Unit group of some finite semisimple group algebras

Abstract

We provide the structure of the unit group of ${\mathbb {F}}_{p^k}S_n$, where $p>n$ is a prime and $S_n$ denotes the symmetric group on n letters. We also provide the complete characterization of the unit group of the group algebra ${\mathbb {F}}_{p^k}A_6$ for $p\ge 7$, where $A_6$ is the alternating group on 6 letters.

Introduction

Let $q=p^k$ for some prime p and $k\in \mathbb {N}$. Let $\mathbb {F}_q$ denote the finite field of cardinality q. For any group G, let $\mathbb {F}_qG$ denote the group algebra of G over $\mathbb {F}_q$. We will follow [14] for basic notations. The group of units of $\mathbb {F}_qG$ has applications in different areas, including the construction of convolutional codes (see [5,6,7]) and solving problems in combinatorial number theory (see [3]) et cetera. This necessitates finding the explicit structure of the group of units of $\mathbb {F}_qG$.

In [15], the author has described units of $\mathbb {F}_qG$, where G is a p-group. The authors of [13] complete the study of unit groups of semisimple group algebras of all groups up to order 120, except that of the symmetric group $S_5$ and groups of order 96. In Remark 3.7 of this article, we complete the characterization for the group $S_5$. However, the complexity of the problem increases if the group has a larger size, as that requires solving an equation of type $\displaystyle \sum \limits _{i=1}^nk_in_i^2=|G|$. The most commonly used tricks include identifying a normal subgroup H of G and considering the algebra $\mathbb {F}_q(G/H)$ inside $\mathbb {F}_qG$. This is not possible if the group G does not have any normal subgroup. This is why we will be using representations of the group $A_6$ to solve the problem in case of the same. See [16, 17] et cetera for more exposition about other groups.

The objective of this article is twofold. We start by investigation of $\mathbb {F}_qS_n$ where $p>n$. This is mainly a consequence of the representation theory of $S_n$ over ${\mathbb {C}}$ and the connection between the Brauer characters of the group when $p>n$ and the ordinary characters over ${\mathbb {C}}$. We state the characterization in this case in Theorem 3.6. The group of units of the semisimple algebras $\mathbb {F}_qA_5$ and $\mathbb {F}_q\text {SL}(3,2)$ have been characterized in [1, 12], respectively. In this article, we look at the next non-Abelian simple group $A_6$, the alternating group on six letters. We give a complete characterization of $\mathbb {F}_qA_6$ for the case $p\ge 7$ in Theorem 4.8.

The rest of the article is organized as follows: in “Preliminaries” section, we give some basic definitions and results. “Units of $\mathbb {F}_{p^k}S_n$ for $p \not \mid n$3” section is about the general description of representations of $S_n$ over an arbitrary field of characteristic $p>n$ and deducing the structure of ${\mathcal {U}}(\mathbb {F}_{p^k}S_n)$ for $p>n$. In section 4, we present the result about $\mathbb {F}_{p^k}A_6$ where $p\ge 7$.

Preliminaries

We start by fixing some notations. Already mentioned notations from section 1 are adopted. For a field extension $E/\mathbb {F}_q$, $\text {Gal}(E/\mathbb {F}_q)$ will denote the Galois group of the extension. For $m\in \mathbb {N}$, the notation $\text {M}(m,R)$ denotes the ring of $m\times m$ matrices over R and $\text {GL}(m,R)$ will denote the set of all invertible matrices in $\text {M}(m,R)$. For a ring R, the set of units of R will be denoted by $R^\times$. Let Z(R) and J(R) denote the center and the Jacobson radical, respectively. If G is a group and $x\in G$, then [x] will denote the conjugacy class of x in G. For the group ring $\mathbb {F}_q G$, the group of units will be denoted as $\mathcal {U}(\mathbb {F}_q G)$. For the notations on projective spaces, we follow [4].

