New York Journal of Mathematics

New York J. Math. 14 (2008) 225–233.

Galois module structure of Milnor K-theory mod p^{s} in characteristic p

Ján Mináč, Andrew Schultz and John Swallow

Abstract. Let E be a cyclic extension of pth-power degree of a field F of characteristic p. For all m, s ∈ ℕ, we determine K_{m}E/p^{s}K_{m}E as a (ℤ/p^{s}ℤ)[Gal(E/F)]-module. We also provide examples of extensions for which all of the possible nonzero summands in the decomposition are indeed nonzero.

Contents

1. Proof of the Theorem 226 2. Examples of E/F with Y_{i} ≠ {0} for all i 229

2.1. The case m = 1 229 2.2. The case m > 1 231

References 233

Let F be a field of characteristic p. Let K_{m}F denote the mth Milnor K-group of F and k_{m}F = K_{m}F/pK_{m}F (see, for instance, [Mi] and [Ma, Chapter 14]). If E/F is a Galois extension of fields, let G = Gal(E/F) denote the associated Galois group. In [BLMS] the structure of k_{m}E as an 𝔽_{p}G-module was determined when G is cyclic of pth-power order. In this paper we determine the Galois module structure of K_{m}E modulo p^{s} for s ∈ ℕ and these same G. We also provide examples of extensions for which the possible free summands in the decomposition are all nonzero. These examples together with the results in [BLMS] show that the dimensions over 𝔽_{p} of indecomposable 𝔽_{p}[Gal(E/F)]-modules occuring as direct summands

225^{ISSN 1076-9803/08}

Received March 6, 2008; revised June 6, 2008.

Mathematics Subject Classification. 19D45, 12F10.

Key words and phrases. Milnor K-theory, cyclic extension, Galois module, norm.

The first author’s research was supported in part by NSERC grant R0370A01, and by a Distinguished Research Professorship at the University of Western Ontario. He was also supported by the Mathematical Sciences Research Institute, Berkeley.

The third author’s research was supported in part by NSF grant DMS-0600122.

of k_{m}E are all powers of p and that all dimensions p^{i}, i = 0, 1, …, n, indeed occur in suitable examples.

Since the classification problem of (ℤ/p^{s}ℤ)G-modules for cyclic G is nontrivial and has not been completely solved — see, for instance, [T] for results and references — it is a pleasant surprise that the (ℤ/p^{s}ℤ)G-modules K_{m}E/p^{s}K_{m}E have a simple description. The main ingredients we use to obtain this description are the lack of p-torsion in K_{m}E, due to Izhboldin [I], together with the result [BLMS] for the case s = 1 — which also depends on Izhboldin’s result.

Suppose that E/F is cyclic of degree p^{n}, and for i = 0, …, n, let E_{i}/F be the subextension of degree p^{i} of E/F and G_{i} := Gal(E_{i}/F). Set R_{s} := ℤ/p^{s}ℤ and let ℤ_{p} be the ring of p-adic integers. We write ι_{F,E}: K_{m}F → K_{m}E and N_{E/F} : K_{m}E → K_{m}F for the natural inclusion and norm maps, and we use the same notation for the induced maps between K_{m}F/p^{s}K_{m}F and K_{m}E/p^{s}K_{m}E.

Theorem. Let s ∈ ℕ. There exists an isomorphism of R_{s}G-modules

K_{m}E/p^{s}K_{m}E ≃ ⊕_{i=0}^{n}Y_{i}

where:

• Y_{n} is a free R_{s}G-module of rank dim_{𝔽_{p}} N_{E/F} k_{m}E.

• Y_{i}, 0 ≤ i < n, is a free R_{s}G_{i}-module of rank

dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1}.

• For each 0 ≤ i ≤ n, Y_{i} ⊆ ι_{E_{i},E}(K_{m}E_{i}/p^{s}K_{m}E_{i}).

Passing to the projective limit with respect to s, we obtain

{ \atop K_{m}E} := \underleftarrow{lim}_{s} K_{m}E/p^{s}K_{m}E ≃ ⊕_{i=0}^{n} Ŷ_{i},

where each Ŷ_{i} is a free ℤ_{p}G_{i}-module of rank

dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1}.

