2.1. Lattices and orders

For further background, we refer the reader to [Reference Reiner35] or [Reference Curtis and Reiner9]. Let $R$ be a Dedekind domain with field of fractions $F$ . An $R$ -lattice $M$ is a finitely generated torsion-free $R$ -module, or equivalently, a finitely generated projective $R$ -module. Note that any $R$ -submodule of an $R$ -lattice is again an $R$ -lattice. For any finite-dimensional $F$ -vector space $V$ , an $R$ -lattice in $V$ is a finitely generated $R$ -submodule $M$ in $V$ . We define a $F$ -vector subspace of $V$ by

\begin{equation*} FM \;:\!=\; \{ \alpha _{1} m_{1} + \alpha _{2} m_{2} + \cdots + \alpha _{r} m_{r} \mid r \in \mathbb {Z}_{\geq 0}, \alpha _{i} \in F, m_{i} \in M \} \end{equation*}

and say that $M$ is a full $R$ -lattice in $V$ if $FM=V$ . We may identify $FM$ with $F \otimes _{R} M$ .

Let $A$ be a finite-dimensional $F$ -algebra. An $R$ -order in $A$ is a subring $\Lambda$ of $A$ (so in particular has the same unity element as $A$ ) such that $\Lambda$ is a full $R$ -lattice in $A$ . A $\Lambda$ -lattice is a $\Lambda$ -module which is also an $R$ -lattice. For $\Lambda$ -lattices $M$ and $N$ , a homomorphism of $\Lambda$ -modules $f\;:\; M \rightarrow N$ is called a homomorphism of $\Lambda$ -lattices.

The following well-known lemma follows from [Reference Curtis and Reiner9, Exercise 23.2].

2.2. Associated orders

Let $\Lambda$ be an $R$ -order in a finite-dimensional $F$ -algebra $A$ . Let $M$ be a full $R$ -lattice in a free $A$ -module of rank $1$ (thus $FM \cong A$ as $A$ -modules). The associated order of $M$ is defined to be

\begin{equation*} \mathfrak {A}(A, M) = \{ \lambda \in A \;:\; \lambda M \subseteq M \}. \end{equation*}

Note that $\mathfrak{A}(A, M)$ is an $R$ -order (see [Reference Reiner35, §8]). In particular, it is the largest order $\Lambda$ over which $M$ has a structure of $\Lambda$ -module. The following well-known result says that $\mathfrak{A}(A, M)$ is the only $R$ -order in $A$ over which $M$ can possibly be free.

Proof. By hypothesis there exists $\alpha \in M$ such that $M=\Lambda \alpha$ is a free $\Lambda$ -module. Thus, $FM=A\alpha$ is free over $A$ . Let $x \in \mathfrak{A}(A, M)$ . Then, $x \alpha \in M = \Lambda \alpha$ , so $x\alpha = y\alpha$ for some $y \in \Lambda$ . Since $FM$ is freely generated by $\alpha$ , we must have $x=y$ . Hence, $\mathfrak{A}(A, M) \subseteq \Lambda$ . The reverse inclusion is trivial and therefore $\Lambda = \mathfrak{A}(A, M)$ .

2.3. Completion and local freeness

Let $\mathfrak{p}$ be any maximal ideal of $R$ . Let $F_{\mathfrak{p}}$ denote the completion of $F$ with respect to a valuation defined by $\mathfrak{p}$ and let $R_{\mathfrak{p}}$ be the corresponding valuation ring. For any $R$ -module $M$ we write $M_{\mathfrak{p}}$ for $R_{\mathfrak{p}} \otimes _{R} M$ and $V_{\mathfrak{p}}= F_{\mathfrak{p}} \otimes _{F} V$ for any $F$ -vector space $V$ . These two notations are consistent as the map $\lambda \otimes _{\mathcal{O}_{F}} v \mapsto \lambda \otimes _{F} v$ ( $v \in V$ , $\lambda \in \mathcal{O}_{F_{\mathfrak{p}}}$ ) is an isomorphism (see [Reference Fröhlich and Taylor16, p. 93]).

Let $\Lambda$ be an $R$ -order and let $M$ be a $\Lambda$ -lattice in some $A$ -module $V$ . Then, $\Lambda _{\mathfrak{p}}$ is an $R_{\mathfrak{p}}$ -order in $A_{\mathfrak{p}}$ and $M_{\mathfrak{p}}$ is a $\Lambda _{\mathfrak{p}}$ -lattice in $V_{\mathfrak{p}}$ . We say that $M$ is locally free over $\Lambda$ if $M_{\mathfrak{p}}$ is free over $\Lambda _{\mathfrak{p}}$ for every $\mathfrak{p}$ .

