1 Introduction

Let \(R := \mathbb {Z}_p C_p\) denote the group ring of the cyclic group of order p over the localisation of \(\mathbb {Z}\) at the prime p. The present paper considers free R-lattices \(L\cong R^a\). The main observation in this situation is Theorem 2.2: Given two free R-modules M and L with \(pM \subseteq L \subseteq M \), there is an R-basis \((g_1,\ldots , g_a)\) of M and \(0\le t\le a\) such that \((g_1,\ldots , g_t, p g_{t+1},\ldots , p g_a )\) 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 R-lattices 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.

2 Existence of compatible bases

For a prime p, we denote by

$$\begin{aligned} \mathbb {Z}_{p} := \{ \frac{a}{b} \in \mathbb {Q}\mid p \text{ does } \text{ not } \text{ divide } b \} \end{aligned}$$

the localisation of \(\mathbb {Z}\) at the prime p. The following arguments also apply accordingly to the completion of this discrete valuation ring. Let \(R:=\mathbb {Z}_{p} C_p\) denote the group ring of the cyclic group \(C_p = \langle \sigma \rangle \) of order p. Then \(e_1:=\frac{1}{p} (1+\sigma + \cdots + \sigma ^{p-1}) \in \mathbb {Q}C_p \) and \(e_{\zeta }:=1-e_1\) are the primitive idempotents in the group algebra \(\mathbb {Q}C_p\) with \(\mathbb {Q}C_p = \mathbb {Q}C_p e_1 \oplus \mathbb {Q}C_p e_{\zeta } \cong \mathbb {Q}\oplus \mathbb {Q}[\zeta _p]\), where \(\zeta _p\) is a primitive p-th root of unity. The ring \(T:=\mathbb {Z}_p[\zeta _p]\) is a discrete valuation ring in the p-th cyclotomic field \(\mathbb {Q}[\zeta _p]\) with prime element \(\pi := (1-\zeta _p)\) and hence

$$\begin{aligned} Re_1 \oplus R e_{\zeta } \cong \mathbb {Z}_{p} \oplus \mathbb {Z}_{p} [\zeta _p] =: S \oplus T \end{aligned}$$

is the unique maximal \(\mathbb {Z}_p\)-order in \(\mathbb {Q}C_p\).

Remark 2.1

With the notation above, \(T/(\pi ) \cong \mathbb {Z}_p / (p) \cong \mathbb {F}_p \) and via this natural ring epimorphism,

$$\begin{aligned} R = \{ (x,y) \in \mathbb {Z}_{p} \oplus \mathbb {Z}_{p} [\zeta _p] \mid x + p \mathbb {Z}_p = y + \pi \mathbb {Z}_{p} [\zeta _p] \} . \end{aligned}$$

R is generated as \(\mathbb {Z}_p\)-algebra by \(1=(1,1)\) and \(1-\sigma = (0,\pi )\). Moreover \(Re_1 \cap R = p R e_1 = p S\) and \(Re_{\zeta } \cap R = \pi R e_{\zeta } = \pi T\) and the radical \(J(R) := pS \oplus \pi T\) 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 \(\mathbb {Z}_p = Re_1=S\), and the lattice \(\mathbb {Z}_{p} [\zeta _p]=Re_{\zeta } =T\) 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 R-lattice L is a direct sum of indecomposable R-lattices

$$\begin{aligned} L \cong R^a \oplus T^b \oplus S^c . \end{aligned}$$

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:

Theorem 2.2

Let \(M\cong R^a\) be a free R-lattice of rank a. Assume that L is a free R-lattice with \(pM \subseteq L \subseteq M\). Then there is an R-basis \((g_1,\ldots , g_a)\) of \(M=Rg_1\oplus \cdots \oplus Rg_a\) and \(0\le t\le a\) such that

$$\begin{aligned} L = Rg_1\oplus \cdots \oplus Rg_t \oplus pRg_{t+1} \oplus \cdots \oplus p R g_a . \end{aligned}$$

Proof

Let \({\tilde{S}} := M e_1 \) and \({\tilde{T}} := M e_{\zeta }\). Now \(M \cong R^a\) is a free R-lattice, so, as in Remark 2.1, M is a sublattice of \({\tilde{S}}\oplus {\tilde{T}}\) of index \(p^a\), \({\tilde{S}} \cap M = p {\tilde{S}} \), and \({\tilde{T}} \cap M = \pi {\tilde{T}} \). The Jacobson radical is \(J(M) = J(R) M = p{\tilde{S}} \oplus \pi {\tilde{T}} \) and of index \(p^a\) in M. We proceed by induction on a.

