On Automorphism Groups of Finite Chain Rings

Abstract

A finite ring with an identity is a chain ring if its lattice of left ideals forms a unique chain. Let R be a finite chain ring with invaraints $p,n,r,k,k^{\prime },m.$ If $n=1,$ the automorphism group $Aut\left(R\right)$ of R is known. The main purpose of this article is to study the structure of $Aut\left(R\right)$ when $n>1.$ First, we prove that $Aut\left(R\right)$ is determined by the automorphism group of a certain commutative chain subring. Then we use this fact to find the automorphism group of R when $p\nmid k.$ In addition, $Aut\left(R\right)$ is investigated under a more general condition; that is, R is very pure and p need not divide $k.$ Based on the j-diagram introduced by Ayoub, we were able to give the automorphism group in terms of a particular group of matrices. The structure of the automorphism group of a finite chain ring depends essentially on its invaraints and the associated j-diagram.

1. Introduction

All rings considered in this paper are finite and have an identity. Chain rings are rings whose left (right) ideals form a unique chain under inclusion. These rings have been used in geometry as coordinatizing rings of Klingenberg planes and Pappian Hjelmslev planes [1,2]. In recent years, chain rings have found significant applications in new places: in coding theory for creating more compact codes with higher capabilities of error correction [3,4]; in combinatorics for constructing bent functions, partial difference sets and relative difference sets [5,6,7]. In addition, such rings arise in the p-adic fields as factor-rings of rings of integers in suitable finite extensions of the field of p-adic numbers, ${\mathbb{Q}}_p$ [8]. The purpose of the present paper is to investigate the automorphism groups of chain rings. The results of this work give immediate corollaries for the applications mentioned above.

For a general background of finite chain rings, we refer to [8,9,10,11,12,13]. Let R denote a chain ring of characteristic $p^n$ with non-zero (Jacobson) radical $J\left(R\right)$ of nilpotencey index $m.$ The case when $J\left(R\right)=0,$R is a field. The residue field $F=R/J\left(R\right)$ is a finite field of order $p^r.$ R contains a subring (coefficient subring) $R_0$ of the form $R_0=GR(p^n,r)\cong Z_{p^n}\left[a\right],$ where a is an element of $R_0$ of the multiplicative order $p^r-1.$ Moreover, there exist $\pi \in J\left(R\right)\backslash J^2\left(R\right)$ and $\sigma \in Aut\left(R_0\right)$ such that $J\left(R\right)=\pi R$ and $\pi u=\sigma \left(u\right)\pi ,$ for each $u\in R_0$. If k is the greatest integer $i\le m$ such that $p\in J^i\left(R\right)$, R can be written as:

$$R={\oplus }_{i=0}^{k-1}{R}_0{\pi }^i$$

(1)

(as $R_0—$module). This implies ${\pi }^k=p{\sum }_{i=0}^{k-1}{u}_i{\pi }^i,$ where $u_i\in R_0$ and $u_0$ is a unit, i.e., $\pi$ is a root of an Eisenstein polynomial over $R_0,$

$$g\left(x\right)=x^k-p\sum _{i=0}^{k-1}{u}_i{x}^i.$$

(2)

If there exists $\pi \in R$ such that ${\pi }^k=p\beta ,$ where $\beta \in <a>$. Then, R is called a very pure chain ring. It is also known that $\sigma$ is uniquely determined by R and $R_0,$ and thus it is called the associated automorphism of R with respect to $R_0.$ If $n>1,$ then ${\sigma }^k=id$, and if $k^{\prime }$ is the order of $\sigma ,$ $k^{\prime }\mid k.$ The integers $p,n,r,k,k^{\prime },m$ are called the invariants of $R.$ Furthermore, there is $t,$ $1\le t\le k$ such that $m=(n-1)k+t.$

When $n=1,$ the automorphism group $Aut\left(R\right)$ is determined in [14]. If $n>1,$ relatively little is known about $Aut\left(R\right).$ Indeed, a special class of automorphisms has been considered in [15]. Our main goal in this article is to investigate the structure of $Aut\left(R\right)$ when $n>1.$ We first show that $Aut\left(R\right)$ is given in terms of the automorphism group of a certain commutative chain subring; thus, it suffices to find $Aut\left(R\right),$ R is commutative ($k^{\prime }=1$). Next, we use this idea to determine $Aut\left(R\right)$ when R is a chain ring with $(p,k)=1.$ In addition, $Aut\left(R\right)$ is studied under a more general condition; R is very pure. If $p\nmid k$ or R is complete, we manage to give $Aut\left(R\right)$ in terms of a specific group of matrices.

2. Preliminaries and Notations

In this section, we mention some facts and introduce notations that will be used in the subsequent discussions.

In the sequel, R is a finite chain ring with invariants $p,n,r,k,k^{\prime },m.$ Let $R_1$ be the centralizer of $R_0$ in $R,$ then

$$R_1={\oplus }_{i=0}^{k_1-1}{R}_0{\pi }^{i{k}^{\prime }},$$

(3)

where $k_1=\frac{k}{k^{\prime }}$ and $m_1=\lfloor \frac{m}{k^{\prime }}\rfloor +1$. From (3), the radical of $R_1,$ $J\left(R_1\right)={\pi }^{k^{\prime }}{R}_1.$ However, $R_1$ is a commutative chain ring with invariants $p,n,r,k_1,m_1.$ Moreover, $R_1$ is the only maximal commutative subring of R containing $R_0$ and it is unique up to the inner automorphisms of R [9].

Let $S=R_0^{\sigma }$ be the fixed subring of $R_0$ by $\sigma$ which is of the form $GR(p^n,r_0)=$$Z_{p^n}\left[b\right],$ b is an element in $<a>$ of the multiplicative order $p^{r_0}-1,$ where $r_0=r/k^{\prime }.$ Denote $Z\left(R\right)$ the center of R, then,

$$Z\left(R\right)={\oplus }_{i=0}^{k_1-1}S{\pi }^{k^{\prime }i}+\Omega ,$$

(4)

where $\Omega =J^{m_1-1}\left(R_1\right)=J^{m-1}\left(R\right)$ if $k^{\prime }|(m-1)$ and $\Omega =0$ otherwise. Let

$$R_2={\oplus }_{i=0}^{k_1-1}S{\pi }^{k^{\prime }i}.$$

(5)

It is easy to check that $R_2=Z\left(R\right)$ if $k^{\prime }\nmid (m-1),$ and hence $Z\left(R\right)$ in such a case is a commutative chain ring with invariants p, n, $r_0$, $k_1$, $m_1$. When $k^{\prime }\mid (m-1),$ then $Z\left(R\right)=R_2+\Omega ,$ which is not a chain subring of $R.$ However, $Z\left(R\right)/\Omega$ is a commutative chain ring with invariants p, n, $r_0$, $k_1$, $m_1-1$ [9]. Note that ${\pi }^k\in Z\left(R\right),$ i.e., ${\pi }^k$ can be written as:

$${\pi }^k=\sum _{i=0}^{k_1-1}{u}_i{\pi }^{k^{\prime }i}+u_{m-1}{\pi }^{m-1}=p{\beta }_1{h}_1+u_{m-1}{\pi }^{m-1},$$

(6)

where ${\beta }_1\in <b>,$ $h_1=1+{\sum }_{i=1}^{k_1-1}{u}_i^{\prime }{\pi }^{k^{\prime }i},$ $p{u}_i^{\prime }={\beta }_1^{-1}{u}_i,u_i\in S$ for $0\le i\le k_1-1$, and $u_{m-1}\in <a>$.