We say an element g $\in$ G is a $p'$-element if the order of g is not divisible by p. Let e be the exponent of the group G and $\eta$ be a primitive rth root of unity, where $e=p^fr$ and $p\not \mid r$. Let

$$\begin{aligned} I_{\mathbb {F}_q} = \left\{ l~(\text {mod }e):\text { there exists } \sigma \in \text {Gal}(\mathbb {F}_q(\eta )/\mathbb {F}_q)\text { satisfying } \sigma (\eta ) = \eta ^l\right\} . \end{aligned}$$

Definition 2.1

For a $p'$-element $g \in G$, the cyclotomic $\mathbb {F}_q$-class of g, denoted by $S_{\mathbb {F}_q}(\gamma _g)$, is defined as $\left\{ \gamma _{g^l} : l \in I_{\mathbb {F}_q} \right\}$, where $\gamma _{g^l}\in \mathbb {F}_q G$ is the sum of all conjugates of $g^l$ in G.

Then, we have the following results, which are crucial in determining the Artin–Wedderburn decomposition of $\mathbb {F}_qG$.

Lemma 2.2

[2, Proposition 1.2] The number of simple components of $\mathbb {F}_qG/J(\mathbb {F}_qG)$ is equal to the number of cyclotomic $\mathbb {F}_q$-classes in G.

Definition 2.3

Let $\pi$ be a representation of a group G over a field F. $\pi$ is said to be absolutely irreducible if $\pi ^E$ is irreducible for every field $F\subseteq E$, where $\pi ^E$ is the representation $\pi \otimes E$ over E.

Definition 2.4

A field F is a splitting field for G if every irreducible representation of G over F is absolutely irreducible.

Lemma 2.5

[2, Theorem 1.3] Let n be the number of cyclotomic $\mathbb {F}_q$-classes in G. If $L_1,L_2,\cdots ,L_n$ are the simple components of $Z(\mathbb {F}_qG/J(\mathbb {F}_qG))$ and $S_1,S_2,\cdots ,S_n$ are the cyclotomic $\mathbb {F}_q$-classes of G, then with a suitable reordering of the indices,

$$\begin{aligned} |S_i| = [L_i:\mathbb {F}_q]. \end{aligned}$$

Lemma 2.6

[11, Lemma 2.5] Let K be a field of characteristic p and let $A_1$, $A_2$ be two finite dimensional K-algebras. Assume $A_1$ to be semisimple. If g : $A_2$ $\longrightarrow$ $A_1$ is a surjective homomorphism of K-algebras, then there exists a semisimple K-algebra l such that $A_2/J(A_2)$ $=$ $l\oplus A_1$.

We need the following lemmas from our previous work to compute some components of the Artin–Wedderburn decomposition of $\mathbb {F}_qG$, for a finite group G under consideration.

Lemma 2.7

[1, Lemma 3.1] Let G be a group of order n and $\mathbb {F}$ be a field of characteristic $p>0$. Let G acts on $\{1,2,\cdots ,k\}$ doubly transitively. Set $G_{i}=\{g\in G:g\cdot i=i\}$ and $G_{i,j}=\{g\in G:g\cdot i=i,g\cdot j=j\}$. Then, the $\mathbb {F}G$ module

$$\begin{aligned} W=\left\{ x\in \mathbb {F}^k:\displaystyle \sum \limits _{i=1}^kx_i=0\right\} \end{aligned}$$

is an irreducible $\mathbb {F}G$ module if $p\not \mid k,p\not \mid |G_{1,2}|$.

Lemma 2.8

[1, Corollary 3.8] Let G be a finite group, K be a finite field of characteristic $p>0$, $p\not \mid |G|$. Suppose there exists an n dimensional irreducible representations of G over k. Then, M(n, k) appears as one of the components of the Artin-Wedderburn decomposition of the semisimple algebra $\mathbb {F}_qG$.

Units of $\mathbb {F}_{p^k}S_n$ for $p \not \mid n$

We start the section by talking about representations of $S_n$ over a finite field. We define the Brauer character and state some important results about representations over an arbitrary field. See [8] for further details.

