(n−1)}(z, x), which is the next to the last term in the Taylor expansion, and construct the previous terms inductively, essentially by integration, finally ending up with a polynomial of degree n with the required properties.

Before continuing the discussion, we recall some facts about the coefficients which arise in the Taylor expansion and introduce some notation. Let f(T) = ∑_{k=0}^{n} a_{k}T^{n−k} as above. Then

\frac{1}{(n − i)!}f^{(n−i)}(T) = ∑_{j=0}^{i} \binom{n − j}{i − j}a_{j}T^{i−j}.  (1)

To see this, simply compute both sides for f(T) = a_{j}T^{n−j}: in computing the left hand side we obtain

\frac{(n − j)(n − j − 1) ⋯ (n − j − (n − i) + 1)}{(n − i)!}a_{j}T^{n−j−(n−i)} =

= \binom{n − j}{n − i}a_{j}T^{i−j} = \binom{n − j}{i − j}a_{j}T^{i−j}.

While this fact is quite straightforward, we mention it explicitly to explain our

interest in binomial coefficients of the form \binom{n − j}{i − j}, for example in Lemma 3.

Applying equation (1) to the polynomial F (S, U) evaluated at z, x, we have

\frac{1}{(n − i)!}F^{(n−i)}(z, x) = ∑_{j=0}^{i} \binom{n − j}{i − j}a_{j}z^{i−j}x^{j}.

Since the main step in the construction involves integration and adjusting constants, we introduce special notation for it. Given a homogeneous polynomial G(S, U) = ∑_{j=0}^{i} a_{j}S^{i−j}U^{j} of degree i and an integer n > i, we let

Int_{n}(G)(S, U) = (n − i) ∑_{j=0}^{i} a_{j} (\frac{S^{i−j+1}}{i − j + 1}) U^{j}.

This definition depends not only on G(S, U), but also on n; in fact, we are thinking of G(S, U) as the (n−i)th term in the Taylor expansion of a polynomial of degree n, and Int_{n}(G)(S, U) is then a candidate for the previous term; it is in fact the unique choice with coefficient of U^{i+1} equal to zero. That is the reason for including the factor n − i.

We now return to the main discussion. We have a system of parameters p, x, y for R and an element z with p^{N}z ∈ (x, y) for some integer N. Thus there exists an element a of R such that p^{N}z + ax ∈ yR. We want to find a homogeneous polynomial F (S, U) of degree n = p^{L} for some L such that

\frac{1}{(n − i)!}F^{(n−i)}(z, x) ∈ y^{i}R

for i = 0, …, n. Since n = p^{L}, for i = 1 this expression becomes

\frac{1}{(n − 1)!}F^{(n−1)}(z, x) = \binom{p^{L}}{1}z + \binom{p^{L} − 1}{0}a_{1}x = p^{L}z + a_{1}x.

From the above condition on z, if we let L = N and a_{1} = a, we have

p^{L}z + a_{1}x ∈ yR,

which is the required condition for i = 1. We let F_{1}(S, U) = p^{L}S + a_{1}U.

As stated above, we wish to construct F (S, U) by repeated integration. If we knew that Int_{p^{L}}(F_{1})(z, x) were in the ideal (x^{2}, y^{2}), we could find an element a_{2} such that

Int_{p^{L}}(F_{1})(z, x) + a_{2}x^{2} = \binom{p^{L}}{2}z^{2} + \binom{p^{L} − 1}{1}a_{1}zx + \binom{p^{L} − 2}{0}a_{2}x^{2} ∈ y^{2}R.

and continue the construction. Unfortunately, this is not necessarily possible (we recall that although we are discussing the question of trying to show that z ∈ (x, y)R′ for a finite extension R′ of R, the actual result we prove is somewhat weaker). However, the next lemma shows that although we cannot solve this problem in R, we can solve it in R[p^{−1}].

Lemma 1. Let R̃ be a ring which contains a field of characteristic zero, and let H(S, T) be a homogeneous polynomial of degree n with coefficients in R̃. Suppose that x, y, z are elements of R̃ such that z ∈ (x, y) and such that

H^{(n−i)}(z, x) ∈ y^{i}R̃ for i = 0, …, n − 1.

Then

H(z, x) ∈ (x^{n}, y^{n}).

Proof. Since z ∈ (x, y), we can write z = cx − dy for some c and d in R̃. Using Taylor’s formula, we have

H(S + T, U) = ∑_{i=0}^{n} \frac{1}{(n − i)!}H^{(n−i)}(S, U)T^{n−i}.

We apply this formula with S = z, U = x, and T = dy, giving

H(z + dy, x) = ∑_{i=0}^{n} \frac{1}{(n − i)!}H^{(n−i)}(z, x)(dy)^{n−i}.

Since H^{(n−i)}(z, x) ∈ y^{i}R̃ for i = 0, …, n − 1, each term in the sum on the right is in y^{n}R̃ except possibly the term with i = n, which is H(z, x). Hence we can write

H(z + dy, x) = H(z, x) + ay^{n}

for some a ∈ R̃. Since z = cx − dy, we also have

H(z + dy, x) = H(cx, x) = x^{n}H(c, 1) ∈ x^{n}R̃.

Thus

H(z, x) = H(z + dy, x) − ay^{n} ∈ (x^{n}, y^{n}).