Let $G$ be a finite group and let $M$ be a full $R[G]$ -lattice in a free $A$ -module of rank $1$ . Then, $R[G] \subseteq \mathfrak{A}(F[G], M)$ and $R_{\mathfrak{p}}[G] \subseteq \mathfrak{A}(F_{\mathfrak{p}}[G], M_{\mathfrak{p}}) \cong \mathfrak{A}(F[G], M)_{\mathfrak{p}}$ . Moreover, $M$ is locally free over $\mathfrak{A}(F[G], M)$ if $M_{\mathfrak{p}}$ is free over $\mathfrak{A}(F_{\mathfrak{p}}[G], M_{\mathfrak{p}})$ for every $\mathfrak{p}$ .

2.4. Associated orders of rings of integers and decomposition obstructions

Let $K/F$ be a finite Galois extension of number fields and let $G=\textrm{Gal}(K/F)$ . We consider the behaviour of the associated order $\mathfrak{A}_{K/F}\;:\!=\;\mathfrak{A}(F[G],\mathcal{O}_{K})$ with respect to localisation and induction.

Let $\mathfrak{p}$ be a maximal ideal of $\mathcal{O}_{F}$ . Then, we have decompositions

\begin{equation*} K_{\mathfrak {p}} \;:\!=\; F_{\mathfrak {p}} \otimes _{F} K \cong \prod _{\mathfrak {P}' \mid \mathfrak {p}} K_{\mathfrak {P}'} \quad \textrm { and } \quad \mathcal {O}_{K, \mathfrak {p}} \;:\!=\; \mathcal {O}_{F_{\mathfrak {p}}} \otimes _{\mathcal {O}_{F}} \mathcal {O}_{K} \cong \prod _{\mathfrak {P}' \mid \mathfrak {p}} \mathcal {O}_{K_{\mathfrak {P}'}}, \end{equation*}

where $\{ \mathfrak{P}' \mid \mathfrak{p} \}$ consists of the primes of $\mathcal{O}_{K}$ above $\mathfrak{p}$ (see [Reference Fröhlich and Taylor16, p. 109]). Fix a prime $\mathfrak{P}$ above $\mathfrak{p}$ and let $D$ be its decomposition group in $G$ . Then, as $G$ acts transitively on $\{ \mathfrak{P}' \mid \mathfrak{p} \}$ we have

\begin{equation*} K_{\mathfrak {p}} \cong \prod _{s \in G/D} sK_{\mathfrak {P}} \quad \textrm { and } \quad \mathcal {O}_{K, \mathfrak {p}} \cong \prod _{s \in G/D} s \mathcal {O}_{K_{\mathfrak {P}}}, \end{equation*}

where the products run over a system of representatives of the left cosets $G/D$ . Hence,

\begin{equation*} \mathcal {O}_{K, \mathfrak {p}} \cong \mathrm {Ind}_{D}^{G}\mathcal {O}_{K_{\mathfrak {P}}} \;:\!=\; \mathcal {O}_{F_{\mathfrak {p}}}[G] \otimes _{\mathcal {O}_{F_{\mathfrak {p}}}[D]} \mathcal {O}_{K_{\mathfrak {P}}}, \end{equation*}

and

\begin{equation*} \mathfrak {A}_{K/F,\mathfrak {p}}=\mathfrak {A}(F[G],\mathcal {O}_{K})_{\mathfrak {p}} \cong \mathfrak {A}(F_{\mathfrak {p}}[G],\mathrm {Ind}_{D}^{G}\mathcal {O}_{K_{\mathfrak {P}}}), \end{equation*}

where the last isomorphism follows from [Reference Curtis and Reiner9, Exercise 24.2], for instance. Thus, $\mathcal{O}_{K}$ is locally free over $\mathfrak{A}_{K/F}$ at $\mathfrak{p}$ if and only if $\textrm{Ind}_{D}^{G}\mathcal{O}_{K_{\mathfrak{P}}}$ is free over $\mathfrak{A}(F_{\mathfrak{p}}[G],\textrm{Ind}_{D}^{G}\mathcal{O}_{K_{\mathfrak{P}}})$ .