Remark. Observe that Y_{0} is a trivial R_{s}G-module. Moreover, the statements about the ranks of the Y_{i} are immediate consequences of the fact that K_{m}E/p^{s}K_{m}E is a direct sum of free R_{s}G_{i}-modules, i = 0, 1, …, n — which is the main point of the theorem.

1. Proof of the Theorem

We prove the result by induction on s. The case s = 1 is [BLMS, Theorem 2]. Assume therefore that s > 1 and the result holds for s − 1:

K_{m}E/p^{s−1}K_{m}E = ⊕Ỹ_{i}

with each Ỹ_{i} a free R_{s−1}G_{i}-module in the image of ι_{E_{i},E}, 0 ≤ i ≤ n.

Galois module structure of Milnor K-theory 227

For each i with 0 ≤ i ≤ n, let B_{s−1,i} ⊆ i_{E_{i},E}K_{m}E_{i}/p^{s−1}K_{m}E_{i} be an R_{s−1}G_{i}-base for the free R_{s−1}G_{i}-module Ỹ_{i}. By induction the cardinality of B_{s−1,i} is

|B_{s−1,i}| = \begin{cases}dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1}, & i < n \\ dim_{𝔽_{p}} N_{E/F} k_{m}E, & i = n. \end{cases}

Set

B_{s−1} := ∪_{0≤i≤n}B_{s−1,i} ⊆ K_{m}E/p^{s−1}K_{m}E.

For each i, let B_{i} ⊆ ι_{E_{i},E}(K_{m}E_{i}) be a set of representatives for the elements of B_{s−1,i}, and let B_{s,i} ⊆ K_{m}E/p^{s}K_{m}E be chosen to make the following first diagram commutative. The second diagram simply recalls where our B_{i}, B_{s,i} and B_{s−1,i} are located.

\xymatrix{B_{i} \ar[r]^{mod p^{s}} \ar[d]_{mod p^{s−1}} & B_{s,i} \ar[dl]^{mod p^{s−1}} & K_{m}E \ar[r] \ar[d] & K_{m}E/p^{s}K_{m}E \ar[dl] \\ B_{s−1,i} &  & K_{m}E/p^{s−1}K_{m}E}

Hence for each i we have bijections

B_{i} ↔ B_{s,i} ↔ B_{s−1,i}

and |B_{s,i}| = |B_{s−1,i}|.

First we observe that every nonzero ideal V of R_{s}G_{i} contains

p^{s−1}(τ − 1)^{p^{i}−1},

where τ is any fixed generator of G_{i}. Indeed consider 0 ≠ β ∈ V. By multiplying by an appropriate power of p, we may assume 0 ≠ β ∈ p^{s−1}R_{s}G_{i}. Let us write

β = ∑_{j=k}^{p^{i}−1} c_{j}(τ − 1)^{j},

where each c_{j} ∈ p^{s−1}R_{s}, j = k, …, p^{i} − 1, and c_{k} ∉ p^{s}R_{s} = {0}, say c_{k} = p^{s−1}c̃_{k} with c̃_{k} ∉ pR_{s}. Using the fact that p^{s−1}(τ − 1)^{p^{i}} = 0 in R_{s}G_{i} we see that we can multiply β by (τ − 1)^{p^{i}−k−1} to obtain

0 ≠ c̃_{k}p^{s−1}(τ − 1)^{p^{i}−1} ∈ V.

Since c̃_{k} ∈ U(R_{s}), the units of R_{s}, we see that p^{s−1}(τ − 1)^{p^{i}−1} ∈ V as asserted.

Set Y_{i} to be the R_{s}G-submodule of K_{m}E/p^{s}K_{m}E generated by B_{s,i}. It is clear that Y_{i} ⊆ ι_{E_{i},E}(K_{m}E_{i}/p^{s}K_{m}E_{i}) and hence Y_{i} is an R_{s}G_{i}-module.