If $\mu \in Aut\left(R_0\right),$ $\lambda \in U\left(R_1\right)$ and $\xi \in U\left(R\right).$ Then, we define the following correspondences:

$$\begin{array}{cc}\hfill & {\varphi }_{\lambda }^{\mu }(\sum u_i{\pi }^i)=\sum \mu \left(u_i\right){\left(\lambda \pi \right)}^i,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & {\psi }_{\xi }(\sum u_i{\pi }^i)=\xi \sum u_i{\pi }^i{\xi }^{-1},\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & {\alpha }_{\sigma }(\sum u_i{\pi }^i)=\sum \sigma \left(u_i\right){\pi }^i.\hfill \end{array}$$

(9)

The following statements are related to the commutative case [16,17]. Let $U\left(R\right)$ denote the unit group of $R,$ then $U\left(R\right)=<a>\otimes H,$ where a is an element of order $p^r-1$ and $H=1+J\left(R\right)$ is the $p—$Sylow subgroup of $U\left(R\right).$ Let $H_s=1+J^s\left(R\right),$ $s\in P_m=\{1,2,\cdots ,m\}.$ Consider

$$H=H_1>H_2>H_3>\cdots >H_m=<1>,$$

(10)

joined with a function j defined by:

$$j\left(s\right)=\left\{\begin{array}{cc}\min (ps,m),\hfill & s\le u,\hfill \\ \min (s+k,m),\hfill & s>u,\hfill \end{array}\right.$$

(11)

where $u=\lfloor \frac{k}{p-1}\rfloor .$ We refer to the series (10) when we mention the j-diagram. We call R an incomplete (complete) chain ring if H has an incomplete (complete) j-diagram in the sense that is given by Ayoub. For more details about j-diagrams, see [16,17].

Definition 1.

Let R be a chain ring, then we call R complete (incomplete) if its $R_1$ is complete (incomplete).

Let $e_{iq}$ denote the matrix with the identity of F in the (i,q) position and zeros elsewhere. The group $M_d$ of all 1-triangular matrices:

$$1+\sum {\alpha }_{iq}{e}_{iq},$$

where ${\alpha }_{iq}\in F$ is the $p—$Sylow subgroup of $G{L}_d\left(F\right),$ which is the general linear group over F [18]. Let $E_d$ be the set of all matrices of the form $\left[a_{iq}\right]$:

$$a_{iq}=\sum _{e+f=i}{a}_{e,1}{a}_{f,q-1}.$$

Note that $E_d$ is a subgroup of $M_d$ of order $p^{rd}.$

All symbols shall retain their meanings throughout the article as stated above, in addition, for a given chain ring $R,$ we denote $T_R$ all pairs ($R_0,\pi$) which fulfill the aforementioned conditions.

3. The Automorphism Group $\mathbf{Aut}\left(\mathbf{R}\right)$

Throughout the section, R is a chain ring with invariants $p,n,r,k,k^{\prime },m.$ If $k=1,$ then $R=R_0$ and $Aut\left(R\right)$ is cyclic of order r generated by $\rho ,$ the Frobenius map. The case when $n=1$ (${\pi }^k=0),$ i.e., $m=k$, $Aut\left(R\right)$ is known [14]. Thus, we assume that $k>1$ and $n>1$ in our discussions.

Proposition 1.

Let R be a chain ring with $p,n,r,k,k^{\prime },m.$ Then,

$${\pi }^k=p\beta h,$$

(12)

$$\left\{\begin{array}{cc}\beta \in <a>,h=1,\hfill & \mathrm{if}m-1=k,\hfill \\ \beta \in <b>,h=h_1+{\alpha }_0{\pi }^{m-k-1},\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

where $h_1\in R_2\cap H\left(R_1\right)\mathrm{and}{\alpha }_0\in R_0.$

Proof. 

First, if $m-1=k,$ i.e., $n=2$ and $t=1.$ Then, by (6), $p{\beta }_1{h}_1=p{\beta }_1.$ This means, ${\pi }^k=p{\beta }_1+u_{m-1}{\pi }^k,$ i.e., $(1-u_{m-1}){\pi }^k=p{\beta }_1,$ and this implies ${\pi }^k=p\beta ,$ where $\beta ={\beta }_1{(1-u_{m-1})}^{-1}\in <a>$. On the other hand, if $m-1>k,$ we consider two cases. The case when $k^{\prime }\nmid (m-1),$ it is easy to prove the result since $Z\left(R\right)=R_2.$ Now, assume that $k^{\prime }\mid (m-1),$ then $k^{\prime }\mid (t-1)$ because $m-1=(n-1)k+t-1.$ Let $t-1=t_1{k}^{\prime }$ for some $t_1$ positive integer, then,

$$\begin{array}{cc}\hfill u_{m-1}{\pi }^{m-1}& =u_{m-1}{\pi }^{(n-1)k}{\pi }^{t_1{k}^{\prime }}\hfill \\ \hfill & =u_{m-1}{(p{\beta }_1{h}_1+u_{m-1}{\pi }^{m-1})}^{n-1}{\pi }^{t_1{k}^{\prime }}\hfill \\ \hfill & =u_{m-1}\left(p^{n-1}{\beta }_1^{n-1}{h}_1^{n-1}\right){\pi }^{t_1{k}^{\prime }}\hfill \\ \hfill & =u_{m-1}{p}^{n-1}{\beta }_1^{n-1}{\pi }^{t_1{k}^{\prime }},\hfill \end{array}$$

where ${\beta }_1\in <b>$. Moreover, since $t_1<k,$ then Equation (6) yields,

$$\begin{array}{cc}\hfill {\pi }^k& =p{\beta }_1{h}_1+u_{m-1}{p}^{n-1}{\beta }_1^{n-1}{\pi }^{t_1{k}^{\prime }}\hfill \\ \hfill & =p\beta (h_1+p^{n-2}{u}_{m-1}{\beta }^{n-1}{\beta }^{-1}{\pi }^{t_1{k}^{\prime }})(n>2)\hfill \\ \hfill & =p\beta h,\hfill \end{array}$$

where $\beta ={\beta }_1,$ $h=h_1+{\alpha }_0{\pi }^{m-k-1}$ and ${\alpha }_0=u_{m-1}{\beta }^{n-1}{\beta }^{-1}$. □

Remark 1.