Let E be a field of characteristic p. We choose a ring of algebraic integers A in ${\mathbb {C}}$ such that $E = A/M$, where M is a maximal ideal of A containing pA. Take f to be the natural map $A \longrightarrow E$. Take $W=\{z\in {\mathbb {C}} | z^m=1\text { for some } m \in {\mathbb {Z}}\text { with } p\not \mid m \}$ (note that $W \subseteq A$). Now let $\pi$ be a representation of a finite group G over E. Let S be the set of $p'$ elements of G. For $\alpha \in S$, let $\epsilon _1,\epsilon _2,\ldots ,\epsilon _l\in E^\times$ be the eigenvalues of $\pi (\alpha )$ with multiplicities. Then, for every i, there exists a unique $u_i \in W$ such that $f(u_i)= \epsilon _i$. Define $\phi$ : $S \longrightarrow {\mathbb {C}}$ as $\phi (\alpha ) = \Sigma u_i$. Then, $\phi$ is called the Brauer character of G afforded by $\pi$.

Remark 3.1

The description of Brauer character comes along with a choice of a maximal ideal M of A.

Suppose $\pi _1,\pi _2,\ldots ,\pi _k$ are all the non-isomorphic irreducible representations of G over E up to isomorphism. Let $\phi _i$ be the Brauer character afforded by $\pi _i$. Then, $\phi _i's$ are called irreducible Brauer characters and we denote by IBr(G) the set {$\phi _i$}. We denote by Irr(G) the set of irreducible characters of G over $\mathbb {C}$. We have the following results.

Lemma 3.2

[8, Theorem 15.13] We have $IBr(G) = Irr(G)$ whenever $p\not | |G|$.

For the rest of this section, take $G= S_n$, the symmetric group on n letters. We say a partition $\lambda = (\lambda _1,\lambda _2,\cdots ,\lambda _l)$ of n is p-singular if for some j we have $\lambda _{j+1}=\lambda _{j+2}= \ldots =\lambda _{j+p}$. If a partition is not p-singular, it is called p-regular. Then we have the following.

Lemma 3.3

[9, Theorem 11.5] If F is a field of characteristic p, then as $\lambda$ varies over the p-regular partitions, $D^{\lambda }$ varies over the complete set of inequivalent irreducible $FS_n$-modules, where $D^\lambda =\dfrac{S^\lambda }{S^\lambda \cap (S^{\lambda })^{\perp }}$ and $S^\lambda$ denotes the Specht module corresponding to the partition $\lambda$. Moreover, every field is a splitting field for $S_n$.

Proof

The proof follows immediately from the fact that every partition of n is a p-regular partition. $\square$

Lemma 3.4

The dimensions of non-isomorphic irreducible representations of $S_n$ over E coincide with the dimensions of non-isomorphic irreducible representations of $S_n$ over $\mathbb {C}$ when characteristic of the field E is greater than n.

Proof

Since the dimension of a representation is as same as the value of the corresponding character $\chi$ at the identity element of the group, the result follows from Lemma 3.2. $\square$

Proposition 3.5

Let $S_n$ denote the symmetric group on n letters and $\mathbb {F}_{p^k}$ be a finite field where $p>n$. Then,

$$\begin{aligned} \mathbb {F}_{p^k}S_n\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {M}(\chi (1),\mathbb {F}_{p^k}). \end{aligned}$$

Proof

Since being a semisimple algebra $\mathbb {C}S_n\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {M}(\chi (1),\mathbb {C})$, the result follows from Lemmas 2.8, 3.2 and 3.4. $\square$

Theorem 3.6

Let $S_n$ denote the symmetric group on n letters and $\mathbb {F}_{p^k}$ be a finite field where $p>n$. Then,

$$\begin{aligned} \mathcal {U}(\mathbb {F}_{p^k}S_n)\cong \displaystyle \bigoplus _{\chi \in Irr(G)}\text {GL}(\chi (1),\mathbb {F}_{p^k}). \end{aligned}$$

Proof

This follows immediately from Proposition 3.5 and the fact that given two rings $R_1,R_2$, we have $(R_1\times R_2)^\times =R_1^\times \times R_2^\times$. $\square$