Each element b ∈ B_{s,i} generates in K_{m}E/p^{s}K_{m}E a free R_{s}G_{i}-module M_{b}, as follows. Suppose that M_{b} is not a free R_{s}G_{i}-module. Then the annihilator of b in R_{s}G_{i} is a nonzero ideal of R_{s}G_{i}. Let b̂ ∈ B_{i} and b̃ ∈ B_{s−1,i} correspond to b under the bijection above. Let also σ be a generator of G and σ its image in G_{i}. Since every nonzero ideal of R_{s}G_{i} contains p^{s−1}(σ − 1)^{p^{i}−1}, for some ĉ ∈ K_{m}E we have

p^{s−1}(σ − 1)^{p^{i}−1}b̂ = p^{s}ĉ.

Since K_{m}E has no p-torsion [I, Theorem A], we obtain

p^{s−2}(σ − 1)^{p^{i}−1}b̂ = p^{s−1}ĉ.

Then in K_{m}E/p^{s−1}K_{m}E

p^{s−2}(σ − 1)^{p^{i}−1}b̃ = 0,

contradicting the fact that b̃ lies in the R_{s−1}G_{i}-base B_{s−1,i} for Ỹ_{i}. (Alternatively, we could use [T, Theorem 5.1] to show that M_{b} is a free R_{s}G_{i}-module.)

Now set B_{s} := ∪_{0≤i≤n}B_{s,i} ⊆ K_{m}E/p^{s}K_{m}E. Suppose we have a relation

∑_{b∈B_{s}} r(b)b = 0,

where for b ∈ B_{s,i} we have r(b) ∈ R_{s}G_{i}, and all but a finite number of r(b) are zero. We write r(b) ∈ R_{s}G_{i(b)}. Let b̃ ∈ B_{s−1} correspond to b under the natural bijection, and similarly let r̃(b) ∈ R_{s−1}G be the image of r(b) ∈ R_{s}G.

Working mod p^{s−1} we have ∑ r̃(b)b̃ = 0. Since each b̃ lies in B_{s−1}, we deduce that r(b) ∈ p^{s−1}R_{s}G_{i(b)}. Write r(b) = ps(b) for elements s(b) ∈ R_{s}G_{i(b)}. We rewrite the original relation as

∑ ps(b)b = 0.

Just as before we divide by p to obtain in K_{m}E/p^{s−1}K_{m}E

∑ s̃(b)b̃ = 0.

Again since each b̃ ∈ B_{s−1}, we deduce that s(b) ∈ p^{s−1}R_{s}G_{i(b)}. But then r(b) = ps(b) = 0 ∈ R_{s}G_{i(b)} for each r(b), as desired.

Hence for each i in 0 ≤ i ≤ n we have that Y_{i} is a direct sum of free R_{s}G_{i}- modules M_{b} for b ∈ B_{s,i}, and moreover that ∑ Y_{i} = ⊕Y_{i}. By Nakayama’s Lemma, since B generates K_{m}E/p^{s−1}K_{m}E it also generates K_{m}E/p^{s}K_{m}E, and hence ⊕Y_{i} = K_{m}E/p^{s}K_{m}E. (More explicitly, choose α ∈ K_{m}E/p^{s}K_{m}E and α̂ ∈ K_{m}E a lift of α. Since B_{s−1} spans K_{m}E/p^{s−1}K_{m}E we have

α̂ = ∑_{b∈B} f_{b}b + p^{s−1}γ̂

for some γ̂ ∈ K_{m}E, where each f_{b} ∈ R_{s}G and all but finitely many f_{b} = 0. We also have γ̂ = ∑ g_{b}b + p^{s−1}δ̂ for some δ̂ ∈ K_{m}E, where again each g_{b} ∈ R_{s}G and all but finitely many g_{b} = 0. Therefore

α = ∑_{b∈B} (f_{b} + p^{s−1}g_{b}) b

as elements of K_{m}E/p^{s}K_{m}E.) □

2. Examples of E/F with Y_{i} ≠ {0} for all i

Let p be an arbitrary prime number, and let q be an arbitrary prime number or 0. We show that for each n, m ∈ ℕ there exists a cyclic field extension E/F of degree p^{n} and characteristic q such that for each i, 0 ≤ i < n,

dim_{𝔽_{p}} N_{E_{i}/F} k_{m}E_{i}/N_{E_{i+1}/F} k_{m}E_{i+1} ≠ 0

and

dim_{𝔽_{p}} N_{E/F} k_{m}E ≠ 0.