We recall from the introduction that it may be the case that we encounter a prime $\mathfrak{p}$ that is a decomposition obstruction for $K$ , i.e. such that $\mathcal{O}_{K_{\mathfrak{P}}}$ is free over $\mathfrak{A}_{K_{\mathfrak{P}}/F_{\mathfrak{p}}}$ but $\mathcal{O}_{K}$ is not locally free over $\mathfrak{A}_{K/F}$ at $\mathfrak{p}$ (indeed, Definition 1.4 can be easily generalised to base fields other than $\mathbb{Q}$ ). In Section 6, we will consider the relationship between $\mathfrak{A}(F_{\mathfrak{p}}[G],\textrm{Ind}_{D}^{G}\mathcal{O}_{K_{\mathfrak{P}}})$ and $\textrm{Ind}_{D}^{G}\mathfrak{A}_{K_{\mathfrak{P}}/F_{\mathfrak{p}}}$ , as well as conditions under which the implication ‘if $\mathcal{O}_{K_{\mathfrak{P}}}$ is free over $\mathfrak{A}_{K_{\mathfrak{P}}/F_{\mathfrak{p}}}$ then $\mathcal{O}_{K}$ is locally free over $\mathfrak{A}_{K/F}$ at $\mathfrak{p}$ ’ holds.

2.5. Reduction to the study of local freeness

We define locally free class groups. Background material on locally free class groups, including a discussion of equivalent definitions, can be found in [Reference Curtis and Reiners10, §49].

The following proposition underpins the proofs of all the new theorems stated in the introduction.

Proof. One implication is trivial. By [Reference Curtis and Reiners10, Theorem (49.25)], the inclusion $\mathbb{Z}[G]\subseteq \mathfrak{A}_{K/\mathbb{Q}}$ induces a surjection $\text{Cl}(\mathbb{Z}[G]) \twoheadrightarrow \text{Cl}(\mathfrak{A}_{K/\mathbb{Q}})$ , and so $\text{Cl}(\mathfrak{A}_{K/\mathbb{Q}})$ is also trivial. Moreover, by [Reference Endô and Hironaka12] the triviality of $\textrm{Cl}(\mathbb{Z}[G])$ implies that $G$ must be abelian, dihedral, or isomorphic to $A_{4}$ , $S_{4}$ or $A_{5}$ (see also [Reference Curtis and Reiners10, Theorem (50.29)]). In each of these cases, $\mathbb{Q}[G]$ is isomorphic to a finite direct product of matrix rings over number fields. Hence by [Reference Curtis and Reiners10, Proposition (51.2)] $\mathbb{Q}[G]$ satisfies the Eichler condition (see [Reference Curtis and Reiners10, Definitions (45.4) or §51A]). Thus the Jacobinski cancellation theorem [Reference Jacobinski19] (see also [Reference Curtis and Reiners10, Theorem (51.24)]) implies that $\mathfrak{A}_{K/\mathbb{Q}}$ has the locally free cancellation property. The non-trivial implication now follows easily.

Proof. Let $G=\textrm{Gal}(K/\mathbb{Q})$ . In each case $\text{Cl}(\mathbb{Z}[G])$ is trivial, as shown in [Reference Reiner and Ullom37].

3.1. Freeness results for Galois extensions of $p$ -adic fields

Many of the results and definitions of this subsection also hold for local fields of positive characteristic, but for simplicity we restrict to the case of $p$ -adic fields. We fix a prime number $p$ .

Theorem 3.7. [Reference Jaulent20] Let $p,n$ and $r$ be positive integers such that $p$ is an odd prime, $n$ divides $p-1$ and $r$ is a primitive $n$ th root modulo $p$ . Let $G$ be the metacyclic group with the following structure:

(3.1) \begin{equation} G= \langle x,y\;:\;x^{p}=1,y^{n}=1,yxy^{-1}=x^{r} \rangle \cong C_{p} \rtimes C_{n}. \end{equation}

Let $K/\mathbb{Q}_{p}$ be a Galois extension with $\textrm{Gal}(K/\mathbb{Q}_p) \cong G$ . Then, $\mathcal{O}_{K}$ is free over $\mathfrak{A}_{K/\mathbb{Q}_p}$ .