By the proof of the previous proposition, we can write:

$$Z\left(R\right)=R_2+p^{n-1}{R}_0{\pi }^{t_1{k}^{\prime }}.$$

(13)

Remark 2.

Note that if $k\ne m-1$ and $k^{\prime }\mid (m-1),$ then by the proof of Proposition 1,

$$h=\sum _{i=0}^{k_1-1}{u}_i{\pi }^{i{k}^{\prime }},$$

(14)

where $u_i\in S$ ($u_0=1$) if $i\ne t_1$ and $u_{t_1}\in R_0.$

Definition 2.

Let $T_1$ be a commutative chain ring, which is cyclic Galois over a commutative chain ring $T_2.$ Let $G=<\chi >$ be the group of all $T_2—$automorphisms of $T_1.$ Define $N_{\chi }\left(y\right):U\left(T_1\right)\to U\left(T_2\right)$ as:

$$N_{\chi }\left(y\right)=\prod _{\eta \in G}\eta \left(y\right),$$

(15)

where $N_{\chi }$ is called the norm function.

Lemma 1.

Let R be a chain ring. Then, the homomorphism ψ defined from $U\left(T_1\right)$ into $U\left(T_2\right)$ by: $\psi \left(\lambda \right)=N_{\sigma }\left(\lambda \right)$ is surjective, where $T_1=R_1/J^{m_1-1}\left(R_1\right)$ and $T_2=Z\left(R\right)/\Omega .$

Proof. 

Note that every finite commutative chain ring is a quotient ring of the ring of integers of an appropriate extension of ${\mathbb{Q}}_p$ [8]. In addition, the extension $T_1$ over $T_2$ corresponds to an unramified extension over ${\mathbb{Q}}_p,$ then it is a Galois extension with $Au{t}_{T_2}\left(T_1\right)=<{\alpha }_{\sigma }>$. Thus, by (Proposition 3, page 82 [19]), $N_{\sigma }\left(U\left(T_1\right)\right)=U\left(T_2\right).$ □

Proposition 2.

With the same assumptions of Lemma 1,

$$ker{N}_{\sigma }=\{{\alpha }_{\sigma }\left(\lambda \right){\lambda }^{-1}:\lambda \in U\left(T_1\right)\}.$$

(16)

Moreover, $ker{N}_{\sigma }\cong U\left(T_1\right)/U\left(T_2\right).$

Proof. 

The first claim is trivial. Consider the map $\psi :U\left(T_1\right)\to ker{N}_{\sigma },$ defined by $\psi \left(\lambda \right)={\alpha }_{\sigma }\left(\lambda \right){\lambda }^{-1}.$ Let ${\lambda }_1,{\lambda }_2\in U\left(T_1\right),$ then by the definition of ${\alpha }_{\sigma },$ ${\alpha }_{\sigma }\left({\lambda }_1{\lambda }_2\right)={\alpha }_{\sigma }\left({\lambda }_1\right){\alpha }_{\sigma }\left({\lambda }_2\right).$ This implies that $\psi$ is a group homomorphism. Moreover, by the first claim, $\psi$ is surjective with $U\left(T_2\right)$ as its kernel. □

Remark 3.

If $R_2$ is a subring, we consider $T_1=R_1$ and $T_2=R_2$ in Lemma 1 and Proposition 2.

Proposition 3.

Let R be a chain ring with invariants $p,n,r,k,k^{\prime },m.$ Then, R is very pure if and only if its $R_1$ is very pure.

Proof. 

Let $R_1$ be very pure. If $k^{\prime }=1,$ then $R=R_1,$ and this ends the proof. Now, if $k^{\prime }>1$, let ($R_0,{\pi }_1$) be an element of $T_{R_1}$ such that ${\pi }_1^{k_1}=p\beta ,$ where $\beta \in <a>.$ Hence, $\left({\pi }_1\right)=\left({\pi }^{k^{\prime }}\right)=J^{k^{\prime }}\left(R\right)$. Moreover, ${\pi }_1={\beta }_1\delta {\pi }^{k^{\prime }},$ where ${\beta }_1\in <b>$ and $\delta \in H\left(T_2\right)$ since ${\pi }^{k^{\prime }}\in Z\left(R\right).$ By Lemma 1, there exist ${\beta }_2\in <a>$ and $\zeta \in H\left(T_1\right)$ such that ${\beta }_1=N_{\sigma }\left({\beta }_2\right)$ and $\delta =N_{\sigma }\left(\zeta \right).$ Now, let $\theta ={\beta }_2\zeta \pi ,$ then it is easy to verify that $(R_0,\theta )$ is an element of $T_R$ and ${\theta }^k=p\beta .$ Therefore, R is very pure. The converse is trivial. □

Corollary 1

([9]). A chain ring R with $p\nmid k$ is very pure.

Corollary 2.

If R is a chain ring with $p\nmid k$ and $m>k+1,$ then $R_2$ is a subring.

Remark 4.

If σ can be extended to an automorphism ψ of R fixing ${\pi }^{k^{\prime }},$ then $\psi \left({\pi }^k\right)={\pi }^k,$ and thus ${\pi }^k\in R_2,$ i.e., $R_2$ is a subring. Conversely, if $R_2$ is a subring, then it is clear that ${\alpha }_{\sigma }$ is the required automorphism.

Proposition 4.

Assume that R is a very pure chain ring with invariants $p,n,r,k,k^{\prime },m$. Then, σ can be extended to an automorphism of R if and only if $\sigma \left(\beta \right){\beta }^{-1}\in <b^e>,$ where $e=(k_1,p^{r_0}-1).$

Proof. 

