WEAK SUBINTEGRALITY AND INVERTIBLE MODULES IN GRADED RINGS

LESLIE G. ROBERTS AND BALWANT SINGH

Abstract. We show that the isomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) introduced by the second author for a subintegral extension A ⊆ B of positively graded rings with A_{0} = B_{0} is also an isomorphism if A ⊆ B is only a weakly subintegral extension.

1. Introduction

In [5] the second author showed that if A ⊆ B is a subintegral extension of positively graded rings with A_{0} = B_{0}, then there exists a functorial isomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}). Here A_{⋆} is the divided power algebra associated with A, which is the ring having the same additive group structure as A but with multiplication (denoted ⋆) given on homogeneous elements by f ⋆ g = \binom{m+n}{n}fg, where m =deg(f) and n =deg(g). It was also remarked in [5] that ζ is still defined and a homomorphism if A ⊆ B is only weakly subintegral. The goal of this paper is to verify that indeed if A ⊆ B is a weakly subintegral extension of positively graded rings with A_{0} = B_{0}, then the homomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) is an isomorphism. A key new observation used in the proof is that if A ⊆ B is weakly subintegral then A_{⋆} ⊆ B_{⋆} is subintegral.

It was also shown in [5] that in the subintegral case there is a natural isomorphism gr_{ℕ}ℐ (A, B) ≅ ℐ (A_{⋆}, B_{⋆}), where gr_{ℕ} is the associated graded corresponding to a filtration induced on ℐ (A, B) by the gradings on A and B. In the weakly subintegral but not subintegral case this isomorphism need no longer hold, as noted in Remark 6.2, so the relation between B/A and ℐ (A, B) is largely lost.

All the rings that we consider are commutative with 1, and all ring homomorphisms are unitary. Throughout ℤ denotes the integers, ℕ the positive integers, ℤ_{+} the non-negative integers, and ℚ the rational numbers.

For an extension of rings A ⊆ B, ℐ (A, B) denotes the multiplicative group of invertible A-submodules of B. It is easily seen that ℐ is a functor from extensions of rings to abelian groups. We will denote the units of a ring A by A^{×}. Given an extension A ⊆ B there is an exact sequence

1 → A^{×} → B^{×} → ℐ (A, B) → Pic(A) → Pic(B). (1.1)

This and other properties of ℐ can be found in [3, Section 2].

Now we recall the definition and some of the properties of subintegral and weakly subintegral extensions.

1991 Mathematics Subject Classification. Primary: 13B22; Secondary: 19A49.

This work was supported by the NSERC grant of the first author. The second author thanks Queen’s University for its hospitality.

A ring extension A ⊆ B is subintegral if it is integral, the induced map Spec(B) → Spec(A) is a bijection, and the induced maps on residue field extensions are isomorphisms. An elementary subintegral extension is an extension of the form A ⊆ B with B = A[b] for some b ∈ B such that b^{2}, b^{3} ∈ A. An extension A ⊆ B is subintegral if and only if it is the filtered union of elementary subintegral extensions, that is, for each b ∈ B there is a finite sequence A = C_{(0)} ⊆ C_{(1)} ⊆ ⋯ ⊆ C_{(r)} ⊆ B of ring extensions such that b ∈ C_{(r)} and C_{(i−1)} ⊆ C_{(i)} is elementary subintegral for each i, 1 ≤ i ≤ r. These and other properties of subintegral extensions can be found in [6] or various papers by the present authors.

A ring extension A ⊆ B is weakly subintegral if it is integral, the induced map Spec(B) → Spec(A) is a bijection, and the induced maps on residue field extensions are purely inseparable field extensions. An elementary weakly subintegral extension is an extension of the form A ⊆ B with B = A[b] for some b ∈ B such that b^{p}, pb ∈ A for some positive prime p. An extension A ⊆ B is weakly subintegral if and only if it is the filtered union of elementary subintegral extensions and elementary weakly subintegral extensions. That is, for each b ∈ B there is a finite sequence A = C_{(0)} ⊆ C_{(1)} ⊆ ⋯ ⊆ C_{(r)} ⊆ B of ring extensions such that b ∈ C_{(r)} and C_{(i−1)} ⊆ C_{(i)} is elementary subintegral or elementary weakly subintegral for each i, 1 ≤ i ≤ r. These and other properties of weakly subintegral extensions can be found in [7] or various papers by the present authors.

If b ∈ B and A ⊆ A[b] is subintegral then we say that b is subintegral over A. Similarly if b ∈ B and A ⊆ A[b] is weakly subintegral then we say that b is weakly subintegral over A.

For a positively graded ring A = ⨁_{n≥0} A_{n}, denote by Â the completion of A with respect to the ideal A_{+} = ⨁_{n≥1} A_{n}. Note that Â consists precisely of possibly infinite sums ∑_{n≥0} f_{n} with f_{n} ∈ A_{n} for every n.

Now we recall some notation introduced in [5, Section 1]. Denote by Â_{⋆} the ring with the same additive group as Â but with multiplication (denoted by ⋆) given on homogeneous elements by f ⋆ g = \binom{m+n}{n}fg, where m = deg(f), n = deg(g) and fg is the product in A, and extended by bilinearity and completion to Â. We denote by A_{⋆} the ring with the same additive group as A, and multiplication ⋆ as defined on Â_{⋆}. If f ∈ A_{d} then f^{n} will denote the n^{th} power of f in the multiplication of A and f^{⋆n} will denote the n^{th} power of f in the multiplication of A_{⋆}. It is easily seen that if f ∈ A_{d} then

f^{⋆n} = \frac{(nd)!}{(d!)^{n}}f^{n}.