Remark 3.7

Theorem 3.6 improves the result of [10] and proves that when $p>5$, unit group of $\mathbb {F}_{p^k}S_5$ is $\mathcal {U}(\mathbb {F}_{p^k}S_5)$ given by

$$\begin{aligned} \mathbb {F}_{p^k}^\times \oplus \mathbb {F}_{p^k}^\times \oplus \text {GL}(4,\mathbb {F}_{p^k})\oplus \text {GL}(4,\mathbb {F}_{p^k})\oplus \text {GL}(5,\mathbb {F}_{p^k}) \oplus \text {GL}(5,\mathbb {F}_{p^k})\oplus \text {GL}(6,\mathbb {F}_{p^k}) \end{aligned}$$

Remark 3.8

For an irreducible representation $\chi$ of $S_n$ over a field of characteristic $p>n$, this is characterized by a partition $\lambda$ of n. The value $\chi (1)$ can be calculated as the number of standard Young tableaux of shape $\lambda$.

Units of $\mathbb {F}_{p^k}A_6$ for $p \ge 7$

We start with the description of the conjugacy classes in $A_6$. Using [18], the group has 7 conjugacy classes, of which the representatives are given by $(1),a=(1,2)(3,4),b=(1,2,3),c=(1,2,3)(4,5,6),d=(1,2,3,4)(5,6),e=(1,2,3,4,5)$ and $f=(1,2,3,4,6).$ We have the following relations:

$$\begin{aligned} \text {for all }&g\not \in [e]\cup [f],~[g]=[g^{-1}],\end{aligned}$$

(4.1)

$$\begin{aligned} \text {and }&[e]=[e^4],[e^2]=[e^3]=[f]. \end{aligned}$$

(4.2)

Proposition 4.1

Let $\mathbb {F}_{q}$ be a field of characteristic $p\ge 7$ and $G=A_6$. Then, the Artin–Wedderburn decomposition of $\mathbb {F}_qG$ is one of the following:

$\mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q)$,

$\mathbb {F}_q\oplus \bigoplus \limits _{i=1}^4M(n_i,\mathbb {F}_q)\oplus M(n_5,\mathbb {F}_{q^2})$

Proof

Since $p\ge 7$, we have $p\not \mid |A_6|$; by Maschke’s theorem we have $J(\mathbb {F}_qG)=0$ . Hence, Wedderburn decomposition of $\mathbb {F}_qG$ is isomorphic to $\bigoplus \limits _{i=1}^nM({n_i},K_i)$, where for all $1\le i\le n$, we have $n_i > 0$ and $K_i$ is a finite extension of $\mathbb {F}_q$.

Firstly, from Lemma 2.6, we have

$$\begin{aligned} \mathbb {F}_qG \cong \mathbb {F}_q \bigoplus \limits _{i=1}^{n-1}M({n_i},K_i), \end{aligned}$$

(4.3)

taking g to be the map $g(\sum \limits _{x\in A_6}\alpha _xx)=\sum \limits _{x\in A_6}\alpha _x$. Now to compute these $n_i$’s and $K_i$’s we calculate the cyclotomic $\mathbb {F}_q$ classes of G. Note that $p^k\equiv \pm 1\mod 4,p^k\equiv \pm 1\mod 3$ for any prime p. Hence, $S_{\mathbb {F}_q}(\gamma _g)=\{\gamma _g\}$ whenever $g\not \in [e]\cup [f]$ (by Equation 4.1). Hence, we have to consider $S_{\mathbb {F}_q}(\gamma _g)$ in the other cases.

When $p\equiv \pm 1\mod 5$, $S_{\mathbb {F}_q}(\gamma _e)=\{\gamma _e\}$ and $S_{\mathbb {F}_q}(\gamma _f)=\{\gamma _f\}$, by Eq. 4.2 and the fact that $p^k\equiv \pm 1\mod 5$. Thus, by Lemmas 2.2 and 2.5, there are seven cyclotomic $\mathbb {F}_q$-classes and $[K_i:\mathbb {F}_q]=1$ for all $1\le i\le 6$. This gives that in this case the Artin–Wedderburn decomposition is

$$\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q). \end{aligned}$$

When $p\equiv \pm 2\mod 5$ and k is even, then $p^k\equiv - 1\mod 5$. Similarly, in this case the Artin–Wedderburn decomposition is

$$\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^6M(n_i,\mathbb {F}_q). \end{aligned}$$