Note that if $m>k+1,$ then $R_2$ is a subring. Thus, the proof is obvious by Remark 4. Assume that $m=k+1$ and $\varphi$ is an extension of $\sigma$ to $R.$ Note that $\varphi \left(R_1\right)=\xi R_1{\xi }^{-1}$ for some $\xi \in U\left(R\right).$ Let $\psi =\varphi {\varphi }^{\prime },$ where ${\varphi }^{\prime }$ is the conjugation by ${\xi }^{-1}.$ Then, $\psi \left(R_1\right)=R_1$ and $\psi \left(\pi \right)=\alpha \zeta \pi ,$ where $\alpha \in <a>$ and $\zeta \in H\left(R_1\right).$ It follows that,

$$\begin{array}{cc}\hfill p\sigma \left(\beta \right)=\psi \left({\pi }^k\right)& =\psi {\left(\pi \right)}^k\hfill \\ \hfill & ={\left(\alpha \zeta \pi \right)}^k\hfill \\ \hfill & =N_{\sigma }^{k_1}\left(\alpha \right)N_{\sigma }^{k_1}\left(\zeta \right){\pi }^k\hfill \\ \hfill & =p{N}_{\sigma }^{k_1}\left(\alpha \right)\beta N_{\sigma }^{k_1}\left(\zeta \right).\hfill \end{array}$$

This implies $\sigma \left(\beta \right){\beta }^{-1}\in <N_{\sigma }^{k_1}\left(\alpha \right)>;$ consequently, $\sigma \left(\beta \right){\beta }^{-1}\in <b^e>,$ where $e=(k_1,p^{r_0}-1).$ For the other direction, consider the correspondence $\psi ,$ defined as:

$$\psi (\sum _{i=0}^{k-1}{u}_i{\pi }^i)=\sum _{i=0}^{k-1}\sigma \left(u_i\right){\left(\alpha \pi \right)}^i,$$

(17)

where $\alpha$ is an element of $<a>$ with $\sigma \left(\beta \right){\beta }^{-1}\in <N_{\sigma }{\left(\alpha \right)}^{k_1}>$. Clearly, $\psi$ is an automorphism which is an extension of $\sigma$ to $R.$ □

Remark 5.

In the light of Lemma 1 and Proposition 2,

$$\begin{array}{cc}\hfill & N_{\sigma }(<a>)=<b>,\hfill \\ \hfill & N_{\sigma }\left(H\left(T_1\right)\right)=H\left(T_2\right).\hfill \end{array}$$

Let ${\alpha }_1,\cdots ,{\alpha }_{r_0},{\alpha }_{r_0+1}\cdots ,{\alpha }_r$ be a representative system in $T_1$ for a basis of $F=GF\left(p^r\right)$ over ${\mathbb{Z}}_p$ such that ${\alpha }_1,\cdots ,{\alpha }_{r_0}$ is a basis of $K=GF\left(p^{r_0}\right)$ over ${\mathbb{Z}}_p.$ Then, ${\{1+{\alpha }_i{\pi }^{s{k}^{\prime }}\}}_{1\le i\le r}$ are generators of $U_s\left(T_1\right),$ and ${\{1+{\alpha }_i{\pi }^{s{k}^{\prime }}\}}_{1\le i\le r_0}$ are generators of $U_s\left(T_2\right),$ where $1\le s\le m_1$ [16]. In addition,

$$H\left(T_1\right)=H\left(T_2\right)\otimes ker{N}_{\sigma }.$$

(18)

Hence, we can consider $1+{\alpha }_i{\pi }^{s{k}^{\prime }},$ $r_0+1\le i\le r$ as generators of

$$U_s\left(ker{N}_{\sigma }\right)=U_s\left(T_1\right)/U_s\left(T_2\right).$$

(19)

Lemma 2.

For $1\le s\le m_1-1,$ let $N_{\sigma s}$ be the restriction of $N_{\sigma }$ on $H_s\left(T_1\right).$ Then,

(i)

$N_{\sigma }\left(H_s\left(T_1\right)\right)=H_s\left(T_2\right).$

(ii)

$N_{\sigma s}$ is a surjective homomorphism, defined from $H_s\left(T_1\right)$ into $H_s\left(T_2\right).$ Moreover,

$$ker{N}_{\sigma s}=ker{N}_{\sigma }\cap H_s\left(T_1\right)=H_s\left(ker{N}_{\sigma }\right).$$

(20)

Proof. 

(i) It is clear that $N_{\sigma }\left(H_s\left(T_1\right)\right)\subseteq H_s\left(T_2\right),$ and thus by Proposition 2 and Remark 5, we have

$$N_{\sigma }^{-1}\left(H_s\left(T_2\right)\right)=L\otimes ker{N}_{\sigma },$$

where L is a subgroup of $H_s\left(T_1\right).$ Moreover, if we take the restriction $\chi$ of $N_{\sigma }$ on $L\otimes ker{N}_{\sigma }$, then it follows that $L\otimes ker{N}_{\sigma }/ker\chi \cong Im\chi .$ As $ker\chi =ker{N}_{\sigma },$ then $Im\chi =L.$ Furthermore, since $\chi$ is surjective, $L=H_s\left(T_2\right).$ Thus, (i) is proved. For (ii), the result follows from (i) and Remark 5. □

Proposition 5.

Let R be a chain ring with invariants $p,n,r,k,k^{\prime },m.$ Then, ϕ is an automorphism of R if and only if

$$\varphi (\sum _{i=0}^{k-1}{u}_i{\pi }^i)=\zeta \sum _{i=0}^{k-1}\mu \left(u_i\right){\left(\alpha w\pi \right)}^i{\zeta }^{-1},$$

(21)

where μ is an automorphism of $R_0,$ $\zeta \in U\left(R\right),$ $\alpha \in <a>$ and $w\in H\left(R_1\right)$ such that

$$\begin{array}{cc}\hfill & {N_{\sigma }\left(\alpha \right)}^{k_1}=\mu \left(\beta \right){\beta }^{-1},\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & p{N_{\sigma }\left(w\right)}^{k_1}=p\varphi \left(h\right)h^{-1}.\hfill \end{array}$$

(23)

Proof. 

Let $\varphi \in Aut\left(R\right),$ then $\varphi \left(R_1\right)$ is the centralizer of $\varphi \left(R_0\right)$ in $R.$ Hence, there exists $\zeta \in U\left(R\right)$ such that $\varphi \left(R_1\right)=\zeta R_1{\zeta }^{-1}.$ Let $\psi$ be the composition of the conjugation by ${\zeta }^{-1}$ and $\varphi .$ It follows that $\psi \left(R_1\right)=R_1.$ Then, $\psi \left(\pi \right)=\lambda \pi ,$ and this means that $\varphi \left(\pi \right)=\zeta \lambda \pi {\zeta }^{-1},$ where $\lambda \in U\left(R_1\right)$. Note that $\lambda =\alpha w,$ where $\alpha \in <a>$ and $w\in H\left(R_1\right).$ If ${\pi }^k=p\beta h,$ where $\beta \in <a>$ and $h\in H\left(R_1\right),$ then as in the proof of Proposition 4,

$$\begin{array}{cc}\hfill p\mu \left(\beta \right)\varphi \left(h\right)=\varphi \left({\pi }^k\right)=\varphi {\left(\pi \right)}^k& ={\left(\alpha w\pi \right)}^k\hfill \\ \hfill & ={N_{\sigma }\left(\alpha \right)}^{k_1}{N_{\sigma }\left(w\right)}^{k_1}{\pi }^k\hfill \\ \hfill & =p\beta {N_{\sigma }\left(\alpha \right)}^{k_1}h{N_{\sigma }\left(w\right)}^{k_1},\hfill \end{array}$$

where $\mu$ is the restriction of $\varphi$ on $R_0.$ Thus, ${N_{\sigma }\left(\alpha \right)}^{k_1}=\mu \left(\beta \right){\beta }^{-1}$ and $p{N_{\sigma }\left(w\right)}^{k_1}=p\varphi \left(h\right)h^{-1}.$ Moreover, $\varphi ={\psi }_{\zeta }{\varphi }_{\lambda }^{\mu }.$ Conversely, if $\varphi$ is defined as in (21), then, it is clear that $\varphi$ is an automorphism of R if and only if $\varphi \left({\pi }^k\right)=\varphi {\left(\pi \right)}^k.$ However, the conditions in Equation (22) and Equation (23) guarantee $\varphi \left({\pi }^k\right)=\varphi {\left(\pi \right)}^k.$ Thus, $\varphi$ is an automorphism of $R.$ □

Corollary 3.