Let $K/F$ be a Galois extension of $p$ -adic fields and let $G=\textrm{Gal}(K/F)$ . We recall that for an integer $t \geq -1$ the $t$ th ramification group is defined to be

\begin{equation*} G_{t} \;:\!=\; \{ \sigma \in G \;:\; v_{K}(\sigma (x)-x)\geq t+1\text { }\forall x\in \mathcal {O}_K \}, \end{equation*}

where $v_{K}$ is the normalised valuation on $K$ (i.e. with image $\mathbb{Z}$ ). When it is not obvious which extension we are referring to we will use the notation ‘ $G_t(K/F)$ ’ or similar. Thus, $K/F$ is unramified if and only if $G_{0}$ is trivial and is tamely ramified if and only if $G_{1}$ is trivial. We say that the extension is weakly ramified if $G_{2}$ is trivial.

For a subgroup $H$ of $G$ define $\textrm{Tr}_{H}=\sum _{h\in H}h \in F[G]$ and $e_{H}=\frac{1}{|H|}\textrm{Tr}_{H} \in F[G]$ . Note that $e_{H}$ is an idempotent. We say that $K/F$ is almost-maximally ramified if $e_{H} \in \mathfrak{A}_{K/F}$ for every subgroup $H$ of $G$ such that $G_{t+1} \subseteq H \subseteq G_{t}$ for some $t \geq 1$ .

Remark 3.12. Let $H$ be a subgroup of $G$ and let $r$ be a positive integer. We now show how to determine whether $\frac{1}{r}\textrm{Tr}_{H}\in \mathfrak{A}_{K/F}$ . For example, when $r=|H|$ this can be used to check for almost-maximal ramification. Let $M$ be the subfield of $K$ fixed by $H$ . We denote by $\mathfrak{D}_{K/M}$ the different of the extension $K/M$ (see [ Reference Serre38, III§3]) and by $v_{p}(x)$ the $p$ -adic valuation of an integer $x$ (thus, $v_{p}$ is the restriction of $v_{\mathbb{Q}_p}$ to $\mathbb{Z}$ ).

\begin{align*} \begin{split} \frac{1}{r}\text{Tr}_H\in \mathfrak{A}_{K/F} & \Longleftrightarrow \frac{1}{r}\text{Tr}_{K/M}(\mathcal{O}_K)\subseteq \mathcal{O}_M\\[5pt] &\Longleftrightarrow \text{Tr}_{K/M}(\mathcal{O}_{K})\subseteq r\mathcal{O}_{M} \\[5pt] &\Longleftrightarrow \mathcal{O}_K \subseteq r\mathcal{O}_M \mathfrak{D}_{K/M}^{-1}\text{ by [Ser79, III Proposition 7]}\\[5pt] &\Longleftrightarrow \mathfrak{D}_{K/M} \subseteq r\mathcal{O}_K \\[5pt] &\Longleftrightarrow v_K(\mathfrak{D}_{K/M})\geq e(K/\mathbb{Q}_p)v_p(r)\\[5pt] &\Longleftrightarrow \sum _{i=0}^\infty (|G_i(K/M)|-1)\geq e(K/\mathbb{Q}_p) v_p(r)\text{ by [Ser79, IV Proposition 4]}. \end{split} \end{align*}

3.2. Local freeness results for Galois extensions of number fields

Let $K/F$ be a finite Galois extension of number fields. If $\mathcal{O}_{K}$ is free over $\mathfrak{A}_{F/K}$ , then it is clear that $\mathcal{O}_{K}$ is locally free over $\mathfrak{A}_{F/K}$ . In particular, the analogues of Theorems 1.1, 1.2 and 1.3 all hold, with ‘locally free’ in place of ‘free’. Theorems 3.15 and 3.17 below are generalisations of the first two of these analogues.

4.1. Hybrid group rings

Let $R$ be a discrete valuation ring with fraction field $F$ and let $G$ be a finite group. Let $M$ be a full $R[G]$ -lattice in $F[G]$ . Note that $R[G]\subseteq \mathfrak{A}(F[G], M)$ .

For a normal subgroup $N$ of $G$ define $e_{N}=\frac{1}{|N|}\sum _{n\in N}n \in F[G]$ to be the central idempotent associated with $N$ .

Proof. Since $|N| \in R^{\times }$ we have $e_{N} \in R[G]$ . Moreover, it is straightforward to show that $e_{N}R[G] \cong R[G/N]$ . Thus, we have established (i) and (ii), and it remains to prove (iii).