Recall that we index E_{i} such that F ⊆ E_{i} ⊆ E and [E_{i}: F] = p^{i}.

2.1. The case m = 1.

Fix p, q, and n as above and set m = 1. We construct a field extension E/F as above together with elements

x_{i} ∈ N_{E_{i}/F} (E_{i}^{×}) \ N_{E_{i+1}/F} (E_{i+1}^{×})F^{×p}, 0 ≤ i < n,
x_{n} ∈ N_{E/F} (E^{×}) \ F^{×p}.

Let A/B be a field extension of degree p^{n} such that the characteristic of B is q. Index the subfields A_{i} of A/B such that [A_{i} : B] = p^{i}, and denote by ι_{B,A_{i}} : K_{1}B ↪ K_{1}A_{i} the natural inclusion. Let σ be a generator of Gal(A/B) and set σ_{i} = σ|_{A_{i}}, the restricted map. Finally, assume that there exist elements x_{0}, x_{1}, …, x_{n} ∈ B^{×} such that the following condition holds:

\begin{aligned}[ι_{B,A_{j}}(x_{j})]^{p^{n−j−1}} & ∉ ⟨[ι_{B,A_{j}}(x_{1})]^{p^{n−1}}, [ι_{B,A_{j}}(x_{2})]^{p^{n−2}}, …, [ι_{B,A_{j}}(x_{n})]⟩ \\ & ⊆ A_{j}^{×}/N_{A/A_{j}}(A^{×}), 0 ≤ j < n, \\ x_{n} & ∉ B^{×p}, \end{aligned} ; (∗)

where [x] denotes the class of x and ⟨S⟩ the subgroup generated by a set S in the named factor group. At the end of this section we shall create an example where condition (∗) holds.

Now consider cyclic algebras

A_{j} = (A/B, σ, x_{j}^{p^{n−j}}), 1 ≤ j ≤ n.

Observe that

[A_{j}] = [(A_{j}/B, σ_{j}, x_{j})] ∈ Br(B), 1 ≤ j ≤ n

([P, Chapter 15, Corollary b]), where Br(B) denotes the Brauer group of B. Let F be the function field of the product of the Brauer–Severi varieties of A_{1}, …, A_{n} (see [SV, page 735]; see also [J, Chapter 3] for basic properties of Brauer–Severi varieties).

Let E = A ⊗_{B} F. Since F is a regular extension of B, we see that E/F is a cyclic extension of degree p^{n}. We denote again as σ the generator of Gal(E/F) which restricts to σ ∈ Gal(A/B), and we write E_{k} = A_{k} ⊗_{B} F for k = 0, 1, …, n. Now [A_{j} ⊗_{B} F] = 0 ∈ Br(F), j = 1, …, n, because F splits each A_{j}. Hence

0 = [(E/F, σ, x_{j}^{p^{n−j}})] = [(E_{j}/F, σ_{j}, x_{j})],

and so x_{j} ∈ N_{E_{j}/F} (E_{j}^{×}) as desired (see [P, Chapter 15, page 278]). However, we claim that

x_{j} ∉ (N_{E_{j+1}/F} (E_{j+1}^{×}))F^{×p}, 0 ≤ j < n,
x_{n} ∉ F^{×p}.

Since x_{n} ∉ B^{×p} by hypothesis and F/B is a regular extension, we have x_{n} ∉ F^{×p}. Assume then that 0 ≤ j < n and, contrary to our statement,

x_{j} ∈ (N_{E_{j+1}/F} (E_{j+1}^{×}))F^{×p}.

Then we have x_{j}f^{p} ∈ N_{E_{j+1}/F} (E_{j+1}^{×}) for some f ∈ F^{×}. Hence

[(E_{j+1}/F, σ_{j+1}, x_{j}f^{p})] = 0 ∈ Br(F)

and so

[(E_{j+1}/F, σ_{j+1}, x_{j})] = −[(E_{j+1}/F, σ_{j+1}, f^{p})]
= −[(E_{j}/F, σ_{j}, f)].

(In the case j = 0, we use (E_{0}/F, σ_{0}, f) to denote the zero element in Br(F).) Consequently (E_{j+1}/F, σ_{j+1}, x_{j}) is split by E_{j} (see [P, Chapter 15, Proposition b]).