Let u_{2} = Int_{p^{L}}(F_{1})(z, x) as above. Then Lemma 1 implies that u_{2} ∈ (x^{2}, y^{2})R[p^{−1}], so there is an integer D such that p^{D}u_{2} ∈ (x^{2}, y^{2}). The main part of the proof consists of adjusting u_{2} so that it lies in (x^{2}, y^{2}) (and similarly adjusting the corresponding element at the ith stage so that it lies in (x^{i}, y^{i}). There are essentially three possibilities:

(1) The element u_{2} itself lies in (x^{2}, y^{2}). In this case, as outlined above, there is no problem.

(2) There is an element c ∈ R such that u_{2} ≡ cxyz modulo(x^{2}, y^{2}). We recall that we had p^{L}z ∈ (x, y), so p^{L}xyz ∈ (x^{2}, y^{2}). In this case, we can add (−c)xyz to u_{2} and proceed as before (we will explain this in detail below).

(3) The first two possibilities do not apply. In this case we have to multiply u_{2} by a power of p. We recall that L was chosen so that p^{L}z ∈ (x, y). Now we are raising the power of p needed, and we have to increase L. Thus we are now looking for a polynomial of larger degree.

These three procedures form the basis of the construction. One of the main facts which makes this construction work is that after a certain point the third case will no longer occur. The lemma we use here is the following.

Lemma 2. Let R be a ring as above. For each i, let

J_{i} = {r ∈ R|p^{N}r ∈ (x^{i}, y^{i}) for some N}.

Let Q_{i} = J_{i}/(x^{i}, y^{i}). Then (1) Q_{i} has finite length for all i.

(2) There exists an integer n such that for all k ≥ n, the map from Q_{n} to Q_{k} induced by multiplication by x^{k−n}y^{k−n} is an isomorphism.

Proof. The fact that Q_{i} has finite length follows immediately from the fact that p, x, y is a system of parameters.

Since x, y form part of a system of parameters and R is assumed to be integrally closed, x, y form a regular sequence. It follows that the map from Q_{n} to Q_{k} induced by multiplication by x^{k−n}y^{k−n} is always injective. To prove the second statement it must be shown that this map is also surjective for large enough n.

To prove this assertion we use the fact that the local cohomology module H_{𝔪}^{2}(R) has finite length, which follows from the assumptions we have made on R. More specifically, since R is an integrally closed domain of dimension three, its non-Cohen-Macaulay locus is supported at the maximal ideal. It then follows from local duality and the assumption that R is a homomorphic image of a regular local ring that the local cohomology module H_{𝔪}^{2}(R) has finite length (see [2], section 3.5). We remark that this is the only place where the assumption that R is a homomorphic image of a regular local ring is used.

The local cohomology module can be computed as the homology of the sequence

R_{p} × R_{x} × R_{y} →^{d_{1}} R_{px} × R_{py} × R_{xy} →^{d_{2}} R_{pxy},

where the first map sends (α_{1}, α_{2}, α_{3}) to (α_{1} −α_{2}, α_{3} −α_{1}, α_{2} −α_{3}), and the second map sends (β_{1}, β_{2}, β_{3}) to β_{1} + β_{2} + β_{3}. Since H_{𝔪}^{2}(R) is finitely generated, there is an integer n such that H_{𝔪}^{2}(R) is generated by elements which can be written in the form

(\frac{a}{(px)^{n}} , \frac{b}{(py)^{n}} , \frac{c}{(xy)^{n}} ) .

Let k ≥ n. Let r be an element of R and N an integer such that

p^{N}r = sx^{k} + ty^{k}

for s and t in R. We then have that the element

(\frac{−t}{p^{N}x^{k}} , \frac{−s}{p^{N}y^{k}} , \frac{r}{x^{k}y^{k}} ) ∈ R_{px} × R_{py} × R_{xy}

is in the kernel of d_{2}. By our choice of n, there is an element of the form

(\frac{a}{(px)^{n}} , \frac{b}{(py)^{n}} , \frac{c}{(xy)^{n}} )

such that the difference between this element and our original one is in the image of d_{1}. Looking at the third component, we can thus find an integer m and e, f ∈ R with

\frac{r}{x^{k}y^{k}} − \frac{c}{(xy)^{n}} = \frac{e}{x^{m}} − \frac{f}{y^{m}} .

Multiplying this equation by x^{k+m}y^{k+m}, we obtain

x^{m}y^{m}(r − cx^{k−n}y^{k−n}) ∈ (x^{k+m}, y^{k+m}).

Since the map induced by multiplication by x^{m}y^{m} from R/(x^{k}, y^{k}) to R/(x^{k+m}y^{k+m}) is injective, we can thus conclude that the image of r in R/(x^{k}, y^{k}) = Q_{k} is equal to x^{k−n}y^{k−n} times the image of c in R/(x^{n}, y^{n}) = Q_{n}. Thus the map induced by multiplication by x^{k−n}y^{k−n} from Q_{n} to Q_{k} is surjective.

The proof of the main theorem of this paper proceeds by repeating the process of integrating as we have outlined in the construction of F_{2} from F_{1} above. In the process, we obtain elements z_{i} at stage i and attempt to add a U^{i} term in such a way to assure that if F_{i}(S, U) is the new polynomial, we have F_{i}(z, x) ∈ y^{i}R.

At each stage there are three cases as above. If the first case holds, there is no problem. If the second case holds, we adjust z_{i} and show that we can still continue as before. If we are in the third case, we have to multiply by a power of p and it is necessary to increase L. Using lemma 2, we can show that eventually the third case will not occur.