The last isomorphism of (iii) is immediate from (ii). We now prove the first equality, that is, $e_{N}\mathfrak{A}(F[G],M)=\mathfrak{A}(F[G],M)\cap e_{N}F[G]$ . Since $e_{N}\in R[G]\subseteq \mathfrak{A}(F[G],M)$ , we easily have that $e_{N}\mathfrak{A}(F[G],M)\subseteq \mathfrak{A}(F[G],M)\cap e_{N}F[G]$ . The other containment follows from the fact that $e_{N}^{2}=e_{N}$ ; hence, any element in $\mathfrak{A}(F[G],M)\cap e_{N}F[G]$ , with the harmless multiplication by $e_{N}$ , can be written as an element in $e_{N}\mathfrak{A}(F[G],M)$ .

We now prove that $\mathfrak{A}(F[G],M)\cap e_{N}F[G]=\mathfrak{A}(e_NF[G],e_{N}M)$ . Consider $e_Nx\in \mathfrak{A}(F[G],M)\cap e_NF[G]$ for a certain $x\in F[G]$ ; we have to prove that $e_{N}x$ preserves $e_{N}M$ . Since $e_{N}M\subseteq M$ and $e_{N}x\in \mathfrak{A}(F[G],M)$ , we have that $e_{N}xe_{N}M\subseteq M$ . Hence, $e_{N}xe_{N}M=e_{N}e_{N}xe_{N}M\subseteq e_{N}M$ . Conversely, let us consider an element $e_{N}x\in \mathfrak{A}(e_NF[G],e_NM)$ , thus such that $e_{N}xe_{N}M\subseteq e_{N}M$ . We must prove that $e_{N}x\in \mathfrak{A}(F[G],M)$ , which is automatic since $e_{N}xM=e_{N}e_{N}xM=e_{N}xe_{N}M\subseteq e_{N}M\subseteq M$ .

We now recall the notion of hybrid group ring introduced in [Reference Johnston and Nickel21, §2] and further developed in [Reference Johnston and Nickel22, §2].

Example 4.4. Let $G=A_{4}$ or $S_{4}$ and let $N$ be its unique normal subgroup of order $4$ . Then, $\mathbb{Z}_{3}[G]$ is $N$ -hybrid as shown in [ Reference Johnston and Nickel21, Examples 2.16 and 2.18]. Indeed, we have

\begin{equation*} \mathbb {Z}_{3}[A_{4}] \cong \mathbb {Z}_{3}[C_{3}] \times M_{3 \times 3}(\mathbb {Z}_{3}) \quad \text {and} \quad \mathbb {Z}_{3}[S_{4}] \cong \mathbb {Z}_{3}[S_{3}] \times M_{3 \times 3}(\mathbb {Z}_{3}) \times M_{3 \times 3}(\mathbb {Z}_{3}), \end{equation*}

where $M_{3 \times 3}(\mathbb{Z}_{p})$ is a maximal $\mathbb{Z}_{p}$ -order by [ Reference Reiner35, Theorem (8.7)].

Proposition 4.6. Suppose $R[G]$ is $N$ -hybrid. Then,

\begin{equation*} \mathfrak {A}(F[G],M) = e_{N}\mathfrak {A}(F[G],M) \times (1-e_N)R[G] \cong \mathfrak {A}(F[G/N],e_NM) \times (1-e_N)R[G]. \end{equation*}

Moreover, $M$ is free over $\mathfrak{A}(F[G],M)$ if and only if $e_{N}M$ is free over $\mathfrak{A}(F[G/N],e_{N}M)$ .

Proof. The first claim follows from Proposition 4.1, Definition 4.2, and the fact that $R[G] \subseteq \mathfrak{A}(F[G],M)$ . Since $(1-e_{N})R[G]$ is a maximal $R$ -order, $(1-e_{N})M$ is free over $(1-e_{N})R[G]$ by [Reference Reiner35, Theorem (18.10)]. The second claim now follows from the decomposition $M \cong e_{N} M \oplus (1-e_{N})M$ .