But then

[(E_{j+1}/E_{j}, σ_{j+1}^{p^{j}}, ι_{F,E_{j}}(x_{j}))] = 0 ∈ Br(E_{j})

(see [D, page 74]). Hence [(E/E_{j}, σ^{p^{j}}, ι_{F,E_{j}}(x_{j}^{p^{n−j−1}}))] = 0 ∈ Br(E_{j}). But E_{j} = A_{j} ⊗_{B} F is the function field of the product of the Brauer–Severi varieties of A_{k} ⊗_{B} A_{j} for k = 1, …, n. Therefore (see [SV, Theorem 2.3])

[(A/A_{j}, σ^{p^{j}}, ι_{B,A_{j}}(x_{j}^{p^{n−j−1}}))] ∈ ⟨[A_{k} ⊗_{B} A_{j}], k = 1, …, n⟩ ⊆ Br(A_{j}).

Consequently

[ι_{B,A_{j}}(x_{j}^{p^{n−j−1}})] ∈ ⟨[ι_{B,A_{j}}(x_{1})]^{p^{n−1}}, …, [ι_{B,A_{j}}(x_{n})]⟩
∈ A_{j}^{×}/N_{A/A_{j}}(A^{×}),

a contradiction to condition (∗).

Thus we have shown that a required extension E/F exists with elements x_{0}, x_{1}, …, x_{n}, provided that we can produce a field extension A/B and elements x_{0}, x_{1}, …, x_{n} ∈ B^{×} such that condition (∗) is valid. Now we show that such an extension and elements exist.

Let B := C(x_{0}, x_{1}, …, x_{n}), where C is a field of characteristic q and x_{0}, x_{1}, …, x_{n} are algebraically independent elements over C. Assume also that there exists a cyclic extension D/C of degree p^{n} with Galois group G = ⟨σ⟩. Finally, let A := D(x_{0}, x_{1}, …, x_{n}). Thus A/B is a cyclic extension of degree p^{n}.

We claim that condition (∗) holds. Clearly x_{n} ∉ B^{×p}. Contrary to our claim assume that

x_{j}^{p^{n−j−1}} = x_{1}^{c_{1}p^{n−1}} ⋯ x_{n}^{c_{n}}N_{A/A_{j}}(γ)

where c_{i} ∈ ℤ, γ ∈ A^{×}, and 0 ≤ j < n.

Consider the discrete valuation v_{j} on A such that v_{j}(x_{j}) = 1, v_{j}(x_{k}) = 0 if j ≠ k, and v_{j}(α) = 0 for α ∈ D. Let H_{j} = Gal(A/A_{j}). Then |H_{j}| = p^{n−j} and v_{j}(σ(γ)) = v_{j}(γ) for each σ ∈ H_{j}. Thus we see that

v_{j}(x_{j}^{p^{n−j−1}}) = p^{n−j−1},

while v_{j} of the right-hand side of the equation is a multiple of p^{n−j}. (Recall that N_{A/A_{j}}(γ) is the product of p^{n−j} conjugates of γ.) Hence the valuations, one equal to p^{n−j−1} and one equal to a multiple of p^{n−j}, disagree, and we have a contradiction.

Thus we see that the equation above is impossible. Hence condition (∗) is valid and we have established the desired example of E/F in the case m = 1.

2.2. The case m > 1.

Fix m, n, and q as above and let L/K be a field extension satisfying the case m = 1 with the same n and q. Let x_{0}, x_{1}, …, x_{n} ∈ K^{×} such that

x_{i} ∈ N_{L_{i}/K}(L_{i}^{×}) \ N_{L_{i+1}/K}(L_{i+1}^{×})K^{×p}, 0 ≤ i < n,
x_{n} ∈ N_{L/K}(L^{×}) \ K^{×p}.

Consider the field of the iterated power series F := K((y_{1})) ⋯ ((y_{m−1})). Then E := L ⊗_{B} F is a cyclic extension of degree p^{n} over F. For each j ∈ {0, 1, …, n} consider the element

α_{j} = {x_{j}, y_{1}, …, y_{m−1}} ∈ k_{m}F.

(If m = 1 then α_{j} = {x_{j}}.) By our hypothesis and the projection formula for the norm map in K-theory, we have

α_{j} ∈ N_{E_{j}/F} k_{m}E_{j}.