The other thing we have to keep track of is the divisibility of binomial coefficients by powers of p. For this we introduce the function τ defined as follows. Let \overline{τ}(n) equal the sum of digits of n−1 in its p-adic expansion. Then let τ(n) = \overline{τ}(n)/(p−1). We let τ(1) = 0. We note that τ(n) is a rational number with denominator dividing p − 1, and we emphasize the fact that τ(n) is defined by the p-adic expansion of n − 1 rather than that of n.

As an example. we compute τ(p^{n}) and τ(p^{n} + 1). Since the p-adic expansion of p^{n} −1 consists of n digits each equal to p−1, we have τ(p^{n}) = n(p−1)/(p−1) = n. The p-adic expansion of p^{n} consists of a 1 followed by n zeros, so τ(p^{n} + 1) = 1/(p − 1).

We next compute the difference τ(n) − τ(n − 1) for any n > 1. Let a be the largest integer for which p^{a} divides n − 1 (a will be zero if p does not divide n − 1). Then the last a digits of the p-adic expansion of n − 2 are equal to p − 1 and the a + 1 digit is less than p − 1. Thus when we add 1 to n − 2, the last a digits are replaced with zeros and the a + 1 digit is increased by 1. Hence we have

τ(n) − τ(n − 1) = \frac{1}{p − 1} − a.  (1)

The connection between τ and binomial coefficients comes from the following lemma. In this lemma and the following discussion we use the notation p^{k}∥n to mean that p^{k} is the highest power of p which divides n.

Lemma 3. Let 0 < j ≤ i ≤ p^{L} be integers. Then p^{k}∥\binom{p^{L} − j}{i − j}, where

k = τ(j) + τ(i − j + 1) − τ(i).

Proof We fix j and L and prove the result by induction on i. If j = i, then

\binom{p^{L} − j}{i − j} = 1 and τ(j) + τ(i − j + 1) − τ(i) = τ(j) + τ(1) − τ(j) = 0; since the

highest power of p which divides 1 is p^{0}, the result is true in this case.

Now assume that j < i ≤ p^{L} and that the result is true for i − 1. We compute how each side of the equation changes when we go from i − 1 to i.

For the left hand side, we compute

\binom{p^{L} − j}{i − j} = \binom{p^{L} − j}{(i − 1) − j} × [\frac{p^{L} − j − (i − j) + 1}{i − j}] =

= \binom{p^{L} − j}{(i − 1) − j} × [\frac{p^{L} − i + 1}{i − j}] .

Hence the highest power of p which divides this binomial coefficient remains the same unless p divides p^{L} − i + 1 or p divides i − j. Let p^{a}∥(p^{L} − i + 1) and

p^{b}∥(i − j). Then we obtain the highest power of p dividing \binom{p^{L} − j}{i − j} from that for

\binom{p^{L} − j}{(i − 1) − j} by adding a − b.

We now compute how τ(j) + τ(i − j + 1) − τ(i) changes as we pass from i − 1 to i. We have

[τ(j) + τ(i − j + 1) − τ(i)] − [τ(j) + τ((i − 1) − j + 1) − τ(i − 1)]

= [τ(j) + τ(i − j + 1) − τ(i)] − [τ(j) + τ(i − j) − τ(i − 1)]
= [τ(i − j + 1) − τ(i − j)] − [τ(i) − τ(i − 1)].

We will now express this difference using the a and b defined earlier in the proof. We have defined b by the condition p^{b}∥(i−j) and a by the condition p^{a}∥(p^{L} −i+1). However, since 0 < j < i ≤ p^{L}, we have 0 < i − 1 < p^{L}, and thus we also have p^{a}∥(i − 1). Hence from equation 1 we obtain

[τ(i − j + 1) − τ(i − j)] − [τ(i) − τ(i − 1)]

= \frac{1}{p − 1} − b − (\frac{1}{p − 1} − a) = a − b.

This is the same quantity as we computed for the left hand side of the equation, so this proves the lemma by induction.

The above proof of Lemma 3 is the shortest way to obtain the result we need for binomial coefficients that appear in Taylor expansions but it may appear somewhat unmotivated. We outline the steps of a more natural approach to this result.

We let σ_{p} denote the function given by σ_{p}(n) = the sum of the digits in the p-adic expansion of n for an integer n ≥ 0; note that we thus have τ(n) = σ_{p}(n−1)/(p−1).

(1) Show by induction that p^{a}∥n! where a = (n − σ_{p}(n))/(p − 1).

(2) Use step 1 to derive a formula for the highest power of p that divides the

binomial coefficient \binom{n}{k} for integers 0 ≤ k ≤ n.

(3) Apply the formula from step 2 to compute the highest power of p that

divides \binom{p^{L} − j}{i − j} and prove Lemma 3.

2. Proof of the Main Theorem.

Theorem 1. (Heitmann) Let R be an integrally closed local domain of mixed characteristic p and of dimension 3. Let p, x, y be a system of parameters for R. Assume that R contains an element σ such that σ^{p−1} = p. Let z be an element of R and N a positive integer such that p^{N}z ∈ (x, y). Then for any rational number ϵ > 0, there exists a finite extension R′ of R in which

p^{ϵ}z ∈ (x, y)R′.