If ℚ ⊆ A and f ∈ A_{d}, d ≥ 1, then we can define the exponential of f in Â_{⋆} by the usual formula

e^{⋆f} = ∑_{n=0}^{∞} \frac{f^{⋆n}}{n!} = ∑_{n=0}^{∞} \frac{(nd)!}{(d!)^{n}n!}f^{n}.

It is easily checked that

\frac{(nd)!}{n!(d!)^{n}} ∈ ℤ.

Motivated by this, the numbers s(d, n) and the series E_{d}(f) are defined in [5] by

s(d, n) = \frac{(nd)!}{n!(d!)^{n}} and E_{d}(f) = ∑_{n=0}^{∞} s(d, n)f^{n}

for d ∈ ℕ and f ∈ A_{d}. We have E_{d}(f ⋆ g) = E_{d}(f) + E_{d}(g) because this property of E_{d} can be checked universally over ℤ. Note that s(d, 0) = s(d, 1) = s(1, n) = 1.

2. Numerical lemmas

In this section we prove some numerical properties of s(d, n).

2.1. Lemma. Let A = ⨁_{n≥0} A_{n} be a positively graded ring, and let a be a homogeneous element of A of positive degree d. Then a^{⋆n} = n!s(d, n)a^{n} and, in particular, a^{⋆n} ∈ ℤn!a^{n} for every n ≥ 0.

Proof. It follows from the definition of the product ⋆ that if a is homogeneous of degree d then

a^{⋆n} = \frac{(nd)!}{(d!)^{n}}a^{n} = n!s(d, n)a^{n}.

The latter element is in ℤn!a^{n} since s(d, n) ∈ ℤ.

We follow the usual convention that \binom{n}{i} = 0 if i < 0. We also make the convention that s(d, i) = 0 if i < 0.

2.2. Lemma. Let d, i, j be integers with d ≥ 1. Then

\binom{d(i + j)}{di}s(d, i)s(d, j) = \binom{i + j}{i}s(d, i + j).

Proof. If i or j is negative then both sides are zero. Hence suppose that i, j ≥ 0. Then

\binom{d(i + j)}{di}s(d, i)s(d, j) = \frac{(d(i + j))!}{(di)!(dj)!} \frac{(di)!}{i!(d!)^{i}} \frac{(dj)!}{j!(d!)^{j}}

= \frac{(i + j)!}{i!j!} \frac{(d(i + j))!}{(i + j)!(d!)^{i+j}}

= \binom{i + j}{i}s(d, i + j).

2.3. Lemma. Let p be a positive prime. Let d, n, i be integers with d ≥ 1, n ≥ 0 and 0 ≤ i ≤ p − 1. Then

s(d, np + i) ≡ \binom{d(np + i)}{di}s(d, i)s(d, np) (mod p).

Proof. Let j = np in Lemma 2.2. It then remains only to check that \binom{np+i}{i} ≡ 1 (mod p) for 0 ≤ i ≤ p − 1. But this is clear.

2.4. Lemma. For all integers i, n ≥ 0, j, d ≥ 1 we have

s(jd, i)s(d, j)^{i}s(d, n − ij)\binom{nd}{ijd} = \binom{n}{ij}s(j, i)s(d, n).

Proof. Direct verification from the definitions.

3. Reduction to a very elementary weakly subintegral extension

We use the definitions and notation of [5, Section 3] and prove appropriate analogues for weakly subintegral extensions. Recall from [5] that a very elementary subintegral extension is an extension A ⊆ B such that B = A[b] for some b ∈ B with b^{2} = 0. By analogy we introduce the following:

3.1. Definition. An extension A ⊆ B is said to be very elementary weakly subintegral if B = A[b] for some b ∈ B with b^{p} ∈ A and pB = 0 for some positive prime p.

Let k be a ring. Let 𝒢 denote the category whose objects are positively graded rings A with A_{0} = k and whose morphisms are graded k-algebra homomorphisms. Note that such homomorphisms are necessarily degree-preserving. An extension in 𝒢 is a ring extension A ⊆ B such that A and B are objects in 𝒢 and the inclusion is a morphism in 𝒢.

3.2. Lemma. Let p be a positive prime. Let A ⊆ B be an extension of rings with B = A[b] and b^{p}, pb ∈ A. Then there exists a chain A = C_{(0)} ⊆ C_{(1)} ⊆ ⋯ ⊆ C_{(m)} = B of subrings of B such that for every i, 1 ≤ i ≤ m, we have C_{(i)} = C_{(i−1)}[x_{i}] with x_{i}^{p} ∈ C_{(i−1)} and pC_{(i)} ⊆ C_{(i−1)}. Each x_{i} can be chosen to be of the form p^{α}b^{β} with α, β non-negative integers. In particular, if A ⊆ B is an extension in 𝒢 and b is homogeneous then each x_{i} can be chosen to be homogeneous, so that each C_{(i−1)} ⊆ C_{(i)} is an extension in 𝒢.