If R is a chain ring. Then, $(R_0,\theta )\in T_R$ if and only if there is ${\varphi }_{\lambda }^{\mu }\in Aut\left(R\right)$ such that ${\varphi }_{\lambda }^{\mu }\left(\pi \right)=\theta .$

Notation 1.

Denote $T_1=R_1/J^{m_1-1}\left(R_1\right)$ and $T_2=Z\left(R\right)/\Omega .$ Let

$$\begin{array}{cc}\hfill & G_0\left(R\right)=\{{\varphi }_{\alpha }^{\mu }\in Aut\left(R\right):\alpha \in <a>,\mu \in Aut\left(R_0\right)\},\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill & G_1\left(R\right)=\{{\varphi }_w\in Aut\left(R\right):w\in H\left(T_1\right)\},\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill & G_2\left(R\right)=\{{\varphi }_w\in Aut\left(R\right):w\in H\left(T_2\right)\}.\hfill \end{array}$$

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Theorem 1.

Let R be a chain ring with invariants $p,n,r,k,k^{\prime },m.$ Then,

$$Aut\left(R\right)=\left\{\begin{array}{cc}Inn\left(R\right)⋊Aut\left(R_2\right),\hfill & \mathrm{if}{R}_2\mathrm{is}\mathrm{a}\mathrm{subring},\hfill \\ (Inn\left(R\right)⋊Aut\left(R_1\right))/Au{t}_{R_0}\left(R\right),\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

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Proof. 

(i) If $R_2$ is a subring, then ${\pi }^k=p\beta h\in R_2$ (Proposition 1). As ${\varphi }_{w_1}={\varphi }_{w_2}$ if and only if $w_1=w_2\mathrm{mod}{H}_{m_1-1}\left(R_1\right).$ Then, we can only consider ${\varphi }_w\in G_1\left(R\right).$ Moreover, since $H\left(T_1\right)=ker{N}_{\sigma }\times H\left(T_2\right)$ (Lemma 2), $w=w_1{w}_2,$ where $w_1\in ker{N}_{\sigma }$ and $w_2\in H\left(T_2\right).$ This follows by Proposition 2, $w_1={\alpha }_{\sigma }\left(\xi \right){\xi }^{-1},$ for some $\xi \in H\left(T_1\right).$ Note that ${\varphi }_w\left(\pi \right)={\varphi }_{w_1{w}_2}\left(\pi \right)={\psi }_{{\xi }^{-1}}{\varphi }_{w_2}\left(\pi \right),$ i.e., ${\varphi }_w={\psi }_{{\xi }^{-1}}{\varphi }_{w_2},$ where ${\psi }_{{\xi }^{-1}}\in Inn\left(R\right)$. Thus by Proposition 5, the set of all automorphisms of the form ${\varphi }_w,$ $w\in H\left(T_2\right),$ is $G_2\left(R_2\right).$ Similarly, if $\alpha \in <a>\cap ker{N}_{\sigma }$ and ${\varphi }_{\alpha }^{\mu }\in Aut\left(R\right),$ then ${\varphi }_{\alpha }^{\mu }={\varphi }_{\alpha }^{id}{\varphi }_1^{\mu },$ where ${\varphi }_{\alpha }^{id}\in Inn\left(R\right).$ Hence, $G_0\left(R\right)=G_{\sigma }\times G_0\left(R_2\right),$ where $G_{\sigma }$ is the subgroup of $Inn\left(R\right)$ contains all automorphisms ${\varphi }_{\alpha }^{id},$ $\alpha \in ker{N}_{\sigma }.$ Since $Aut\left(R_2\right)=G_2\left(R_2\right)\times G_0\left(R_2\right),$ then by Proposition 5,

$$Aut\left(R\right)=Inn\left(R\right)\times Aut\left(R_2\right).$$

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Furthermore, note that if $\varphi \in Inn\left(R\right)\cap Aut\left(R_2\right),$ then $\varphi ={\psi }_{\xi },$ where $\xi \in U\left(R\right).$ Moreover, $\varphi \in Aut\left(R_2\right),$ then $\varphi ={\varphi }_{\lambda }^{\mu },$ where $\mu \in Aut\left(S\right)$ and $\lambda \in U\left(R_2\right).$ Thus, ${\psi }_{\xi }={\varphi }_{\lambda }^{\mu },$ and hence $\mu =id$ and $\lambda =\xi {\alpha }_{\sigma }\left({\xi }^{-1}\right)=\sigma \left(\alpha \right){\alpha }^{-1}{\alpha }_{\sigma }\left(w\right)w^{-1}$ is an element of $ker{N}_{\sigma },$ $\alpha \in <a>,$ $w\in H\left(R_1\right)$ satisfying Equations (22) and (23), respectively. Since $ker{N}_{\sigma }\cap H\left(T_2\right)=<1>$ (Remark 5), then $\lambda =1.$ This yields $Inn\left(R\right)\cap Aut\left(R_2\right)=<id>,$ and this ends the proof. (ii) Assume that $R_2$ is not a subring. Note that, in this case, ${\pi }^k\notin R_2.$ By a similar argument, one can check that $G_1\left(R\right)=G_1\left(R_1\right).$ Furthermore, $G_0\left(R\right)=G_0\left(R_1\right).$ Hence, $Aut\left(R\right)=Inn\left(R\right)\times Aut\left(R_1\right).$ Moreover, $Inn\left(R\right)\cap Aut\left(R_1\right)=Au{t}_{R_0}\left(R\right).$ Therefore, the result follows. □

Remark 6.

By Theorem 1, it is enough to determine $Aut\left(R\right)$ when R is a commutative chain ring, i.e., $k^{\prime }=1$.

In what follows, unless otherwise mentioned, R is a very pure chain ring with ${\pi }^k=p\beta$ and $\beta =a^s.$

Proposition 6.