The main idea, as in the analogous situation explained in the previous section, is to prove the existence of a monic polynomial f(T) satisfying conditions which imply that if w is a root of f(T), then p^{ϵ}w will be integral over R, and if we let v be the element such that p^{ϵ}z = p^{ϵ}wx + p^{ϵ}vy, then p^{ϵ}v is also integral over R. We next prove the main lemma which shows that this can be done.

We remark that the reason for requiring that R contain the element σ is that we will need to define powers p^{r} of p where r is a rational number of the form n/(p−1) for some integer n. We can then define p^{r} = p^{n/(p−1)} = σ^{n}.

Lemma 4. Let R, p, x, y, and z be as in the statement of Theorem 1, and let K be a positive integer. Then there exists an integer L and a polynomial

f(T) = T^{p^{L}} + a_{1}T^{p^{L}−1} + ⋯ + a_{p^{L}}

satisfying the following conditions:

(1) a_{j} ∈ p^{K−τ(j)} for j = 1, …, p^{L}.

(2) If we let F (S, U) = S^{p^{L}} + a_{1}S^{p^{L}−1}U + ⋯ + a_{p^{L}}U^{p^{L}} as in the previous section, we have

F^{(k)}(z, x) ∈ y^{p^{L}−k}R[p^{−1}]

for k = 0, …, p^{L}.

Proof. We fix a positive integer K.

As outlined in the previous section, we construct the polynomial F (S, U) inductively starting with degree 1. Throughout the construction we will make use of the quantity E_{i} defined by the formula

E_{i} = K − τ(i) + τ(2).

Note the E_{i} is a rational number with denominator dividing p − 1 and that it can be positive or negative.

We now state the situation we have after the ith step precisely. We assume that we have, for each integer k with 1 ≤ k ≤ i,

(1) A homogeneous polynomial F_{k}(S, U) of degree k,

(2) an integer L_{k} > K with p^{L_{k}} > k for all k < i (we may have p^{L_{i}} = i; in that case L_{i} will be the L of the Lemma and we will be done),

(3) an element z_{k} in R with p^{N_{k}}z_{k} ∈ (x^{k}, y^{k}) for some integer N_{k}. Furthermore, these polynomials, elements of R, and integers will satisfy the following conditions:

(A) The coefficient of S^{k} in F_{k}(S, U) is \binom{p^{L_{k}}}{k}. Thus we can write

F_{k}(S, U) = \binom{p^{L_{k}}}{k}S^{k} + \binom{p^{L_{k}} − 1}{k − 1}a_{k1}S^{k−1}U + … + \binom{p^{L_{k}−k}}{0}a_{kk}U^{k},

where the a_{kj} are elements of R[p^{−1}] but may not be in R. (B) For j = 1, …, k we have a_{kj} ∈ p^{K−τ(j)}. (C) F_{k}^{(m)}(z, x) ∈ y^{k−m}R[p^{−1}] for m = 0, …, k. (D) If we let G_{1}(S, U) = S and G_{k}(S, U) = p^{−E_{k}}Int_{p^{L_{k−1}}} F_{k−1}(S, U) for 2 ≤

k ≤ i, then G_{i}(z, x) = z_{i}. This is a lengthy induction hypothesis; however, the induction step is itself quite

complicated and this information from previous steps is used.

We now do the first step, where i = 1. By hypothesis, there is nonnegative integer N_{1} > 0 such that p^{N_{1}}z ∈ (x, y), so we can find an a ∈ R such that p^{N_{1}}z + ax ∈ yR. Multiplying this expression by p^{K} we have that p^{N_{1}+K}z + p^{K}a ∈ yR. We let L_{1} = N_{1} + K and a_{11} = p^{K}a, and we define

F_{1}(S, U) = p^{N_{1}+K}S + p^{K}aU = p^{L_{1}}S + a_{11}U.

Then F_{1} is a homogeneous polynomial of degree 1 as required. Let z_{1} = z. Then we have L_{1} > K, p^{L_{1}} > 1, and p^{N_{1}}z_{1} ∈ (x, y), so conditions 1 through 3 are satisfied.

The coefficient of S in F_{1} is p^{L_{1}} = \binom{p^{L_{1}}}{1}, so condition A holds. Since \binom{p^{L_{i}}}{0} =

1, we have

F_{1}(S, U) = \binom{p^{L_{1}}}{1}S + \binom{p^{L_{1}}}{0}a_{11}U.

Condition B states that a_{11} ∈ p^{K−τ(1)}R = p^{K}R, which is true since we defined a_{11} = p^{K}a. Condition C states that

F_{1}(z, x) = p^{L_{1}}z + a_{1}x = p^{N+K}z + p^{K}ax ∈ yR[p^{−1}],

which is also true by construction (in fact, we have p^{N+K}z + p^{K}ax ∈ yR in this case). Finally, since G_{1}(S, U) = S, we have z_{1} = z = G_{1}(z, x). Hence all the necessary conditions are satisfied for i = 1.

We now suppose that L_{k}, z_{k}, and F_{k}(S, U) have been defined for 1 ≤ k < i and define F_{i}(S, U), L_{i}, and z_{i}.

As stated in the previous section, the main idea is to integrate F_{i−1}(S, U) and modify it so that it will satisfy the conditions listed above. However, to make the construction work it is also necessary to multiply the integral by a power of p using the number E_{i} = K − τ(i) + τ(2) defined above.

As a first approximation to F_{i}, we let

G_{i}(S, U) = p^{−E_{i}}Int_{p^{L_{i−1}}} (F_{i−1}(S, U)).

(This is the same polynomial as in condition D for i ≥ 2.)