If \(a=1\), then \(M=R\), \({\tilde{S}} \cong S\), \({\tilde{T}} \cong T\). As \(M/pM \cong \mathbb {F}_pC_p \cong \mathbb {F}_p[x]/(x-1)^p \) is a chain ring, the R-sublattices of M that contain pM form a chain:

$$\begin{aligned} M \supset p {\tilde{S}} \oplus \pi {\tilde{T}} \supset p {\tilde{S}} \oplus \pi ^2 {\tilde{T}} \supset \cdots \supset p {\tilde{S}} \oplus \pi ^{p-2} {\tilde{T}} \supset p {\tilde{S}} \oplus p {\tilde{T}} \supset pM . \end{aligned}$$

The only free R-lattices among these are M and pM.

Now assume that \(a>1\). If \( L \not \subseteq J(M)\), then we may choose \(g_1 \in L\setminus J(M)\). As \(g_1\not \in J(M)\), the R-submodule \(Rg_1\) of M is a free submodule of both modules L and M, so \(M=Rg_1 \oplus M'\), \(L=Rg_1 \oplus L'\) where \(M'\) and \(L'=L\cap M'\) 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

$$\begin{aligned} L\subseteq J(M) = p{\tilde{S}} \oplus \pi {\tilde{T}} . \end{aligned}$$
(1)

The element \(e_1\in \mathbb {Q}C_p\) is a central idempotent in \(\text {End}_R(J(M))\) projecting onto \(p{\tilde{S}}= J(M) e_1\). The assumption that \(pM \subseteq L \subseteq J(M)\) implies that

$$\begin{aligned} p{\tilde{S}} = pM e_1 \subseteq L e_1 \subseteq J(M) e_1 = p{\tilde{S}} . \end{aligned}$$

So \(L e_1 = pM e_1 =p{\tilde{S}}\).

To show that \(L=pM\), we first show that \(Le_{\zeta } = pM e_{\zeta }\).

As \(pM\subseteq L\), we clearly have that \(pMe_{\zeta } \subseteq L e_{\zeta } \).

To see the opposite inclusion, put \(K := L\cap L e_{\zeta } \) to be the kernel of the projection \(e_1: L \rightarrow Le_1\). As L is free, we get, as in Remark 2.1, that \(K = \pi Le_{\zeta } \). Let k be maximal such that \( K \subseteq \pi ^{k} {\tilde{T}}\). Then \(k\ge 2\) because \(Le_{\zeta } \subseteq \pi {\tilde{T}}\) (see equation (1)).

Assume that \(k \le p-1\). There is \(\ell \in L\) such that \(y=\ell e_{\zeta } \not \in \pi ^{k} {\tilde{T}} \). As \(pMe_1 = Le_1\), there is \(m \in M \) such that \(pm e_1= \ell e_1\). Now \(pM\subseteq L\), so \(pm \in L\) and \(\ell -pm \in K= Ke_{\zeta }\).

We compute \(\ell -pm = (\ell -pm ) e_{\zeta } = y - pm e_{\zeta }\).

As \(pMe_{\zeta } = p {\tilde{T}} = \pi ^{p-1}{\tilde{T}}\) and \(y\not \in \pi ^{k}{\tilde{T}}\), the assumption that \(k\le p-1\) shows that \(\ell -pm \not \in \pi ^k {\tilde{T}}\), which contradicts the definition of k.

Therefore \(k \ge p\) and \(L e_{\zeta } \subseteq pM e_{\zeta } \).