Proof. The given conditions clearly imply that p^{p−1}B ⊆ A. So we have the chain A = A[p^{p−1}B] ⊆ A[p^{p−2}B] ⊆ ⋯ ⊆ A[p^{i}B] ⊆ A[p^{i−1}B] ⊆ ⋯ ⊆ A[pB] ⊆ B. It is enough to prove the existence of a chain of the required type starting with A[p^{i}B] and ending with A[p^{i−1}B] for each i, 1 ≤ i ≤ p − 1. Fix an i, and put C = A[p^{i}B], D = A[p^{i−1}B]. Clearly pD ⊆ C, and D = C[p^{i−1}b, p^{i−1}b^{2}, ⋯, p^{i−1}b^{p−1}]. So D can be obtained from C by successively adjoining p^{i−1}b, p^{i−1}b^{2}, …, p^{i−1}b^{p−1}. Each of these elements has its p^{th} power in A and, as noted, we have pD ⊆ C. The lemma follows.

Recall from [5, Section 3] that an extension A ⊆ B in 𝒢 is subintegral (respectively elementary subintegral, respectively very elementary subintegral) in 𝒢 if it is subintegral (respectively elementary subintegral, respectively very elementary subintegral) as an extension of rings. In the latter two cases there is an element b ∈ B such that B = A[b] and b^{2}, b^{3} ∈ A (respectively b^{2} = 0). If this element b can be chosen to be homogeneous then the extension is said to be homogeneous elementary subintegral (respectively homogeneous very elementary subintegral). We make analogous definitions for weakly subintegral extensions in 𝒢:

3.3. Definition. (i) An extension A ⊆ B in 𝒢 is said to be weakly subintegral in 𝒢 if A ⊆ B is weakly subintegral as an extension of rings.

(ii) An extension A ⊆ B in 𝒢 is said to be elementary weakly subintegral in 𝒢 if A ⊆ B is elementary weakly subintegral as an extension of rings, i.e. there exists an element b ∈ B such that B = A[b] , b^{p} ∈ A and pb ∈ A for some positive prime p.

(iii) An extension A ⊆ B in 𝒢 is said to be very elementary weakly subintegral in 𝒢 if A ⊆ B is very elementary weakly subintegral as an extension of rings, i.e. there exists an element b ∈ B such that B = A[b], b^{p} ∈ A and pB = 0 for some positive prime p.

(iv) If the element b in parts (ii) or (iii) can be chosen to be homogeneous then the extension is said to be homogeneous elementary weakly subintegral (respectively homogeneous very elementary weakly subintegral) in 𝒢.

3.4. Lemma. Every weakly subintegral extension in 𝒢 is the filtered union of extensions that can be obtained by a finite succession of homogeneous elementary subintegral extensions and homogeneous elementary weakly subintegral extensions.

Proof. Follows from the proof of [1, Lemma 4.1].

3.5. Lemma. Let A ⊆ B be a weakly subintegral extension in 𝒢. Then the extension A_{⋆} ⊆ B_{⋆} is subintegral.

Proof. In view of Lemma 3.4, it is enough to prove the assertion when A ⊆ B is either a homogeneous elementary subintegral extension or a homogeneous elementary weakly subintegral extension. The first case is proved in [5, Lemma 3.2]. Hence assume we are in the second case, i.e. that B = A[b] where b ∈ B is homogeneous and b^{p}, pb ∈ A for some positive prime p. In this case we may use Lemma 3.2 to assume that b^{p} ∈ A and pB ⊆ A. Since A_{0} = B_{0} we may further assume that the degree of b is positive. Let i be any positive integer. We have (b^{i})^{⋆p} ∈ ℤb^{ip} ⊆ A_{⋆} and

(b^{i})^{⋆(p+1)} = (b^{i})^{⋆p} ⋆ b^{i} ∈ (ℤp!b^{ip}) ⋆ b^{i} (by Lemma 2.1) ⊆ (ℤb^{ip}) ⋆ (pb^{i}) ⊆ A_{⋆}.

Therefore (ab^{i})^{⋆p} ∈ A_{⋆} and (ab^{i})^{⋆(p+1)} ∈ A_{⋆} for every homogeneous element a of A. It follows that ab^{i}, regarded as an element of B_{⋆}, is subintegral over A_{⋆}. Now, every element of A[b] (hence of B_{⋆} = A[b]_{⋆}) is a finite sum of elements of the form a_{i}b^{i} with a_{i} homogeneous elements of A, i ≥ 0. It follows that A_{⋆} ⊆ B_{⋆} is subintegral.

Recall from [5, Section 3] that ℰ (𝒢) denotes the category of all extensions in 𝒢, and that 𝒜 b denotes the category of abelian groups.

3.6. Definition. (cf. [5, Definition 3.4]) A functor ℋ : ℰ (𝒢) → 𝒜 b is said to be strongly admissible if it satisfies the following three conditions:

(a) If A ⊆ B is an extension in 𝒢 and 𝔟 is a homogeneous ideal of both A and B then the induced map ℋ (A, B) → ℋ (A/𝔟, B/𝔟) is an isomorphism.

(b) If A ⊆ C ⊆ B are weakly subintegral extensions in 𝒢 then the sequence

0 → ℋ (A, C) → ℋ (A, B) → ℋ (C, B) → 0

of induced maps is exact.