Lastly, when $p\equiv \pm 2\mod 5$ and k is odd, then $p^k\equiv \pm 2\mod 5$ and $S_{\mathbb {F}_q}(\gamma _e)=\{\gamma _e,\gamma _f\}$ by Eq. 4.2. Thus, by Lemmas 2.2 and 2.5, there are six cyclotomic $\mathbb {F}_q$-classes and $[K_i:\mathbb {F}_q]=1$ for all $1\le i\le 4$, $[K_5:\mathbb {F}_q]=2$ . In this case, the Artin–Wedderburn decomposition is

$$\begin{aligned} \mathbb {F}_q\oplus \bigoplus \limits _{i=1}^4M(n_i,\mathbb {F}_q)\oplus M(n_5,\mathbb {F}_{q^2}) . \end{aligned}$$

$\square$

Since $\dim \mathbb {F}_{q}A_6=|A_6|=360$, Proposition 4.1 gives that the $n_i$’s should satisfy $n_1^2+n_2^2+n_3^2+n_4^2+n_5^2+n_6^2=359$ or $n_1^2+n_2^2+n_3^2+n_4^2+2n_5^2=359$. Since these equations do not have a unique solution, we find some of the $n_i$’s using representations of $A_6$ over $\mathbb {F}_q$ and invoke Lemma 2.7 to reach a unique solution for the mentioned equations. We have the following results.

Lemma 4.2

The group $S_6$ has four inequivalent irreducible representations of degree 5, which on restriction on $A_6$ give two inequivalent irreducible representations of $A_6$ over $\mathbb {F}_{p^k}$ for $p\ge 7$. Moreover, these irreducible representations are obtained from two non-isomorphic doubly transitive actions on a set of 6 points.

Proof

Note that $S_6$ acts on $T=\{1,2,3,4,5,6\}$ doubly transitively. Hence, by Lemma 2.7, we get an irreducible representation of degree 5. Since tensoring with sign representation gives irreducible representations, we get two inequivalent irreducible representations of degree 5 of $S_6$, say $\pi _1$ and $\pi _2$.

For the other two irreducible representations of dimension 5, we consider the outer automorphism of $S_6$, say $\varphi$, given on generators as follows:

$$\begin{aligned} \varphi ((1,2))&=(1,2)(3,4)(5,6)\\ \varphi ((2,3))&=(1,3)(2,5)(4,6)\\ \varphi ((3,4))&=(1,5)(2,6)(3,4)\\ \varphi ((4,5))&=(1,3)(2,4)(5,6)\\ \varphi ((5,6))&=(1,6)(2,5)(3,4). \end{aligned}$$

This gives another doubly transitive action on T, which is not isomorphic to the previous action. Thus, we get another 5 dimensional irreducible representation, say $\pi _3$. Tensoring $\pi _3$ with the sign representations, we get $\pi _4$ which is a 5-dimensional irreducible representation of $S_6$ different from $\pi _3$. By considering the characters of the corresponding representations, we see that $\pi _1,\pi _2,\pi _3$ and $\pi _4$ are all distinct.

Since $A_6$ acts doubly transitively on T via the restrictions of these two actions, we obtain two non-isomorphic 5-dimensional irreducible representations of $A_6$. $\square$

Corollary 4.3

The algebra $\mathbb {F}_qA_6$ has two components to be $\text {M}(5,\mathbb {F}_q)$ for $p\ge 7$.

Proof

Immediately follows from Lemmas 4.2 and 2.7. $\square$

Corollary 4.4

There does not exist any 4 dimensional irreducible representations of $A_6$ over $\mathbb {F}_{p^k}$ for $p\ge 7$.

Proof

From Lemma 3.3, we know that any field $\mathbb {F}_{p^k},p\ge 7$ is a splitting field of $S_6$. Hence, by Proposition 3.5, we have degrees of irreducible representations of $S_6$ are $\{1,5,9,10,16\}$.