Let

z_{i} = G_{i}(z, x).

We claim that z_{i} is an element of R. To simplify notation we let a_{j} = a_{i−1,j} for j = 0, …, i − 1 (with a_{0} = 1), so that we have

F_{i−1}(S, U) = ∑_{j=0}^{i−1} \binom{p^{L_{i−1}} − j}{i − 1 − j}a_{j}S^{i−1−j}U^{j}.

We then have that a_{j} ∈ p^{K−τ(j)} for j = 1, …, i − 1 by induction.

By the definitions of z_{i} and G_{i}(S, U), we have that

z_{i} = p^{−E_{i}} ∑_{j=0}^{i−1} \binom{p^{L_{i−1}} − j}{i − j}a_{j}z^{i−j}x^{j}.

We show that each term of this sum is in R; to do so we distinguish two cases.

If j = 0, then a_{j} = 1, and we must show that p^{−E_{i}} \binom{p^{L_{i−1}}}{i} ∈ R. To see this,

we note that the highest power of p dividing \binom{p^{L_{i−1}}}{i} is p^{L_{i−1}−k}, where p^{k}∥i (of

course, k can be zero). Thus to show that p^{−E_{i}} \binom{p^{L_{i−1}}}{i} ∈ R we must show that

−E_{i} + L_{i} − k = −K + τ(i) − τ(2) + L_{i−1} − k ≥ 0.

Since L_{i−1} > K, this reduces to showing that τ(i) + 1 ≥ τ(2) + k. Since p^{k} divides i, the p-adic expansion of i − 1 ends with k digits equal to p − 1. Hence

τ(i) + 1 ≥ k + 1 ≥ k + (1/(p − 1)) = k + τ(2),

so the inequality holds.

We now consider the general case, where i ≤ p^{L_{i−1}} and 0 < j < i. In this case we

use Lemma 3. We must show that p^{−E_{i}} \binom{p^{L_{i−1}} − j}{i − j}a_{j} ∈ R. Putting together the

powers of p dividing the factors in this product, and using the fact that the highest

power of p that divides \binom{p^{L_{i−1}} − j}{i − j} is τ(j) + τ(i − j + 1) − τ(i), the inequality to

be proven is

−K + τ(i) − τ(2) + τ(j) + τ(i − j + 1) − τ(i) + K − τ(j) ≥ 0.

This expression simplifies to

τ(i − j + 1) − τ(2) ≥ 0,

which is true since i > j, so that i − j + 1 ≥ 2 and thus τ(i − j + 1) ≥ τ(2).

Thus we have shown that z_{i} ∈ R. We claim that we also have that p^{N_{i}}z_{i} ∈ (x^{i}, y^{i})R for some N_{i}. If we can show that z_{i} ∈ (x^{i}, y^{i})R[p^{−1}], then we can conclude that p^{N_{i}}z_{i} ∈ (x^{i}, y^{i})R for some N_{i} by clearing denominators. To show that z_{i} ∈ (x^{i}, y^{i})R[p^{−1}] we use the fact that G_{i}(S, U) is, up to a constant which is a unit in R[p^{−1}], the integral of F_{i−1}(S, U), and z_{i} = G_{i}(z, x). Hence, again up to constants which again are units in R[p^{−1}], we have that G_{i}^{(k)}(z, x) agrees with F_{i−1}^{(k−1)}(z, x) for k = 1, …, i. By induction we have that F_{i−1}^{(k−1)}(z, x) ∈ y^{(i−1)−(k−1)})R[p^{−1}] for k = 1, …, i. It thus follows that G_{i}^{(k)}(z, x) ∈ y^{i−k}R[p^{−1}] for k = 1, …, i. Hence Lemma 1 implies that G_{i}(z, x) ∈ (x^{i}, y^{i})R[p^{−1}] as was to be shown.

To summarize the argument up to this point, we have defined

G_{i}(S, U) = p^{−E_{i}}Int_{p^{L_{i−1}}} (F_{i−1}(S, U)),

defined z_{i} = G_{i}(z, x), and shown that z_{i} is an element of R such that p^{N_{i}}z_{i} ∈ (x^{i}, y^{i}) for some integer N_{i}. We now wish to add a U^{i} term to G_{i}(S, U) so that if we evaluate the resulting polynomial at (z, x) the result is in y^{i}R[p^{−1}] and continue the induction. In fact, G_{i}(S, U) may have to be modified first as we show below. We distinguish three cases.

Case 1. The first, and easiest, case, is when z_{i} ∈ (x^{i}, y^{i})R. We then know that there exists an element a such that

z_{i} + ax^{i} ∈ y^{i}R.

We now let

F_{i}(S, U) = p^{E_{i}}(G_{i}(S, U) + aU^{i}).

It is clear that F_{i}(z, x) = p^{E_{i}}(z_{i} + ax^{i}) ∈ y^{i}R[p^{−1}]. Furthermore, since the derivatives of F_{i} with respect to S are, up to a constant which is a unit in R[p^{−1}], the same as those of G_{i}, it follows from the argument in the next to the last paragraph that

F_{i}^{(k)}(z, x) ∈ y^{i−k}R[p^{−1}]

for k = 1, …, i, and hence, since this condition also holds for k = 0, it holds for k = 0, …, i.