If R is a commutative chain ring with invariants $p,n,r,k,m.$ Then, $\varphi \in Aut\left(R\right)$ if and only if

$$\varphi (\sum _{i=0}^{k-1}{u}_i{\pi }^i)=\sum _{i=0}^{k-1}\mu \left(u_i\right){\left(\alpha w\pi \right)}^i,$$

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where $\alpha \in <a>,w\in H$ satisfying:

$$\begin{array}{cc}\hfill & {\alpha }^k=\sigma \left(\beta \right){\beta }^{-1},\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill & w^k=1\mathrm{mod}{\pi }^{m-k},\hfill \end{array}$$

(31)

for some $\mu \in Aut\left(R_0\right).$

Proof. 

The proof is direct from that of Proposition 5 with $k^{\prime }=1$ and $h=1.$ □

Remark 7.

From Proposition 6, it is obvious that if $\varphi \in Aut\left(R\right),$ then $\varphi ={\varphi }_w{\varphi }_{\alpha }^{\mu }.$

Remark 8.

If we consider a commutative chain ring $R^{\prime }$ with invariants $p,n-1,r,k,m,$ i.e., $R^{\prime }=R/J^{m-k}\left(R\right).$ Thus, we can write Equation (31) as $w^k=1.$

Next, we focus on automorphisms ${\varphi }_{\alpha }^{\mu }\in G_0\left(R\right)$ which satisfy Equation (30), where $G_0\left(R\right)$ is defined in (24). Assume that $\alpha =a^z,$ we denote such ${\varphi }_{\alpha }^{\mu }$ by ${\varphi }_z^{\mu },$ and we write $G_0$ instead of $G_0\left(R\right).$ Let

$$\left\{\begin{array}{c}d=(p^r-1,s),c=\frac{p^r-1}{d},\hfill \\ q=(d,k),b=\frac{d}{q},\hfill \\ e=(\frac{p^r-1}{b},k).\hfill \end{array}\right.$$

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Lemma 3.

Let R be a commutative chain ring with invariants $p,n,r,k,m.$

(i) Let ${\varphi }_z^{\mu }$ be an automorphism of $R.$ Then, $a^z=a^{bi}$ for some $i.$

(ii) Let i be any integer. There exists an automorphism ${\varphi }_{bi}^{\mu }$ of R if and only if there exists a non-negative integer f such that $(p^r-1)\mid s(p^f-1)-bik.$

Proof .

(i) If $u\in R_0,$ then ${\varphi }_z^{\mu }\left(u\right)=\mu \left(u\right)=u^{p^f}$ for some $0\le f\le r-1.$ Then, ${\varphi }_z^{\mu }\left({\pi }^k\right)={\pi }^k{a}^{zk}$ implies,

$$p{a}^{s{p}^f}={\varphi }_z^{\mu }\left(p{a}^s\right)={\varphi }_z^{\mu }\left({\pi }^k\right)=p{a}^s{a}^{zk}.$$

Moreover, $a^{zk}\in <a^s>=<a^d>.$ As $(k,d)=q,$ we obtain $a^{az}\in <a^d>,$ and as a result $b\mid z.$

(ii) Assume that such ${\varphi }_{bi}^{\mu }$ exists. Now, ${\varphi }_{bi}^{\mu }\left(u\right)=u^{p^f}$ for some $f\ge 0,$ where $u\in R_0.$ Then, ${\varphi }_{bi}^{\mu }\left({\pi }^k\right)={\pi }^k{a}^{bik}$ leads to

$$p{a}^{s{p}^f}=p{a}^{s+bik}.$$

This means, $a^{s(p^f-1)-bik}=1,$ i.e., $(p^r-1)\mid s(p^f-1)-bik.$ Conversely, let such f exist. Then, we have an automorphism $\mu$ of $R_0$ such that $\mu \left(u\right)=u^{p^f},$ where $u\in R_0.$ Consider ${\pi }_1=\pi a^{bi},$ then ${\pi }_1^k=p\mu \left(a^s\right).$ This gives an automorphism ${\varphi }_{bi}^{\mu }$ that extends $\mu$ and for which ${\varphi }_{bi}^{\mu }\left(\pi \right)={\pi }_1.$ □

Lemma 4.

Let R be a commutative chain ring, then $G_0$ is a subgroup of $Aut\left(R\right)$ of order $\frac{re}{\tau \left(c\right)},$ where $\tau \left(c\right)$ is the multiplicative order of p modulo $c.$

Proof. 

Let ${\varphi }_{z_1}^{\mu }$ and ${\varphi }_{z_2}^{\eta },$ where $\mu ,\eta \in Aut\left(R_0\right).$ Then,

$${\varphi }_{z_1}^{\mu }{\varphi }_{z_2}^{\eta }\left(\pi \right)={\varphi }_{z_1}^{\mu }\left(a^{z_2}\pi \right)=a^{z_1}\mu \left(a^{z_2}\right)\pi =a^{z_1+p^i{z}_2}\pi ={\varphi }_{z_1+z_2{p}^i}^{\mu \eta }\left(\pi \right),$$

where $\mu ={\rho }^i.$ In other words, ${\varphi }_{z_1}^{\mu }{\varphi }_{z_2}^{\eta }={\varphi }_{z_1+z_2{p}^i}^{\mu \eta }.$ Similarly, one can easily find that,

$${\left({\varphi }_z^{\mu }\right)}^{-1}={\varphi }_{-z{p}^i}^{\eta },$$

where $\eta ={\mu }^{-1}={\rho }^i,$ for some i positive integer $1\le i\le r.$ Thus, $G_0$ is a subgroup of $Aut\left(R\right).$ Next, we compute the order of $G_0.$ Note that from Lemma 3, $b\mid z.$ Fix $\sigma ,$ if ${\varphi }_{z_2}^{\mu }\in G_0,$ by (30), ${\varphi }_{z_2}^{\mu }\in G_0$ if and only if $a^{z_1-z_2}\in A,$ where A is the set of all zeros of $x^e-1$ in the group generated by $a^{(p^r-1)/b}.$ Since there are exactly e different zeros of $x^e-1,$ hence there are e of automorphisms of the type ${\varphi }_z^{\mu }.$ On the other hand, let z be fixed. Equation (30) gives $\mu \left(\beta \right){\beta }^{-1}=1$ mod $<a^d>,$ i.e., $\mu \left(\beta \right)=\beta .$ Thus, it is enough to find the stabilizers of $\beta$ modulo $<a^d>$ when $Aut\left(R_0\right)$ acts on $<a>$ in the usual way. It is well known that

$$\mid stab\left(\beta \right)\mid \mid orb\left(\beta \right)\mid =\mid Aut\left(R_0\right)\mid .$$

Moreover, since orb $\left(\beta \right)=\{{\rho }^i\left(\beta \right):1\le i\le r\};$ thus, $orb\left(\beta \right)=\{{\rho }^i\left(\beta \right):1\le i\le \tau \left(c\right)\},$ where $\tau \left(c\right)$ is the multiplicative order of p modulo $c.$ Therefore,

$$\mid stab\left(\beta \right)\mid =\frac{r}{\tau \left(c\right)},$$

which follows that $\mid G_0\mid =\frac{re}{\tau \left(c\right)}.$ □

Proposition 7.

