On free elementary ZpCp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {Z}_p C_p$$\end{document}-lattices

We show that all elementary lattices that are free ZpCp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {Z}_p C_p$$\end{document}-modules admit an orthogonal decomposition into a sum of a free unimodular and a p-modular ZpCp\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {Z}_p C_p$$\end{document}-lattice.


Introduction.
Let R := Z p C p denote the group ring of the cyclic group of order p over the localisation of Z at the prime p. The present paper considers free R-lattices L ∼ = R a . The main observation in this situation is Theorem 2.2: Given two free R-modules M and L with pM ⊆ L ⊆ M , there is an R-basis (g 1 , . . . , g a ) of M and 0 ≤ t ≤ a such that (g 1 , . . . , g t , pg t+1 , . . . , pg a ) is an R-basis of L. So these lattices do admit a compatible basis. Applying this observation to Hermitian R-lattices shows that free elementary Hermitian Rlattices admit an invariant splitting (see Theorem 4.1) as the orthogonal sum of a free unimodular lattice and a free p-modular lattice.
The results of this note have been used in the thesis [1] to study extremal lattices admitting an automorphism of order p in the case that p divides the level of the lattice.

Existence of compatible bases.
For a prime p, we denote by the localisation of Z at the prime p. The following arguments also apply accordingly to the completion of this discrete valuation ring. Let R := Z p C p denote the group ring of the cyclic group C p = σ of order p. Then e 1 := 1 p (1 + σ + · · · + σ p−1 ) ∈ QC p and e ζ := 1 − e 1 are the primitive idempotents in the group algebra QC p with a primitive p-th root of unity. The ring T := Z p [ζ p ] is a discrete valuation ring in the p-th cyclotomic field Q[ζ p ] with prime element π := (1 − ζ p ) and hence Remark 2.1. With the notation above, T /(π) ∼ = Z p /(p) ∼ = F p and via this natural ring epimorphism, R is generated as Z p -algebra by 1 = (1, 1) and 1 − σ = (0, π). Moreover Re 1 ∩ R = pRe 1 = pS and Re ζ ∩ R = πRe ζ = πT and the radical J(R) := pS ⊕ πT of R is the unique maximal ideal of the local ring R.
By [6], the indecomposable R-lattices are the free R-module R, the trivial R-lattice Z p = Re 1 = S, and the lattice Z p [ζ p ] = Re ζ = T in the rational irreducible faithful representation of C p . The theorem by Krull-Remak-Schmidt-Azumaya [2, Chapter 1, Section 11] ensures that any finitely generated Rlattice L is a direct sum of indecomposable R-lattices In this note, we focus on the case of free R-lattices. Though R is not a principal ideal domain, for certain sublattices of free R-lattices, there do exist compatible bases: Then there is an R-basis (g 1 , . . . , g a ) of M = Rg 1 ⊕ · · · ⊕ Rg a and 0 ≤ t ≤ a such that The only free R-lattices among these are M and pM . Now assume that a > 1. If L ⊆ J(M ), then we may choose g 1 ∈ L \ J(M ). As g 1 ∈ J(M ), the R-submodule Rg 1 of M is a free submodule of both modules L and M , so M = Rg 1 ⊕ M , L = Rg 1 ⊕ L where M and L = L ∩ M are free R-lattices of rank a − 1 satisfying the assumption of the theorem and the theorem follows by induction. So we may assume that The element e 1 ∈ QC p is a central idempotent in So Le 1 = pM e 1 = pS.
To show that L = pM , we first show that Le ζ = pM e ζ .
As pM ⊆ L, we clearly have that pM e ζ ⊆ Le ζ . To see the opposite inclusion, put K := L ∩ Le ζ to be the kernel of the projection e 1 : L → Le 1 . As L is free, we get, as in Remark 2.1, that K = πLe ζ . Let k be maximal such that K ⊆ π kT . Then k ≥ 2 because Le ζ ⊆ πT (see equation (1)).
Assume that k ≤ p − 1. There is ∈ L such that y = e ζ ∈ π kT . As pM e 1 As pM e ζ = pT = π p−1T and y ∈ π kT , the assumption that k ≤ p − 1 shows that − pm ∈ π kT , which contradicts the definition of k.
Therefore  of M such that x generates a free direct summand and y, its complement isomorphic to S. Let L be the R-sublattice generated by pxe 1 and x(1 − σ) + y. As x(1 − σ) + y generates a free R-sublattice of M and R(pxe 1 ) ∼ = S, we have L ∼ = S ⊕ R. For p > 2, we compute that pM ⊆ L ⊆ M . Then the fact that |M/L| = p 2 implies that these two modules do not admit a compatible pseudo-basis.