(c) ℋ commutes with direct limits in the second variable.

3.7. Lemma. (a) The functor 𝒜 : ℰ (𝒢) → 𝒜 b taking (A, B) to the quotient B/A of the underlying additive groups is strongly admissible.

(b) The functor 𝒜 : ℰ (𝒢) → 𝒜 b taking (A, B) to ℐ (A_{⋆}, B_{⋆}) is strongly admissible.

Proof. (a) This is clear.

(b) In view of Lemma 3.5, the proof of [5, Lemma(3.5)(c)] works in this case.

3.8. Lemma. Let ℋ_{1}, ℋ_{2} : ℰ (𝒢) → 𝒜 b be strongly admissible functors and let θ : ℋ_{1} → ℋ_{2} be a natural transformation of functors. Let A ⊆ C ⊆ B be weakly subintegral extensions in 𝒢. If θ(A, C) and θ(C, B) are isomorphisms, then so is θ(A, B).

Proof. This follows from condition (b) of Definition 3.6 and the five lemma (as in [5, Lemma 3.6]).

3.9. Theorem. (cf. [5, Theorem 3.7 and Corollary 3.8] Let ℋ_{1}, ℋ_{2} : ℰ (𝒢) → 𝒜 b be strongly admissible functors, and let θ : ℋ_{1} → ℋ_{2} be a natural transformation of functors. Suppose θ(A, B) is an isomorphism for every homogeneous very elementary subintegral extension and every homogeneous very elementary weakly subintegral extension A ⊆ B in 𝒢. Then θ(A, B) is an isomorphism for every weakly subintegral extension A ⊆ B in 𝒢.

Proof. In view of the results proved above the assertion can be proved in several steps, on the lines of the proofs of [5, Theorem 3.7 and Corollary 3.8]:

Step 1: We observe that, by [5, Corollary 3.8], θ(A, B) is an isomorphism for every subintegral extension A ⊆ B in 𝒢.

Step 2: We prove that θ(A, B) is an isomorphism for every homogeneous elementary weakly subintegral extension A ⊆ B in 𝒢. Let B = A[b] with b homogeneous, such that b^{p}, pb ∈ A for some positive prime p. We may assume that the degree of b is positive. By Lemma 3.2 there exists a chain A = C_{(0)} ⊆ C_{(1)} ⊆ ⋯ ⊆ C_{(m)} = B of extensions in 𝒢 such that for every i, 1 ≤ i ≤ m we have C_{(i)} = C_{(i−1)}[x_{i}] with x_{i} homogeneous, x_{i}^{p} ∈ C_{(i−1)} and pC_{(i)} ⊆ C_{(i−1)}. In view of Lemma 3.8 it is enough to show that θ(C_{(i−1)}, C_{(i)}) is an isomorphism for each i. Fix an i, 1 ≤ i ≤ m, and put D = C_{(i−1)} and E = C_{(i)}. Since pE ⊆ D, pE is a (homogeneous) ideal of both D and E. We have E/pE = (D/pE)[y] where y is the natural image of x_{i}. Since y^{p} ∈ D/pE and p(E/pE) = 0, D/pE ⊆ E/pE is a homogeneous very elementary weakly subintegral extension. Therefore θ(D/pE, E/pE) is an isomorphism by assumption. Hence, in view of condition (a) of Definition 3.6, so is θ(D, E). This completes the proof of Step 2.

Step 3: Suppose that A ⊆ B is a weakly subintegral extension in 𝒢 such that B is finitely generated as an A-algebra. Then by Lemma 3.4 there is a finite sequence A = C_{(0)} ⊆ C_{(1)} ⊆ ⋯ ⊆ C_{(r)} = B such that each C_{(i−1)} ⊆ C_{(i)} is a homogeneous elementary subintegral extension in 𝒢 or a homogeneous elementary weakly subintegral extension in 𝒢. It then follows from Steps 1 and 2 and Lemma 3.8 that θ(A, B) is an isomorphism.

Step 4: If A ⊆ B is an arbitrary weakly subintegral extension in 𝒢 then B is the direct limit of finitely generated subextensions of A in 𝒢, so it follows from condition (c) of Definition 3.6 that θ(A, B) is an isomorphism.

4. Recollection of the definition of ζ

Let A = ⨁_{d≥0} A_{d} ⊆ B = ⨁_{d≥0} B_{d} be a weakly subintegral extension of positively graded rings with A_{0} = B_{0}. Under these conditions a functorial group homomorphism

ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆})

is constructed in [5, Section 4; see Remark 4.4]. It is proved there that ζ_{B/A} is an isomorphism if the extension A ⊆ B is subintegral. Our aim is to show that ζ_{B/A} is an isomorphism also for a weakly subintegral extension A ⊆ B.

Some of the notation used here and in the next two sections has been recalled in the Introduction. See [5, Section 4] for further details.

Recall that ζ_{B/A} is defined by first constructing, for each d ≥ 1, a homomorphism

J_{d} : B_{d} → ℐ (A_{⋆}, B_{⋆})

given by J_{d}(b) = (Â_{⋆} ⋆ E_{d}(b)) ∩ B_{⋆}, and then defining

ζ_{B/A} : B/A = B_{+}/A_{+} → ℐ (A_{⋆}, B_{⋆})

by

ζ_{B/A}(\overline{b}) = ∏_{d≥1}_{⋆}J_{d}(b_{d})

for b = ∑_{d≥1} b_{d} with b_{d} ∈ B_{d}.