Let R be a commutative chain ring with invariants $p,n,r,k,m.$ Then, $G_0$ is a solvable group. Moreover, if $d=1,$ then $G_0$ is cyclic.

Proof. 

Let $G_{11}=<{\varphi }_{z_0}^{id}>,$ where $z_0=\frac{p^r-1}{e}.$ Then, $G_{11}$ is clearly a normal subgroup of $G_0$. Moreover, $G_0/G_{11}$ is abelian, and hence $G_0$ is solvable. Now, if $d=1,$ then $c=p^r-1.$ Note that $a^{z_1+z_2{p}^{\tau \left(c\right)}}=a^{z_1+z_2+z_{2}ci}=a^{z_1+z_2}$ consequently,

$${\varphi }_{z_1}^{\mu }{\varphi }_{z_2}^{\eta }={\varphi }_{z_1+z_2}^{\mu \eta }.$$

Moreover, since ${\left(a^{p^r-1/e}\right)}^k={\rho }^{\tau \left(c\right)}\left(\beta \right){\beta }^{-1}=1,$ then ${\varphi }_{z_0}^{\mu }$ lies in $G_0,$ where $\mu ={\rho }^{\tau \left(c\right)}.$ It is clear that the cyclic subgroup of $G_0$ generated by $\chi ={\varphi }_{z_0}^{\mu }$ is of order $\frac{re}{\tau \left(c\right)},$ and then by Lemma 4, $G_0$ is cyclic generated by $\chi .$ However, in this case, $\tau \left(c\right)=r,$ which means $\chi ={\varphi }_{z_0}^{id}$ and the order of $G_0=G_{11}$ is $e.$ □

For every $s\in P_m,$ we define $\nu \left(s\right)$ the least positive integer satisfying $j^{\nu \left(s\right)}\left(s\right)=m.$ Also we write $k=(p-1)u+v,$ $0\le v<p-1.$

Lemma 5.

Let R be a commutative chain ring, and let $s_0$ be the least positive integer such that $\nu \left(s_0\right)\le l,$ where $k=k_0{p}^l$ and $(k_0,p)=1.$ Then,

$$s_0=\left\{\begin{array}{cc}(n-l-1)k+t,\hfill & \mathrm{if}l<n-1,\hfill \\ \max \{t,u\},\hfill & \mathrm{if}l=n-1,\hfill \\ \left\{\begin{array}{cc}{q}_0,\hfill & \mathrm{if}t\ge u,\hfill \\ \lceil \frac{u}{p^{l-n-1}}\rceil ,\hfill & \mathrm{if}t<u,\hfill \end{array}\right.\left(l>n-1\right)\hfill \end{array}\right.$$

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where,

$$q_0=\left\{\begin{array}{cc}{s}_0^{\prime },\hfill & \mathrm{if}{s}_0^{\prime }{p}^{l-n-1}\ge t,\hfill \\ s_0^{\prime }p,\hfill & \mathrm{if}{s}_0^{\prime }{p}^{l-n-1}<t,\hfill \end{array}\right.s_0^{\prime }=\left\{\begin{array}{cc}\lceil \frac{u}{p^{l-n-1}}\rceil ,\hfill & \mathrm{if}{p}^{l-n-1}\nmid u,\hfill \\ \frac{u}{p^{l-n-1}}+1,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Proof. 

First if $l<n-1,$ then $j^l((n-l-1)k+t)=m$ which means that $s_0=(n-l-1)k+t.$ The case when $l=n-1,$ note that if $t\ge u,$ then $j^l\left(t\right)=m,$ and thus $s_0=t.$ Now, when $t<u,$ $j^l\left(u\right)=m$ take $s_0=u$ since $p(u-1)<k+t,$ i.e., $\nu (u-1)>l.$ Next, let $l>n-1.$ It suffices to find $s_0^{\prime }\in P_m$ such that $j^{l-n-2}\left(s_0^{\prime }\right)\le u<j^{l-n-1}\left(s_0^{\prime }\right),$ i.e., $s_0^{\prime }{p}^{l-n-2}\le u<s_0^{\prime }{p}^{l-n-1}.$ This implies $\frac{u}{p^{l-n-1}}<s_0^{\prime }\le \frac{u}{p^{l-n-2}}.$ As $s_0$ is the least, then $s_0^{\prime }=\lceil \frac{u}{p^{l-n-1}}\rceil$ if $p^{l-n-1}\nmid u$ and $s_0^{\prime }=\frac{u}{p^{l-n-1}}+1$ otherwise. Consider two cases. If $t<u,$ then obviously $s_0=\lceil \frac{u}{p^{l-n-1}}\rceil$ is the required number. For the second case $t\ge u,$ observe that if $j^{l-n-1}\left(s_0^{\prime }\right)\ge t,$ then clearly $s_0=s_0^{\prime },$ and if $j^{l-n-1}\left(s_0^{\prime }\right)<t,$ take $s_0=j\left(s_0^{\prime }\right)=p{s}_0^{\prime }.$ □

Remark 9.

Let R be a commutative chain ring. Denote $L\left(R\right)=\{w\in H\left(R\right):w^k=1mod{\pi }^{m-k}\}.$ Note that if $p\nmid k,$ i.e., $l=0$ or $m=k+1,$ then by Lemma 5 applied to $R/\left({\pi }^{m-k}\right),$ $s_0=m-k$ and this means, $L\left(R\right)=H_{m-k}.$ The structure of the subgroup of $H,$ $H_{s_0},$ can be obtained from [16] via the j-diagram:

$$H_{s_0}>H_{s_0+1}>\cdots >H_m=1.$$

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In addition, if $p\nmid k$ or R is a complete chain ring, then $L\left(R\right)=H_{s_0}\left(R\right).$ The case when $m=k+1,$ R is complete [16].

Proposition 8.

If R is a commutative chain ring with invariants $p,n,r,k,m.$ Let $G_1=\{{\varphi }_w:w\in L\left(R\right)\}.$ Then, $G_1$ is a normal subgroup of $Aut\left(R\right).$ In particular, if $p\nmid k$ or R is complete, $G_1\cong E_{m-s_0-1}.$

Proof. 