Lattices in rational quadratic spaces.
From now on, we consider Z p -lattices L in a non-degenerate rational quadratic space (V, B). The dual lattice of L is Following O'Meara [5, Section 82 G], we call a lattice L unimodular if L = L # , and p j -modular if p j L # = L.
We now assume that σ is an automorphism of L of order p, so σ is an orthogonal mapping of (V, B) with Lσ = L. Then also the dual lattice L # is a σ-invariant lattice in V . As the dual basis of a lattice basis of L is a lattice basis of L # , the symmetric bilinear form B yields an identification between L # and the lattice Hom Zp (L, Z p ) of Z p -valued linear forms on L. The σ-invariance of B shows that this is an isomorphism of Z p [σ]-modules. As all indecomposable Z p [σ]-lattices are isomorphic to their homomorphism lattices, we obtain The group ring R comes with a natural involution, the unique Z p -linear map : R → R with σ i = σ −i for all 0 ≤ i ≤ p − 1. This involution is the restriction of the involution on the maximal order S ⊕ T that is trivial on S and the complex conjugation on T . Remark 3.3. The Z p -lattice R is unimodular with respect to the symmetric bilinear form where Tr reg : QC p → Q denotes the regular trace of the p-dimensional Qalgebra QC p . We thus obtain a bijection between the set of σ-invariant Z pvalued symmetric bilinear forms on the R-lattice L and the R-valued Hermitian forms on L: If h : L × L → R is such a Hermitian form, then B = 1 p Tr reg • h is a symmetric bilinear σ-invariant form on L. As R = R # , these forms yield the same notion of duality. In particular, the dual lattice L # of a free lattice L = ⊕ a i=1 Rg i is again free L # = ⊕ a i=1 Rg * i with the Hermitian dual basis (g * 1 , . . . , g * a ) as a lattice basis, giving a constructive argument for Proposition 3.2 for free lattices.

Free elementary lattices.
In this section, we assume that L is an elementary lattice and σ an automorphism of L of prime order p. Recall that R is the commutative ring R := Z p [σ], so L is an R-module. Theorem 4.1. Let p be a prime and let L be an elementary lattice with an automorphism σ such that L ∼ = R a is a free R-module. Then also L # ∼ = R a and there is an R-basis (g 1 , . . . , g a ) of L # and 0 ≤ t ≤ a such that (g 1 , . . . , g t , pg t+1 , . . . , pg a ) is an R-basis of L. In particular, L is the orthogonal sum of the unimodular free R-lattice L 0 := Rg 1 ⊕ · · · ⊕ Rg t and a p-modular free R-lattice L 1 := L ⊥ 0 .
Proof. Under the assumption, both lattices L and M := L # are free R-modules satisfying pM ⊆ L ⊆ M . So, by Theorem 2.2, there is a basis (g 1 , . . . , g a ) of M such that (g 1 , . . . g t , pg t+1 , . . . , pg a ) is a basis of L. Clearly L is an integral lattice and L 0 := Rg 1 ⊕ · · · ⊕ Rg t is a unimodular sublattice of L. By [3, Satz 1.6], unimodular free sublattices split as orthogonal summands, so L = L 0 ⊥ L 1 with L # 1 = 1 p L 1 , i.e. L 1 is p-modular.
Note that the assumption that the lattice is elementary is necessary, as the following example shows. Here we identify R as a subring of S ⊕ T , so (p, 0) = pe 1 = 1 + σ + · · · + σ p−1 and (0, π) = (0, (1 − ζ p )) = 1 − σ ∈ R. Then L is orthogonally indecomposable because Le ζ is an orthogonally indecomposable T -lattice, but L is not modular. Note that the base change matrix between (g 1 , g 2 ) and the dual basis, an Rbasis of L # , is the inverse of the Gram matrix above, so As (1, 0) = e 1 ∈ R, this shows that pL # ⊆ L, so L is not an elementary lattice.
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