Now pM and L both have index \(p^a\) in \(pMe_1 \oplus pM e_{\zeta } = Le_1 \oplus Le_{\zeta }\) (again by Remark 2.1 as L and M are free). So the assumption \(pM \subseteq L\) implies that \(pM = L\). \(\square \)

Remark 2.3

Let \(M\cong T^b\oplus S^c\) and let L be a sublattice of M again isomorphic to \(T^b \oplus S^c\). Then \(M=Me_{\zeta } \oplus Me_1\) and \(L=Le_{\zeta } \oplus Le_1\). By the main theorem for modules over principal ideal domains, there is a T-basis \((x_1,\ldots , x_b)\) of \(Me_{\zeta } \) and a \(\mathbb {Z}_p\)-basis \((y_1,\ldots , y_c) \) of \(Me_1\), as well as \(0\le n_1\le \cdots \le n_b\), \(0\le m_1\le \cdots \le m_c\), such that \(L = \bigoplus _{i=1}^b \pi ^{n_i} T x_i \oplus \bigoplus _{i=1}^c p^{m_i} \mathbb {Z}_p y_i .\)

Example 2.4

For general modules M, however, Theorem 2.2 has no appropriate analogue. To see this, consider \(M\cong R\oplus S \) and choose a pseudo-basis (x, y) 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 \(p x e_1\) and \(x(1-\sigma ) + y\). As \(x(1-\sigma )+y\) generates a free R-sublattice of M and \(R(pxe_1)\cong S\), we have \(L \cong S \oplus R \). For \(p>2\), we compute that \(pM\subseteq L \subseteq M\). Then the fact that \(|M/L|=p^2\) implies that these two modules do not admit a compatible pseudo-basis.

3 Lattices in rational quadratic spaces

From now on, we consider \(\mathbb {Z}_{p} \)-lattices L in a non-degenerate rational quadratic space (V, B). The dual lattice of L is

$$\begin{aligned} L^{\#} := \{ x\in V \mid B(x,\ell ) \in \mathbb {Z}_{p} \text{ for } \text{ all } \ell \in L \}. \end{aligned}$$

The lattice L is called integral if \(L \subseteq L^{\#} \), and elementary if

$$\begin{aligned} pL^{\#} \subseteq L \subseteq L^{\#} . \end{aligned}$$

Following O’Meara [5, Section 82 G], we call a lattice L unimodular if \(L=L^{\#}\), and \(p^j\)-modular if \(p^jL^{\#} = L\).

We now assume that \(\sigma \) is an automorphism of L of order p, so \(\sigma \) is an orthogonal mapping of (V, B) with \(L \sigma = L\). Then also the dual lattice \(L^{\#} \) is a \(\sigma \)-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 \(\text {Hom}_{\mathbb {Z}_p}(L,\mathbb {Z}_p)\) of \(\mathbb {Z}_p\)-valued linear forms on L. The \(\sigma \)-invariance of B shows that this is an isomorphism of \(\mathbb {Z}_p[\sigma ]\)-modules.

Remark 3.1

As a \(\mathbb {Z}_{p}[\sigma ]\)-module, we have \(L^{\#} \cong \text {Hom}_{\mathbb {Z}_{p}} (L,\mathbb {Z}_{p}) \).

As all indecomposable \(\mathbb {Z}_{p}[\sigma ]\)-lattices are isomorphic to their homomorphism lattices, we obtain

Proposition 3.2

([4, Lemma 5.6]). If \(L\cong R^a\oplus T^b \oplus S^c \) as \(\mathbb {Z}_{p}[\sigma ]\)-lattice, then also \(L^{\#} \cong R^a\oplus T^b \oplus S^c \).

The group ring R comes with a natural involution, the unique \(\mathbb {Z}_p\)-linear map with \(\overline{\sigma ^i} = \sigma ^{-i} \) for all \(0\le i\le p-1\). This involution is the restriction of the involution on the maximal order \(S\oplus T\) that is trivial on S and the complex conjugation on T.