5. Isomorphism for a very elementary weakly subintegral extension

Our goal in this section is to prove that ζ_{B/A} is an isomorphism if A ⊆ B is a very elementary weakly subintegral extension in 𝒢. As we shall see, every invertible A_{⋆}-submodule of B_{⋆} is principal in this case and it is possible to describe a generator of the A_{⋆}-module ζ_{B/A}(\overline{b}) explicitly in terms of a map σ defined below. This description enables us to prove that ζ_{B/A} is an isomorphism in this case. The proof works in fact under the following slightly weaker condition:

5.1. Assumption. A ⊆ B is an extension in 𝒢 with char(A) = p, a positive prime, and B^{p} ⊆ A.

Note that the maps J_{d} and ζ_{B/A} of the previous section make sense for the extension A ⊆ B, since the conditions of 5.1 make this extension a weakly subintegral extension.

Throughout this section p denotes a positive prime.

5.2. Lemma. (1) If A is a positively graded ring of characteristic p then (A_{+})^{⋆p} = 0 and (A_{⋆})^{×} = ((A_{0})^{×} ⊕ A_{+})_{⋆}.

(2) If A ⊆ B is an extension satisfying 5.1 then ℐ (A_{⋆}, B_{⋆}) = (B_{⋆})^{×}/(A_{⋆})^{×} = (1 + B_{+})_{⋆}/(1 + A_{+})_{⋆}.

Proof. (1) Let x be a homogeneous element of A of positive degree. Then, by Lemma 2.1, x^{⋆p} ∈ ℤp!x^{p}. Therefore x^{⋆p} = 0. The first part follows, and the second part is an immediate consequence of this.

(2) By (1), nil(A_{⋆}) = nil(A_{0}) ⊕ A_{+}. Therefore Pic(A_{⋆}) = Pic(A_{⋆}/nil(A_{⋆})) = Pic(A_{⋆}/(nil(A_{0})+A_{+})) = Pic(A_{0}/nil(A_{0})) = Pic(A_{0}). Similarly Pic(B_{⋆}) = Pic(B_{0}). Now, since A_{0} = B_{0} the first equality is immediate from (1.1) applied to A_{⋆} ⊆ B_{⋆}. The second equality follows from (1), since A_{0} = B_{0}.

Let B be a positively graded ring of characteristic p. For a positive integer d and for x ∈ B_{d} define

σ_{d}(x) = ∑_{i=0}^{p−1} s(d, i)x^{i}.

Since s(d, 0) = 1, we have σ_{d}(x) ∈ (1 + B_{+})_{⋆} ⊆ (B_{⋆})^{×} by the above lemma. Define σ : B_{+} → (1 + B_{+})_{⋆} by

σ(b) = ∏_{d≥1}_{⋆}σ_{d}(b_{d})

for b = ∑_{d≥1} b_{d} with b_{d} ∈ B_{d}.

Sometimes we write σ_{B} for σ to indicate the ring under consideration.

5.3. Proposition. Let A ⊆ B be an extension satisfying 5.1. Let x be a homogeneous element of B of degree d. Then J_{d}(x) = A_{⋆} ⋆ σ_{d}(x).

Proof. We show first that σ_{d}(x) ∈ J_{d}(x). We have

E_{d}(x) = ∑_{i=0}^{∞} s(d, i)x^{i}

= ∑_{i=0}^{p−1} ∑_{n=0}^{∞} s(d, np + i)x^{np+i}

= ∑_{i=0}^{p−1} ∑_{n=0}^{∞} \binom{d(np + i)}{di}s(d, i)s(d, np)x^{np+i} (by Lemma 2.3)

= f ⋆ σ_{d}(x)

with f = ∑_{n=0}^{∞} s(d, np)x^{np}. Since B^{p} ⊆ A, we have f ∈ Â_{⋆}. Furthermore, since s(d, 0) = 1, f is a unit in Â_{⋆} by Lemma 5.2. Let g ∈ Â_{⋆} be the inverse of f. Then

σ_{d}(x) = g ⋆ E_{d}(x) ∈ (Â_{⋆} ⋆ E_{d}(x)) ∩ B_{⋆} = J_{d}(x).

This proves that A_{⋆}⋆σ_{d}(x) ⊆ J_{d}(x). Now, since σ_{d}(x) ∈ (B_{⋆})^{×}, we have A_{⋆}⋆σ_{d}(x) ∈ ℐ (A_{⋆}, B_{⋆}). Thus both A_{⋆} ⋆ σ_{d}(x) and J_{d}(x) are invertible A_{⋆}-submodules of B_{⋆} and A_{⋆} ⋆ σ_{d}(x) ⊆ J_{d}(x). Therefore, since the extension A_{⋆} ⊆ B_{⋆} is subintegral by Lemma 3.5, it follows from [5, Proposition 3.10] that J_{d}(x) = A_{⋆} ⋆ σ_{d}(x).