We let L_{i} = L_{i−1}. Then L_{i} > K, and conditions (1) through (3) are satisfied except that we may have p^{L_{i}} = i, which would mean that we could not continue the induction to i + 1. However, we claim that in this case the polynomial

f(T) = F_{i}(T, 1)

satisfies the conditions required in the conclusion of Lemma 4 and we are done. However, we first verify that conditions A through D hold.

Conditions A and D hold by construction, and we have already checked condition C. That leaves condition B, that the coefficients a_{ij} are in p^{K−τ(j)}R. For j < i, we have a_{ij} = a_{i−1,j}, and this condition follows by induction. Thus we must check that a_{i} ∈ p^{K−τ(i)}R. Since a ∈ R, we have a_{i} = p^{E_{i}}a ∈ p^{E_{i}}R = p^{K−τ(i)+τ(2)}R ⊆ p^{K−τ(i)}R, which proves that the condition holds in this case as well.

We now return to the case in which p^{L_{i}} = i. If we let f(T) = F_{i}(T, 1), then

the leading coefficient of f(T) is \binom{p^{L_{i}}}{i}, which is 1, so f(T) is monic as required.

Furthermore, conditions (1) and (2) of Lemma 4 follow immediately from conditions B and C of the induction hypothesis. Thus if p^{L_{i}} = i, the proof of Lemma 4 is complete.

If p^{L_{i}} > i, we continue the induction. We remark that the argument that a_{ii} ∈ p^{K−τ(i)}R and the fact that the proof of Lemma 4 is complete when p^{L_{i}} = i will be used in the other cases as well.

Case 2. We assume now that z_{i} is not in (x^{i}, y^{i}), but that we can write z_{i} modulo (x^{i}, y^{i}) in terms of z_{k} for 1 ≤ k < i in the following sense: there exist c_{1}, …, c_{i−1} such that

z_{i} + c_{i−1}xyz_{i−1} + ⋯ + c_{1}x^{i−1}y^{i−1}z_{1} ∈ (x^{i}, y^{i})R.

We recall that for each k = 1, … i − 1 we have

z_{k} = G_{k}(z, x),

where G_{k} is the homogeneous polynomial of degree k defined above. We now let

F̃_{i}(S, U) = G_{i}(S, U) + c_{i−1}yUG_{i−1}(S, U) + ⋯ + c_{1}y^{i−1}U^{i−1}G_{1}(S, U).

From the above equation we thus have

F̃_{i}(z, x) ∈ (x^{i}, y^{i}),

so we can find an element a ∈ R such that

F̃_{i}(z, x) + ax^{i} ∈ y^{i}R.

We define

F_{i}(S, U) = p^{E_{i}}(F̃_{i}(S, U) + aU^{i}).

We must now check that F_{i} has the required properties. We let L_{i} = L_{i−1} as in the previous case. Since the coefficient of S^{i} in F_{i} is the same as that of

p^{E_{i}}G_{i} = Int_{p^{L_{i−1}}} (F_{i−1}), we have by induction that this coefficient is \binom{p^{L_{i}}}{i} as

required. We now check that a_{ij} ∈ p^{K−τ(j)}R for j = 1, …, i − 1; the verification that a_{ii} ∈ p^{K−τ(i)}R is the same as in the previous case. We must show that the coefficients of p^{E_{i}} F̃(S, U) satisfy Condition B, and, since the coefficients of p^{E_{i}}G_{i}(S, U) satisfy this condition, it suffices to show that the contributions of the coefficients of p^{E_{i}}y^{i−k}U^{i−k}G_{k} satisfy the condition for each k = 1, …, i − 1.

Let k be an integer with 1 ≤ k ≤ i − 1. By construction, we can write

G_{k}(S, U) = p^{−E_{k}} (∑_{j=0}^{k−1} \binom{p^{L_{k}} − j}{k − j}a_{kj}S^{k−j}U^{j}) ,

where a_{kj} ∈ p^{K−τ(j)}R. What we have to show is that if we write

p^{E_{i}}U^{i−k}G_{k}(S, U) = ∑_{m=0}^{i−1} \binom{p^{L_{i}} − m}{i − m}a′_{m}S^{i−m}U^{m},

then we have a′_{m} ∈ p^{K−τ(m)}R.

We note that the since we are multiplying by U^{i−k}, the term of G_{k}(S, U) which contains a_{kj} corresponds to the term of p^{E_{i}}U^{i−k}G_{k}(S, U) which contains a′_{j+i−k}. Thus we have to show that if a_{kj} ∈ p^{K−τ(j)}R and

p^{E_{i}}p^{−E_{k}} \binom{p^{L_{k}} − j}{k − j}a_{kj} = \binom{p^{L_{i}} − (j + i − k)}{i − (j + i − k)}a′_{j+i−k},

then a′_{j+i−k} ∈ p^{K−τ(j+k−i)}R. Dividing by the binomial coefficient on the right, this means that

p^{E_{i}}p^{−E_{k}} \binom{p^{L_{k}} − j}{k − j}p^{K−τ(j)}\binom{p^{L_{i}} − (j + i − k)}{i − (j + i − k)}^{−1} ∈ p^{K−τ(j+i−k)}R.

To verify this statement we use Lemma 3 to determine the power of p dividing the binomial coefficients in this expression. We also use the above expressions for E_{i} and E_{k}. What results is that we must prove the inequality

(K − τ(i) + τ(2)) − (K − τ(k) + τ(2)) + (τ(j) + τ(k − j + 1) − τ(k))
+(K − τ(j)) − (τ(j + i − k) + τ(k − j + 1) − τ(i)) ≥ K − τ(j + i − k).