Remark 3.3

The \(\mathbb {Z}_p\)-lattice R is unimodular with respect to the symmetric bilinear form

$$\begin{aligned} R\times R \rightarrow \mathbb {Z}_p , (x,y) \mapsto \frac{1}{p} \text {Tr}_{reg} (x\overline{y} ) \end{aligned}$$

where \(\text {Tr}_{reg} : \mathbb {Q}C_p \rightarrow \mathbb {Q}\) denotes the regular trace of the p-dimensional \(\mathbb {Q}\)-algebra \(\mathbb {Q}C_p\). We thus obtain a bijection between the set of \(\sigma \)-invariant \(\mathbb {Z}_p\)-valued symmetric bilinear forms on the R-lattice L and the R-valued Hermitian forms on L: If \(h:L\times L \rightarrow R \) is such a Hermitian form, then \(B=\frac{1}{p} \text {Tr}_{reg} \circ h \) is a symmetric bilinear \(\sigma \)-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 \(L = \oplus _{i=1}^a R g_i \) is again free \(L^{\#} = \oplus _{i=1}^a R g^*_i \) with the Hermitian dual basis \((g^*_1,\ldots , g^*_a)\) as a lattice basis, giving a constructive argument for Proposition 3.2 for free lattices.

4 Free elementary lattices

In this section, we assume that L is an elementary lattice and \(\sigma \) an automorphism of L of prime order p. Recall that R is the commutative ring \(R:= \mathbb {Z}_{p} [\sigma ] \), so L is an R-module.

Theorem 4.1

Let p be a prime and let L be an elementary lattice with an automorphism \(\sigma \) such that \(L\cong R^a\) is a free R-module. Then also \(L^{\#} \cong R^a\) and there is an R-basis \((g_1,\ldots , g_a)\) of \(L^{\#}\) and \(0\le t \le a\) such that \((g_1,\ldots , g_t, p g_{t+1},\ldots , p g_a ) \) is an R-basis of L. In particular, L is the orthogonal sum of the unimodular free R-lattice \(L_0:=R g_1\oplus \cdots \oplus R g_t\) and a p-modular free R-lattice \(L_1 := L_0^{\perp }\).

Proof

Under the assumption, both lattices L and \(M:=L^{\#}\) are free R-modules satisfying \(pM\subseteq L\subseteq M\). So, by Theorem 2.2, there is a basis \((g_1,\ldots , g_a)\) of M such that \((g_1,\ldots g_t, pg_{t+1}, \ldots , pg_a)\) is a basis of L. Clearly L is an integral lattice and \(L_0:=Rg_1\oplus \cdots \oplus Rg_t\) is a unimodular sublattice of L. By [3, Satz 1.6], unimodular free sublattices split as orthogonal summands, so \(L=L_0 \perp L_1\) with \(L_1^{\#} = \frac{1}{p} L_1\), i.e. \(L_1\) is p-modular. \(\square \)

Note that the assumption that the lattice is elementary is necessary, as the following example shows.

Example 4.2

Let \(L=R g_1 \oplus R g_2 \) be a free lattice of rank 2 with R-valued Hermitian form defined by the Gram matrix

$$\begin{aligned} \left( \begin{array}{cc} (p,0) &{} (0,\pi ) \\ (0,\overline{\pi }) &{} (p,0) \end{array} \right) . \end{aligned}$$

Here we identify R as a subring of \(S\oplus T\), so \((p,0) = p e_1 = 1+\sigma + \cdots + \sigma ^{p-1} \) and \((0,\pi ) = (0,(1-\zeta _p)) = 1-\sigma \in R \). Then L is orthogonally indecomposable because \(Le_{\zeta }\) 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 R-basis of \(L^{\#}\), is the inverse of the Gram matrix above, so

$$\begin{aligned} \left( \begin{array}{cc} (p^{-1},0) &{} (0,-\overline{\pi }^{-1} ) \\ (0,-{\pi }^{-1}) &{} (p^{-1},0) \end{array} \right) . \end{aligned}$$

As \((1,0) = e_1 \not \in R\), this shows that \(pL^{\#} \not \subseteq L\), so L is not an elementary lattice.