Recall that by Frobenius reciprocity we have the following bijection

$$\begin{aligned} \text {Hom}_{\mathbb {F}_qS_6}(\text {Ind}V,W)\cong \text {Hom}_{\mathbb {F}_qA_6}(V,\text {Res}W), \end{aligned}$$

where Ind, Res denote the induction functor, restriction functor, respectively. Here V is an irreducible representation of $A_6$ and W is an irreducible representation of $S_6$. Suppose $A_6$ has an irreducible representation V with $\dim V=4$. Since $[S_6:A_6]=2$, we have that $\dim \text {Ind}V=8$. Since $S_6$ does not have any irreducible representation of dimension 8, the induced representation splits. Being $\dim \text {Ind}V=8$, $\text {Ind}(V)$ does not have any component of dimensions 9, 10 and 16. Now, let us assume that $\dim W=5$, then by Lemma 4.2, $\text {Res}W$ is an irreducible representation. Hence $\text {Hom}_{\mathbb {F}_qA_6}(V,\text {Res}W)=0$, which implies that $\text {Ind}V$ does not have any irreducible component of dimension 5. Similarly, $\text {Ind}V$ does not have any irreducible component of dimension 1. This completes the proof. $\square$

Corollary 4.5

The algebra $\mathbb {F}_qA_6$ has one component to be $\text {M}(9,\mathbb {F}_q)$ for $p\ge 7$.

Proof

The group $A_6$ being isomorphic to $\text {PSL}(2,\mathbb {F}_9)$ acts doubly transitively on a set with 10 points (see [4]), hence the conclusion. $\square$

Corollary 4.6

We have $(n_1,n_2,n_3,n_4,n_5,n_6)=(5,5,9,8,8,10)$ or $(n_1,n_2,n_3,n_4,n_5)=(5,5,9,10,8)$ up to permutation.

Proof

Since $A_6$ has one 1-dimensional, two 5-dimensional and one 9-dimensional irreducible representations, we can assume that $n_1=5,n_2=5,n_3=9$. Hence, we are left with the equation

$$\begin{aligned} n_4^2+n_5^2+n_6^2=228\text { or }n_4^2+2n_5^2=228. \end{aligned}$$

Then, $(n_4,n_5,n_6)\in \{(4,4,14),(8,8,10)\},(n_4,n_5)\in \{(14,4),(10,8)\}$. Hence, the result is obvious from Corollary 4.4. $\square$

Proposition 4.7

Let $\mathbb {F}_{p^k}$ be a field of characteristic $p\ge 7$ and $A_6$ denote the alternating group on six letters. Then, the Artin–Wedderburn decomposition of $\mathbb {F}_{p^k}A_6$ is

$$\begin{aligned} \mathbb {F}_q\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(9,\mathbb {F}_q)\oplus \text {M}(10,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_{q^2}), \end{aligned}$$

when $p\equiv \pm 2\mod 5,k\equiv 1\mod 2$ and

$$\begin{aligned} \mathbb {F}_q\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(5,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_q)\oplus \text {M}(8,\mathbb {F}_q)\oplus \text {M}(9,\mathbb {F}_q)\oplus \text {M}(10,\mathbb {F}_q), \end{aligned}$$

otherwise.

Proof

Follows from Proposition 4.1 and Corollary 4.6. $\square$

Theorem 4.8

Let $\mathbb {F}_{p^k}$ be a field of characteristic $p\ge 7$ and $A_6$ denote the alternating group on six letters. Then, the unit group of the algebra, $\mathcal {U}(\mathbb {F}_{p^k}A_6)$ is

$$\begin{aligned} \mathbb {F}_q^\times \oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(9,\mathbb {F}_q)\oplus \text {GL}(10,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_{q^2}), \end{aligned}$$

(4.4)

when $p\equiv \pm 2\mod 5,k\equiv 1\mod 2$ and

$$\begin{aligned} \mathbb {F}_q^\times \oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(5,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_q)\oplus \text {GL}(8,\mathbb {F}_q)\oplus \text {GL}(9,\mathbb {F}_q)\oplus \text {GL}(10,\mathbb {F}_q), \end{aligned}$$

(4.5)

otherwise.

Proof

This follows immediately from Proposition 4.7 and the fact that given two rings $R_1,R_2$, we have $(R_1\times R_2)^\times =R_1^\times \times R_2^\times$. $\square$