When this is worked out, it is seen that the two sides of the equation are in fact equal.

We must now prove the Condition C on derivatives. By construction, we have

F_{i}(z, x) ∈ y^{i}R[p^{−1}].

To prove that F_{i}^{(k)}(z, x) ∈ y^{i−k}R[p^{−1}] for k ≥ 1 it suffices to prove the corresponding condition for F̃_{i}, and to prove this it suffices to prove it for each term in the sum defining F̃_{i}. For G_{i} the proof is the same as in the previous case. For c_{m}y^{i−m}U^{i−m}G_{m}(S, U), it suffices to show that if we let H(S, U) = y^{i−m}G_{m}(S, U), then H^{(k)}(z, x) ∈ y^{i−k}R[p^{−1}] for k ≥ 0. By construction, we have that G_{m}^{(k)}(z, x) ∈ y^{m−k}R[p^{−1}] for k ≥ 0. Hence

H^{(k)}(z, x) = y^{i−m}G_{m}^{(k)}(z, x) ∈ y^{i−m}y^{m−k}R[p^{−1}] = y^{i−k}R[p^{−1}]

as was to be shown.

This finishes the second case.

Case 3. This is the case in which z_{i} cannot be forced into (x^{i}, y^{i})R even after modifying it by multiples of x^{i−k}y^{i−k}z_{k}. Let N_{i} be a positive integer such that p^{N_{i}}z_{i} ∈ (x^{i}, y^{i})R. We now let L_{i} = L_{i−1} + N_{i}. Note that this is the only case where we increase L_{i}. We now let

F̃_{i}(S, U) = \binom{p^{L_{i}}}{i}\binom{p^{L_{i−1}}}{i}^{−1}G_{i}(S, U).

Regardless of what i is, the highest power of p dividing \binom{p^{L_{i}}}{i}\binom{p^{L_{i−1}}}{i}^{−1} is p^{L_{i}−L_{i−1}} =

p^{N_{i}}. Hence, since p^{N_{i}}z_{i} ∈ (x^{i}, y^{i})R, we have F̃_{i}(z, x) ∈ (x^{i}, y^{i}), and we can find an element a such that F̃_{i}(z, x) + ax^{i} ∈ y^{i}R. We now let

F_{i}(S, U) = p^{E_{i}}(F̃_{i}(S, U) + aU^{i}).

Since F_{i} is, apart from the U^{i} term, a constant multiple of G_{i} by a unit in R[p^{−1}], the condition on derivatives is clear in this case. Furthermore, we have

constructed F_{i} so that the coefficient of S^{i} is \binom{p^{L_{i}}}{i}, so it satisfies the Condition

A. We now check the Condition B on the a_{ij}. In fact, it follows from Lemma 3

that the highest power of p dividing \binom{p^{L_{i}} − j}{i − j} is the same as the highest power of

p dividing \binom{p^{L_{i−1}} − j}{i − j} for all j = 1, …, i − 1. Hence the divisibility conditions on

a_{ij} for G_{i} are the same as those for F_{i}. Thus, since the a_{ij} are, up to units in R, multiples of a_{i−1,j} by p^{N_{i}}, the condition follows. Thus all the conditions hold, and this completes the proof of Case 3.

To finish the proof of the lemma, we invoke Lemma 2. As in that lemma, we let Q_{i} denote the submodule of R/(x^{i}, y^{i}) consisting of elements u annihilated by p^{N} for some N. Lemma 2 states that for large i multiplication by x^{k}y^{k} identifies Q_{i} with Q_{i+k} for all k ≥ 0. Let i be large enough so that this holds. For each n let M_{n} denote the submodule of Q_{n} generated by z_{n}, xyz_{n−1}, …, x^{n−1}y^{n−1}z_{1}, and for each k ≥ 0 let N_{k} be the submodule of Q_{i} corresponding to M_{i+k} under the above identification. Since the N_{k} form an increasing sequence of submodules of Q_{i}, we must have N_{k} = N_{k+1} for sufficiently large k. It follows that multiplication by xy identifies M_{i} with M_{i+1} for large i. Thus for large i it is always possible to write z_{i} in terms of z_{i−1}, … z_{1} as in Case 2, and Case 3 will not occur. Thus we will eventually reach the situation where i = p^{L_{i}}, and at that point F_{i}(S, U) will satisfy the required conditions.

We now prove the theorem. Assume that p^{N}z ∈ (x, y), where p, x, y form a system of parameters. We wish to show that for all rational ϵ > 0, we have p^{ϵ}z ∈ (x, y)R′ for some finite extension R′ of R. Choose K > 0 such that 1/p^{K} < ϵ; we can in fact assume that ϵ = 1/p^{K}. Let f(T) be a polynomial such that

(1) f(T) = T^{n} + a_{1}T^{n−1} + … + a_{n}, where

(2) F^{(k)}(z, x) ∈ y^{n−k}R[p^{−1}] for k = 0, …, n, and

(3) a_{j} ∈ p^{K−τ(j)}R for j = 1, …, n.

We claim that if we let g(T) = ∑_{i=0}^{n} p^{ϵj}a_{j}T^{n−j}, then g(T) has coefficients in R. This amounts to the statement that

jϵ + K − τ(j) = j/p^{K} + K − τ(j) ≥ 0

for all j.