4.2. Applications to local freeness for extensions of number fields

Proposition 4.7. Let $K/F$ be a finite Galois extension of number fields and let $G=\textrm{Gal}(K/F)$ . Let $p$ be a rational prime and let $\mathfrak{p}$ be a prime of $F$ above $p$ . Let $N$ be a normal subgroup of $G$ such that $p \nmid |N|$ and let $M$ be the subfield of $K$ fixed by $N$ . Then, we have an identification $e_{N}\mathcal{O}_{K,\mathfrak{p}}=\mathcal{O}_{M,\mathfrak{p}}$ . Moreover, via this identification, the structure of $e_{N}\mathcal{O}_{K,\mathfrak{p}}$ as an $e_{N}\mathcal{O}_{F_{\mathfrak{p}}}[G]$ -module coincides with the structure of $\mathcal{O}_{M,\mathfrak{p}}$ as an $\mathcal{O}_{F_{\mathfrak{p}}}[G/N]$ -module under the canonical identification $G/N \cong \textrm{Gal}(M/F)$ . In particular,

\begin{equation*} e_{N}\mathfrak {A}(F_{\mathfrak {p}}[G],\mathcal {O}_{K,\mathfrak {p}}) =\mathfrak {A}(F_{\mathfrak {p}}[G],\mathcal {O}_{K,\mathfrak {p}})\cap e_{N}F[G] \cong \mathfrak {A}(F_{\mathfrak {p}}[G/N],\mathcal {O}_{M,\mathfrak {p}}). \end{equation*}

Now further suppose that $\mathcal{O}_{F_{\mathfrak{p}}}[G]$ is $N$ -hybrid. Then,

\begin{equation*} \mathfrak {A}_{K/F,\mathfrak {p}} \cong \mathfrak {A}_{M/F,\mathfrak {p}} \times (1-e_{N})\mathcal {O}_{F_{\mathfrak {p}}}[G], \end{equation*}

and $\mathcal{O}_{K,\mathfrak{p}}$ is free over $\mathfrak{A}_{K/F,\mathfrak{p}}$ if and only if $\mathcal{O}_{M,\mathfrak{p}}$ is free over $\mathfrak{A}_{M/F,\mathfrak{p}}$ .

Proof. The claims regarding the identifications are clear. The remaining claims are then specialisations of Propositions 4.1 and 4.6.

Proof. Let $M$ be the subfield of $K$ fixed by $N$ . Then, $\mathcal{O}_{M,p}$ is free over $\mathfrak{A}_{M/\mathbb{Q},p}$ by Theorem 1.1 or Theorem 3.17. The result now follows from Proposition 4.7.

4.3. Preliminary results on $A_{4}$ and $S_{4}$ -extensions of $\mathbb{Q}$

Proof. By Corollary 2.7, it suffices to show that $\mathcal{O}_{K,p}$ is free over $\mathfrak{A}_{K/\mathbb{Q},p}$ for each rational prime $p \geq 3$ . For $p \geq 5$ , this follows from Theorem 3.14. Let $G=\textrm{Gal}(K/\mathbb{Q})$ and let $N$ be its unique normal subgroup of order $4$ . By Example 4.4, the group ring $\mathbb{Z}_{3}[G]$ is $N$ -hybrid. Moreover, $G/N \cong C_{3}$ or $S_{3}$ (note that $S_{3} \cong D_{6}$ is metacyclic of type (3.1)). Thus by Corollary 4.8, we have that $\mathcal{O}_{K,3}$ is free over $\mathfrak{A}_{K/\mathbb{Q},3}$ .

Proof. This can easily be checked by, for instance, using the database of $p$ -adic fields [Reference Jones and Roberts24] (which is now accessible via the database [30]). Indeed, $L$ is the Galois closure of the extension of $\mathbb{Q}_{2}$ generated by the polynomial $x^{4}+2x^{3}+2x^{2}+2$ .

Proof. By Proposition 4.10, it suffices to show that $\mathcal{O}_{K,2}$ is free over $\mathfrak{A}_{K/\mathbb{Q},2}$ . If $2$ is tamely ramified in $K/\mathbb{Q}$ , then this follows from Theorem 3.14. Now suppose that $2$ has full decomposition group in $G\;:\!=\;\textrm{Gal}(K/\mathbb{Q})$ . Then, $2$ is weakly ramified in $K/\mathbb{Q}$ by Lemma 4.11. Let $\mathfrak{P}$ be the unique prime of $K$ above $2$ . Then,

\begin{equation*} \mathcal {O}_{K,2} \cong \mathrm {Ind}_{G}^{G}\mathcal {O}_{K_{\mathfrak {P}}}=\mathcal {O}_{K_{\mathfrak {P}}} \quad \text {and} \quad \mathfrak {A}_{K/\mathbb {Q},2}\cong \mathfrak {A}_{K_{\mathfrak {P}}/\mathbb {Q}_{2}}, \end{equation*}

so the result now follows from Theorem 3.9.