5.4. Lemma. Let B be a positively graded ring of characteristic p, and let x be a homogeneous element of B of positive degree d. Let j be an integer with 2 ≤ j ≤ p − 1. Suppose that

f = ∑_{i=0}^{p−1} λ_{i}s(d, i)x^{i}

with λ_{i} ∈ ℤ, λ_{0} = λ_{1} = 1 and λ_{i} = 0 for 2 ≤ i ≤ j − 1. Then there exists y ∈ B_{jd} such that

f ⋆ σ_{jd}(y) = ∑_{i=0}^{p−1} μ_{i}s(d, i)x^{i}

with μ_{i} ∈ ℤ, μ_{0} = μ_{1} = 1 and μ_{i} = 0 for 2 ≤ i ≤ j.

Proof. Take y = −λ_{j}s(d, j)x^{j}. Putting λ_{r} equal to zero for r < 0 and r ≥ p, we get

f ⋆ σ_{jd}(y) = ∑_{n≥0} ν_{n}x^{n}

with

ν_{n} = ∑_{i=0}^{p−1}(−1)^{i}λ_{j}^{i}λ_{n−ij}s(jd, i)s(d, j)^{i}s(d, n − ij)\binom{nd}{ijd}.

By Lemma 2.4 ν_{n} = μ_{n}s(d, n), where

μ_{n} = ∑_{i=0}^{p−1}(−1)^{i}λ_{j}^{i}λ_{n−ij}\binom{n}{ij}s(j, i).

Since f = 1 + x + λ_{j}s(d, j)x^{j}+ higher degree terms in x, and σ_{jd}(y) = 1 − λ_{j}s(d, j)x^{j}+ higher degree terms in x, we have f ⋆ σ_{jd}(y) = 1 + x+ terms in x^{j+1} and higher. Thus μ_{0} = μ_{1} = 1 and μ_{i} = 0 for 2 ≤ i ≤ j. So it is enough to prove that μ_{n} = 0 for n ≥ p. We show in fact that if n ≥ p then some factor of λ_{j}^{i}λ_{n−ij}\binom{n}{ij}s(j, i) is zero (modulo p). If ij ≥ p then s(j, i) = (ij)!/i!(j!)^{i} is zero modulo p, since 1 ≤ i, j ≤ p − 1. If n − ij ≥ p then λ_{n−ij} = 0. We are left with the case ij < p, n − ij < p and n ≥ p. In this case \binom{n}{ij} is clearly zero modulo p.

5.5. Corollary. Let B be a positively graded ring of characteristic p, and let x be a homogeneous element of B of positive degree d. Then there exist elements y_{j} ∈ B_{jd}, 1 ≤ j ≤ p − 1, such that

1 + x = ∏_{j=1}^{p−1}_{⋆}σ_{jd}(y_{j}).

Proof. Let y_{1} = x and

f_{1} = σ_{d}(y_{1}) = ∑_{i=0}^{p−1} s(d, i)x^{i}.

Applying the above lemma with f = f_{1} and j = 2 we obtain y_{2} ∈ B_{2d} such that if we write f_{2} = f_{1} ⋆ σ_{2d}(y_{2}) = σ_{d}(y_{1}) ⋆ σ_{2d}(y_{2}) then f_{2} = ∑_{i=0}^{p−1} μ_{i}s(d, i)x^{i} with μ_{i} ∈ ℤ, μ_{0} = μ_{1} = 1 and μ_{2} = 0. Now apply the lemma with f = f_{2} and j = 3. Proceeding thus, we get

1 + x = ∏_{j=1}^{p−1}_{⋆}σ_{jd}(y_{j}).

5.6. Proposition. Let B be a positively graded ring of characteristic p. Then the map σ : B_{+} → (1 + B_{+})_{⋆} is an isomorphism of groups.

Proof. σ is a homomorphism: For d ≥ 1 and x, y ∈ B_{d} we have

σ(x) ⋆ σ(y) = (∑_{i=0}^{p−1} s(d, i)x^{i}) ⋆ (∑_{i=0}^{p−1} s(d, i)y^{i}) = ∑_{\begin{matrix}0≤i≤p−1 \\ 0≤j≤p−1 \end{matrix}} t_{ij}x^{i}y^{j}

where, by Lemma 2.2, t_{ij} = \binom{i+j}{i}s(d, i + j). Noting that \binom{i+j}{i} ≡ 0 (mod p) if 0 ≤ i ≤ p − 1, 0 ≤ j ≤ p − 1 and i + j ≥ p, we get

σ(x) ⋆ σ(y) = ∑_{n=0}^{p−1} s(d, n) ∑_{i+j=n} \binom{n}{i}x^{i}y^{j} = ∑_{n=0}^{p−1} s(d, n)(x + y)^{n} = σ(x + y).

This shows that σ_{d} is a homomorphism for every d . It follows that σ is a homomorphism.

Injectivity of σ: Let x be a nonzero element of B_{+}. Then x has a homogeneous decomposition x = ∑_{i≥r} x_{i} for some r ≥ 1 with x_{r} ≠ 0. So σ(x) = 1 + x_{r} + terms of degree > r, showing that σ(x) ≠ 1.