To prove this inequality, we first note that if j ≤ p^{K}, then τ(j) ≤ K and the inequality is clear. Thus we may assume that j > p^{K}, and this implies that there is a nonnegative integer M such that p^{K+M} ≤ j ≤ p^{K+M+1}.

The inequalities p^{K+M} ≤ j ≤ p^{K+M+1} imply that j/p^{K} ≥ p^{M} and τ(j) ≤ K + M + 1. We thus have

j/p^{K} + K − τ(j) ≥ p^{M} + K − (K + M + 1) = p^{M} − M − 1.

It thus suffices to show that for all nonnegative integers M and all prime numbers p we have p^{M} ≥ M + 1. For M = 0 this states that p^{0} ≥ 1, and for M = 1 it states that p ≥ 2, and both of these statements are true. The general case can be shown by an easy induction on M.

Thus g(T) has coefficients in R. Let w be a root of f(T); we then have that p^{ϵ}w is a root of g(T) so is integral over R. The condition on derivatives in Lemma 4 implies that if we let v = (z − wx)/y, then v is integral over R[p^{−1}], and hence p^{ϵ}v is also integral over R[p^{−1}]. Furthermore, p^{ϵ}v = (p^{ϵ}z − (p^{ϵ}w)x)/y is a quotient of an element integral over R by y so is clearly integral over R[y^{−1}]. Thus p^{ϵ}v is integral over R[p^{−1}] and R[y^{−1}], so, since p and y generate an ideal of height 2 in R and R is a normal domain, p^{ϵ}v is integral over R. We now have

p^{ϵ}z = (p^{ϵ}w)x + (p^{ϵ}v)y,

so p^{ϵ}z is in the ideal generated by x and y in a finite extension of R. This concludes the proof of the theorem.

3. A proof of the Canonical Element Conjecture in dimension 3

In this section we show how Heitmann’s theorem can be used to give a direct proof of the Canonical Element Conjecture in dimension 3. We recall the statement of one version of this conjecture.

Let x_{1}, …, x_{d} be a system of parameters for R. Let K_{•} be the Koszul complex on x_{1}, …, x_{d}, and let F_{•} be a minimal free resolution of R/(x_{1}, …, x_{d}). Since K_{•} is a complex of free modules and F_{•} is exact, there is a map of complexes ϕ_{•} : K_{•} → F_{•} which is the identity map in degrees 0 and 1 (note that the maps K_{1} → K_{0} and F_{1} → F_{0} are the same). The Canonical Element Conjecture states that

ϕ_{d}(1) ∉ 𝔪F_{d}.

In dimension 3 in mixed characteristic, we can assume that the system of parameters is of the form p^{N}, x, y, where x, y form a regular sequence. Then the homology of the Koszul complex K_{•} in degree 1 is isomorphic to Q/(x, y), where Q = {r ∈ R|p^{N}r ∈ (x, y)}. If s ∈ R is such that sQ ⊆ (x, y), then s annihilates the homology of K_{•} in degree 1, and, since K_{•} is exact in degrees 2 and 3, we can construct a map ψ_{•} : F_{•} → K_{•} such that ψ_{•} is multiplication by s in degrees 0 and 1.

Now suppose that we had ϕ_{d}(1) ∈ 𝔪F_{d}. We take s = p^{e} for some small rational number e. Let R′ be a finite integrally closed extension of R containing p^{e} and such that p^{e}(QR′) ⊆ (x, y)R′; the existence of such an extension R′ is guaranteed by Heitmann’s Theorem. By the previous paragraph, there is a map ψ_{•} : F_{•} ⊗ R′ → K_{•} ⊗ R′ which is multiplication by p^{e} in degrees 0 and 1. Thus the composition ψ_{•}ϕ_{•} is a map from K_{•} ⊗ R′ to itself which is also given by multiplication by p^{e} in degrees 0 and 1 (we use the notation ϕ_{•} also to denote the extension of the original ϕ_{•} to R′). Since K_{•} ⊗ R′ is exact in degrees 2 and 3, we can conclude that multiplication by p^{e} is homotopic to ψ_{•}ϕ_{•}. Thus there is a map σ from K_{2} ⊗ R′ to K_{3} ⊗ R′ such that for all a ∈ K_{3} ⊗ R′ we have

p^{e}a = ψ_{•}ϕ_{•}(a) + σ(d_{3}^{K}(a)).

Since we are assuming that ϕ_{d}(a) ∈ 𝔪R′, and d_{3}^{K}(a) is clearly in 𝔪R′, this implies that p^{e} ∈ 𝔪R′. The next lemma shows that this is impossible if we take e small enough.

Lemma 5. Let R be a local integral domain with maximal ideal 𝔪, and let c be a nonzero element of R. Then for sufficiently small rational e > 0, for any finite extension R′ of R, we have c^{e} ∉ 𝔪.

Proof We use the fact that there is a discrete valuation v on R such that v(m) > 0 for all m ∈ 𝔪. Let u be the minimum of v(m) for a set of generators m of 𝔪. Then if we choose e < u/v(c), since v can be extended to any finite extension R′ of R, we have that v(c^{e}) = ev(c) < u ≤ v(m) for all m in a set of generators for 𝔪R′. Thus v(c^{e}) < v(m) for all m ∈ 𝔪R′, so c^{e} ∉ 𝔪R′.

References

[1] R. Heitmann, The direct summand conjecture in dimension three, Annals of Mathematics

156 (2002), 695–712.

[2] W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge University Press, Cambridge