6.1. Associated orders and induction

Let $R$ be a Dedekind domain with field of fractions $F$ . Let $H$ be a subgroup of a finite group $G$ and let $M$ be an $R[H]$ -lattice such that $FM$ is free of rank $1$ over $F[H]$ .

We recall that $\textrm{Ind}_{H}^{G} M$ is the induced module $R[G] \otimes _{R[H]} M \cong \bigoplus _{s\in G/H} sM$ , where on the right-hand side we choose a system of representatives in $G$ of the left cosets $G/H$ and the left $R[G]$ -module structure is given by the relation $gs=th$ for some coset representative $t$ and $h\in H$ . We wish to understand the relationship between $\mathfrak{A}(F[G], \textrm{Ind}_{H}^{G} M)$ and $\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)$ . Note that these both contain the group ring $R[G]$ .

Proposition 6.1 ([Reference Bergé4, §1.3]). We have

\begin{equation*}\mathfrak {A}(F[G], \mathrm {Ind}_{H}^{G} M) = \bigcap _{g \in G} g\mathrm {Ind}_{H}^{G} \mathfrak {A}(F[H], M)g^{-1}. \end{equation*}

Proof. If $\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)$ is a ring, then $g\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)g^{-1} = \textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)$ for all $g \in G$ and thus $\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)=\mathfrak{A}(F[G], \textrm{Ind}_{H}^{G} M)$ by Proposition 6.1. Conversely, if $\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)=\mathfrak{A}(F[G], \textrm{Ind}_{H}^{G} M)$ then the left-hand side is a ring since the right-hand side is an associated order and thus a ring.

Proof. Since $R[H] \subseteq \mathfrak{A}(F[H], M)$ and $M$ is free (necessarily of rank $1$ ) over $\mathfrak{A}(F[H], M)$ , we see that $M$ and $\mathfrak{A}(F[H], M)$ are isomorphic as $R[H]$ -lattices. Extension of scalars gives an isomorphism $\textrm{Ind}_{H}^{G}M \cong \textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)$ of $R[G]$ -lattices. By Lemma 2.1, this is also an isomorphism of $\mathfrak{A}(F[G], \textrm{Ind}_{H}^{G} M)$ -lattices.

Proof. In case (i) this follows immediately from Proposition 6.5. Clearly, (ii) $\Rightarrow$ (i). Moreover, (ii) $\Leftrightarrow$ (iii) by Corollary 6.2.

The following is a partial converse to Corollary 6.6(i).

If $H$ is normal in $G$ , then we define $\mathfrak{A}^{*}(M)=\bigcap _{g\in G}g\mathfrak{A}(F[H], M)g^{-1}$ .

6.2. Clean orders and induction

Let $R$ be a discrete valuation ring with field of fractions $F$ of characteristic zero and suppose that the residue field of $R$ is finite.

Proof. See [Reference Hattori17] or [Reference Roggenkamp36, IX Corollary 1.5].

Proof. (i) $\Rightarrow$ (iv). By Proposition 6.9(iii), $M$ is projective over $\mathfrak{A}^{*}(M)$ . Moreover, $\mathfrak{A}^{*}(M)$ is a clean order by Theorem 6.11 and thus $M$ is in fact free over $\mathfrak{A}^{*}(M)$ . Hence $\mathfrak{A}^{*}(M)=\mathfrak{A}(F[H], M)$ by Proposition 2.2.

(iv) $\Rightarrow$ (iii). We have $\textrm{Ind}^{G}_{H}\mathfrak{A}(F[H], M)=\textrm{Ind}^{G}_{H}\mathfrak{A}^{*}(M)=\mathfrak{A}(F[G], \textrm{Ind}_{H}^{G} M)$ , where the second equality is Proposition 6.9(ii). Thus, $\textrm{Ind}^{G}_{H}\mathfrak{A}(F[H], M)$ is a ring by Corollary 6.2. Hence, $\textrm{Ind}_{H}^{G}M$ is free over $\textrm{Ind}_{H}^{G} \mathfrak{A}(F[H], M)$ by Corollary 6.6(iii).

(iii) $\Rightarrow$ (ii). This follows from Corollary 6.2.

(ii) $\Rightarrow$ (i). This follows from the general fact that every free module is projective.