Surjectivity of σ: Let x ∈ B_{+}. We want to show that 1 + x ∈ im(σ). Suppose first that x is homogeneous, say of degree d. Then by Corollary 5.5 there exist y_{j} ∈ B_{jd}, 1 ≤ j ≤ p − 1, such that

1 + x = ∏_{j=1}^{p−1}⋆σ_{jd}(y_{j}),

showing that 1 + x ∈ im(σ). Now, suppose x is arbitrary. Write x = x_{1} + ⋯ + x_{m} with x_{i} ∈ B_{i}. Let C be the B_{0}-subalgebra of B_{⋆} generated by x_{1}, …, x_{m}. Then C = ⨁_{d≥0} C_{d} is a graded subring of B_{⋆}. For n ≥ 0 let Γ_{n} = ⨁_{d≥n} C_{d}. With this notation it is clearly enough to prove the following two statements:

(1) For every n ≥ 1 there exists y_{n} ∈ Γ_{n} such that 1 + x + y_{n} ∈ im(σ).

(2) Γ_{n} = 0 for n >> 0.

Proof of (1): Induction on n. For n = 1 take y_{1} = −x. Suppose 1+x+y_{n} ∈ im(σ) with y_{n} ∈ Γ_{n} for some n ≥ 1. Write y_{n} = t + z with t ∈ C_{n}, z ∈ Γ_{n+1}. By the homogeneous case already proved, 1 − t belongs to im(σ) whence so does

(1 + x + y_{n}) ⋆ (1 − t).

Now, (1 + x + y_{n}) ⋆ (1 − t) = 1 + x + y_{n+1} with y_{n+1} ∈ Γ_{n+1}, so (1) is proved.

Proof of (2): Suppose n > (p − 1)(1 + 2 + ⋯ + m). If x_{1}^{⋆r_{1}} ⋆ ⋯ ⋆ x_{m}^{⋆r_{m}} is any monomial in Γ_{n} then r_{i} ≥ p for some i whence x_{i}^{⋆r_{i}} = 0 by Lemma 2.1. Therefore Γ_{n} = 0.

5.7. Corollary. Let A ⊆ B be an extension of positively graded rings of characteristic p. Then the map σ_{B} induces an isomorphsim

σ_{B/A} : B_{+}/A_{+} → (1 + B_{+})_{⋆}/(1 + A_{+})_{⋆}.

Proof. σ_{B} : B → (1+B_{+})_{⋆} and σ_{A} : A → (1+A_{+})_{⋆} are isomorphisms by the above proposition and clearly σ_{B} |_{A_{+}}= σ_{A}.

5.8. Corollary. Let A ⊆ B be an extension of positively graded rings of characteristic p. If A_{0} = B_{0} then the map σ_{B} induces an isomorphsim

σ_{B/A} : B/A → (1 + B_{+})_{⋆}/(1 + A_{+})_{⋆}.

Proof. In this case we have B/A = B_{+}/A_{+}.

5.9. Corollary. Let A ⊆ B be an extension satisfying 5.1. Then the homomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) is an isomorphism.

Proof. Identify (1+B_{+})_{⋆}/(1+A_{+})_{⋆} with ℐ (A_{⋆}, B_{⋆}) by Lemma 5.2. Then it follows from Proposition 5.3 that ζ_{B/A} = σ_{B/A}. Therefore the assertion is immediate from the above corollary.

5.10. Corollary. Let A ⊆ B be a very elementary weakly subintegral extension of positively graded rings with A_{0} = B_{0}. Then the homomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) is an isomorphism.

Proof. Such an extension of nonzero rings satisfies the hypotheses of Corollary 5.9 for some positive prime p.

6. Isomorphism for a weakly subintegral extension

6.1. Theorem. Let A ⊆ B be a weakly subintegral extension of positively graded rings with A_{0} = B_{0}. Then the homomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) is an isomorphism.

Proof. By Lemma 3.7, ζ is a natural transformation of strongly admissible functors. By [5, Theorem 4.5], ζ_{B/A} is an isomorphism if the extension A ⊆ B is subintegral. By Corollary 5.10, ζ_{B/A} is an isomorphism if the extension A ⊆ B is very elementary weakly subintegral. Therefore ζ_{B/A} is an isomorphism by Theorem 3.9.

6.2. Remark. (cf. [4, Example (3.1)]) Let A = k[t^{2}] ⊆ B = k[t], where k is a field of characteristic two and t is an indeterminate. Then A ⊆ B is a weakly subintegral extension of positively graded rings with A_{0} = B_{0}. So we have the isomorphism ζ_{B/A} : B/A → ℐ (A_{⋆}, B_{⋆}) given by the above theorem. This implies, in particular, that the group ℐ (A_{⋆}, B_{⋆}) is nontrivial. On the other hand, ℐ (A, B) is the trivial group, by (1.1). Therefore ℐ (A_{⋆}, B_{⋆}) is not isomorphic to gr_{ℕ}ℐ (A, B), unlike the situation for a subintegral extension in which case the two groups are isomorphic as shown in [5, Main Theorem].

7. Some explicit results

Throughout this section p is a positive prime, and B is a positively graded ring of characteristic p . We illustrate some of the previous ideas. In particular, we see that E_{d} = σ_{d} for some, but not all, d, so the properties of σ_{d} proved in Proposition 5.6 are not a consequence of the properties of E_{d}.

7.1. Theorem. (a) If d is not a power of p then E_{d} = σ_{d}.

(b) If d = p^{r}, r ≥ 0 and x ∈ B_{d} then E_{d}(x) = ∑_{i=0}^{∞} x^{i} and E_{d}(x) = (∑_{n=0}^{∞} x^{np})⋆ σ_{d}(x).