Now for 0 ≤ j < n we shall prove by induction on m ∈ ℕ that

α_{j} ∉ N_{E_{j+1}/F} k_{m}E_{j+1}.

If m = 1 then our statement is true by the choice of the field extension L/K and the elements x_{j}. Assume then that m > 1 and that our statement is true for m − 1.

Consider the complete discrete valuation v on F with uniformizer y_{m−1} and residue field F_{v} = K((y_{1})) ⋯ ((y_{m−2})) if m > 2 and F_{v} = K if m = 2.

For the sake of simplicity we denote by E′ the field E_{j+1}, and denote the unique extension of v on E′ again by v. Since we are considering an unramified extension we assume that both valuations are normalized. Let ∂ : k_{m}F → k_{m−1}F_{v} and k_{m}E′ → k_{m−1}E′_{v} be the homomorphisms induced by residue maps in Milnor K-theory. Then applying [K, Lemma 3] we see that the following diagram is commutative:

\xymatrix{k_{m}E′ \ar[r]^{∂} \ar[d]_{N_{E′/F}} & k_{m−1}E′_{v} \ar[d]^{N_{E′_{v}/Fv}} \\ k_{m}F \ar[r]^{∂} & k_{m−1}F_{v}.}

If α_{j} ∈ N_{E′/F} k_{m}E′, then ∂α_{j} ∈ N_{E′_{v}/F_{v}}k_{m−1}E′_{v}.

But ∂α_{j} = {x_{j}, y_{1}, …, y_{m−2}} if m > 2 and ∂α_{j} = {x_{j}} if m = 2, a contradiction in either case. Therefore we have constructed a field extension E/F with the desired properties.

Remarks. (1) In [MSS1] we determined the 𝔽_{p}G-module structure of k_{1}E for all cyclic extensions E/F of degree p^{n}, where G is the Galois group. In particular, the decomposition does not depend upon the characteristic of the base field. The ranks of the free 𝔽_{p}G_{i}-summands appearing in that decomposition are determined by the images of the various N_{E_{i}/F} (E_{i}^{×})/N_{E_{i+1}/F} (E_{i+1}^{×}) in E^{×}/E^{×p}.

(2) When no primitive pth root of unity lies in F, we have that F^{×}/F^{×p} embeds in E^{×}/E^{×p}. Therefore the construction given above for field extensions E/F applies in this case, and these free 𝔽_{p}G_{i}-summands do indeed occur for any characteristic. When a primitive pth root of unity is in F^{×}, the kernel of the homomorphism F^{×}/F^{×p} → E^{×}/E^{×p} is generated by a class [a] ∈ F^{×}/F^{×p}, where E_{1} = F (\sqrt[p]{a}). In this case it is enough to additionally require that a ∈ N_{E/F} (E^{×}) when p = 2 and n = 1 (since the condition is automatic otherwise), and the construction above of field extensions E/F applies again.

(3) If F contains a primitive pth-root of unity and m = 1, the decomposition contains at most one other indecomposable module, a cyclic 𝔽_{p}G-module of dimension p^{k} + 1 for k ∈ {−∞, 0, 1, …, n − 1} (where we set p^{−∞} = 0). In [MSS2] we showed that all of these modules are realizable as well.

Acknowledgments. Andrew Schultz would like to thank Ravi Vakil for his encouragement and direction in this and in all other projects. We would also like to thank Ganesh Bhandari for his useful comments related to this paper. Finally, we would like to thank the referee for such a careful reading of our article, resulting in valuable suggestions for improving our text and the simplification of the proof of condition (∗) for our examples.

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Department of Mathematics, Middlesex College, University of Western Ontario, London, Ontario N6A 5B7, CANADA minac@uwo.ca

Department of Mathematics, University of Illinois at Urbana-Champaign, 273 Altgeld Hall, MC-382, 1409 W. Green Street, Urbana, IL, 61801, USA acs@math.uiuc.edu

Department of Mathematics, Davidson College, Box 7046, Davidson, North Carolina 28035-7046, USA joswallow@davidson.edu

This paper is available via http://nyjm.albany.edu/j/2008/14-10.html.