First, some preliminary lemmas:

7.2. Lemma. Let n, d be positive integers and let p be a positive prime. Then s(d, np) is coprime to p if and only if d is a power of p.

Proof. Put φ(d) = s(d, np). For a nonzero rational number m let ν_{p}(m) denote the p-adic valuation of m. That is, ν_{p}(m) = a − b, where a and b are respectively the highest powers of p dividing the numerator and denominator of m. We have to show that ν_{p}(φ(d)) = 0 if and only if d is a power of p. We make the following claims:

(1) φ(1) = 1 whence ν_{p}(φ(1)) = 0.

(2) ν_{p}(φ(p^{r}d)) = ν_{p}(φ(d)) for every r ≥ 0.

(3) If d > 1 and gcd(d, p) = 1 then ν_{p}(φ(d)) > 0. Granting these for the moment, write d = p^{r}q with r, q nonnegative integers and gcd(q, p) = 1. Then ν_{p}(φ(d)) = ν_{p}(φ(q)) by (2). If d is a power of p then q = 1 so that ν_{p}(φ(d)) = 0 by (1). On the other hand, if d is not a power of p then q > 1 whence ν_{p}(φ(d)) > 0 by (3).

We now prove the claims.

(1) This is clear.

(2) By induction on r it is enough to prove that ν_{p}(φ(pd)) = ν_{p}(φ(d)). We have

\frac{φ(pd)}{φ(d)} = \frac{s(pd, np)}{s(d, np)} = \frac{ψ(pdn)}{ψ(d)^{np}},

where ψ(m) = (mp)!/m!. Among the integers 1, 2, …, ip, …, mp those divisible by p are precisely p, 2p, …, (m − 1)p, mp. It follows that ν_{p}(ψ(m)) = m. Therefore

ν_{p}(\frac{φ(pd)}{φ(d)}) = ν_{p}(\frac{ψ(pdn)}{ψ(d)^{np}}) = ν_{p}(ψ(pdn)) − np ν_{p}(ψ(d)) = pdn − npd = 0

whence ν_{p}(φ(pd)) = ν_{p}(φ(d)).

(3) We have

\frac{φ(d)}{φ(d − 1)} = \frac{s(d, np)}{s(d − 1, np)}
= \frac{(dnp)!(d − 1)!^{np}}{((d − 1)np)!d!^{np}}
= dnp\frac{(dnp − 1)!}{((d − 1)np)!}(\frac{1}{d})^{np}.

Therefore, since φ(d − 1) and \frac{(dnp−1)!}{((d−1)np)!} are integers and gcd(d, p) = 1, ν_{p}(φ(d)) > 0.

Now, we sharpen the result if d is a power of p.

7.3. Lemma. Let r and i be nonnegative integers. Then s(p^{r}, i) ≡ 1 (mod p).

Proof. We have

s(p^{r}, i) = \frac{(ip^{r})!}{i!((p^{r})!)^{i}} = ∏_{j=1}^{i} \frac{1}{j}\binom{jp^{r}}{p^{r}},

so that it suffices to prove that

\frac{1}{j}\binom{jp^{r}}{p^{r}} ≡ 1 (mod p).

But if we pair the factors jp^{r} − c, p^{r} − c, 0 ≤ c ≤ p^{r} − 1 of the numerator and denominator of \binom{jp^{r}}{p^{r}} then the c = 0 factor jp^{r} in the numerator cancels with the j and the c = 0 factor p^{r} in the denominator. For 1 ≤ c ≤ p^{r} − 1 write c = c′p^{s} with ν_{p}(c′) = 0. Note that s < r. Then

\frac{jp^{r} − c}{p^{r} − c} = \frac{jp^{r−s} − c′}{p^{r−s} − c′} ≡ 1 (mod p).

Proof of Theorem 7.1. If n and d are positive integers then s(d, n) divides s(d, n+1). If d is not a power of p then by Lemma 7.2 s(d, p) ≡ 0 (mod p). Part (a) follows. In part (b) the first formula follows from Lemma 7.3, and the second one from the proof of Proposition 5.3.

References

[1] Les Reid and Leslie G. Roberts, Monomial subrings and systems of subintegrality, To appear in J. Pure Appl. Algebra.

[2] Les Reid, Leslie G. Roberts and Balwant Singh, On weak subintegrality, J. Pure Appl. Algebra 114 (1996), 93-109.

[3] Leslie G. Roberts and Balwant Singh, Subintegrality, invertible modules, and the Picard group, Compositio Math. 85 (1993), 249-279.

[4] Balwant Singh, The Picard group and subintegrality in positive characteristic, Compositio Math. 95 (1995), 309-321.

[5] Balwant Singh, Subintegrality and invertible modules in graded rings, J. Algebra 217 (1999), 73-94.

[6] R. G. Swan, On seminormality, J.Algebra 67 (1980), 210-229.

[7] H. Yanagihara, On an intrinsic definition of weakly normal rings, Kobe J. Math., 2 (1985), 89-98.

Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, Canada, K7L 3N6

   E-mail address: robertsl@mast.queensu.ca

Department of Mathematics, Indian Institute of Technology, Bombay, Powai, Mumbai 400076, India

   E-mail address: balwant@math.iitb.ernet.in