It is clear that $G_1$ is subgroup of $Aut\left(R\right).$ Every element $\varphi$ of $Aut\left(R\right)$ can be written as $\varphi ={\varphi }_w{\varphi }_z^{\mu },$ where ${\varphi }_z^{\mu }\in G_0$ and ${\varphi }_w\in G_1,$ i.e.,

$$Aut\left(R\right)=G_1\times G_0.$$

Moreover, if ${\varphi }_{\lambda }$ in $G_1$ and $\varphi ={\varphi }_z^{\mu },$ then,

$${\varphi }^{-1}{\varphi }_{\lambda }\varphi ={\left({\varphi }_z^{\mu }\right)}^{-1}{\varphi }_{\lambda }{\varphi }_z^{\mu }={\varphi }_{\psi \left(\lambda \right)},$$

where $\psi ={\left({\varphi }_z^{\mu }\right)}^{-1}.$ Consequently, $G_1$ is a normal subgroup of $Aut\left(R\right).$ Also observe that if $w_1,$ $w_2\in L\left(R\right),$ then ${\varphi }_{w_1}={\varphi }_{w_2}$ if and only if $\pi w_1=\pi w_2$ if and only if $1-w_1{w}_2\in \left({\pi }^{m-1}\right)$ if and only if $w_1{H}_{m-1}=w_2{H}_{m-1}.$ Hence, the order of $G_1$ is $\frac{\mid L\left(R\right)\mid }{p^r}.$ Moreover, if $p\nmid k$ or R is complete, then by Lemma 5 and Remark 9, $L\left(R\right)=H_{s_0}.$ This implies that,

$$\mid G_1\mid =p^{r(m-s_0-1)}.$$

Furthermore, if we define a map g from $G_1$ into $E_{m-s_0-1}$ by $g\left({\varphi }_w\right)=\left[a_{iq}\right],$ where $a_{iq}={\sum }_{e+f=i}{a}_{e,1}{a}_{f,q-1},$ $w=1+{\sum }_{i=1}^{m-s_0-1}{a}_{i+1,1}{\pi }^i$ and $a_{i+1,1}\in F.$ It can be verified that g is a group isomorphism. □

Corollary 4.

Assume that R is as in Proposition 8. Then, $(p,k)=1$ or $m-1=k$ if and only if $G_1=\{{\varphi }_w:w\in H_{m-k}\}.$ Moreover, $G_1\cong E_{k-1}$ and, in this case,

$$\mid G_1\mid =p^{r(k-1)}.$$

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Theorem 2.

Let R be a commutative chain ring, which is very pure with invariants $p,n,r,k,m$. Then,

$$Aut\left(R\right)=G_1⋊G_0.$$

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In particular, if R is complete or $(p,k)=1,$

$$Aut\left(R\right)\cong E_{m-s_0-1}⋊G_0.$$

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Proof. 

First, note that from Proposition 8, $Aut\left(R\right)=G_1\times G_0.$ Now, suppose that $\varphi \in G_1\cap G_0,$ then as $\varphi \in G_1,$ $\varphi \left(u\right)=u,$ where $u\in R_0.$ Thus, $\varphi ={\varphi }_z^{id},$ i.e., $\varphi \left(\pi \right)=\pi a^z={\pi }^{bi}$ for some $i.$ Moreover, $\varphi \in G_1$ yields that $\varphi \left(\pi \right)=\pi w,$ $w\in H.$ This implies that $\varphi =id,$ and thus $G_1\cap G_0=<id>$. The last assertion follows from Proposition 8. □

Corollary 5.

If R is as in Theorem 2. Then, $Au{t}_{R_0}\left(R\right)=G_1⋊<\chi >,$ where $\chi ={\varphi }_{z_0}^{id}.$

For the following results, denote $G_{qi}=G_q\left(R_i\right),$ $q=0,1$ and $i=1,2.$

Corollary 6.

Let R be a very pure chain ring with invariants $p,n,r,k,k^{\prime },m.$ Then,

$$Aut\left(R\right)=\left\{\begin{array}{cc}Inn\left(R\right)⋊G_{12}⋊G_{02},\hfill & \mathrm{if}{R}_2\mathrm{is}\mathrm{a}\mathrm{subring},\hfill \\ (Inn\left(R\right)⋊G_{11}⋊G_{01})/Au{t}_{R_0}\left(R\right),\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

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In particular, if $(p,k)=1$ or R is complete,

$$Aut\left(R\right)\cong \left\{\begin{array}{cc}Inn\left(R\right)⋊E_{m_1-s_0-1}⋊G_{02},\hfill & \mathrm{if}{R}_2\mathrm{is}\mathrm{a}\mathrm{subring},\hfill \\ (Inn\left(R\right)⋊E_{m_1-s_0-1}⋊G_{01})/Au{t}_{R_0}\left(R\right),\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

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Proof. 

The proof is just a direct application of Theorems 1 and 2. □

Corollary 7.

Let R be a chain ring with invariants $p,n,r,k,k^{\prime },m$ such that $(p,k)=1$ and $m>k+1.$ If $d=1,$ then

$$Aut\left(R\right)\cong Inn\left(R\right)⋊E_{k_1-1}⋊C_e,$$

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where $C_e$ is a cyclic group of order e.

Remark 10.

Note that by Remark 9, Corollarys 6 and 7, the structure of $Aut\left(R\right)$ is strongly dependent on the invariants $p,n,r,k,k^{\prime },m$ and on the associated j-diagram.

Next, we introduce some results on the subgroup $Au{t}_{R_0}\left(R\right).$

Proposition 9.

Let R be a commutative chain ring with invariants $p,n,r,k,m.$ Then, $\mid Au{t}_{R_0}\left(R\right)\mid =p^{(k-1)r}$ if and only if $(k,p)=1$ and $(k,p^r-1)=1$ or $m-1=k$ and $(k,p^r-1)=1.$

Proof. 

From Corollary 5, $Au{t}_{R_0}\left(R\right)=G_1⋊<{\varphi }_{z_0}^{id}>$ and $\mid G_1\mid =\frac{\mid L\left(R\right)\mid }{p^r}$$\mid <{\varphi }_{z_0}^{id}>\mid =e.$ Thus by Corollary 4, $\mid Au{t}_{R_0}\left(R\right)\mid =e{p}^{(k-1)r}$ if and only if $G_1=\{{\varphi }_w:w\in H_{m-k}\}$ if and only if $(p,k)=1$ or $m-1=k.$ Moreover, $\mid Au{t}_{R_0}\left(R\right)\mid =p^{(k-1)r}$ if and only if $e=1$ if and only if $(k,p^r-1)=1$ Therefore, the proof follows. □

Remark 11.

Consider a complete chain ring R with $k^{\prime }>1,$ i.e., R is non-commutative. Now, if $w=1+{\sum }_{i=1}^{m-s_0}{a}_{i+1,1}{\pi }^i$ and $a_{i+1,1}\in F.$ Let $D_1$ be the subgroup of $E_{m-s_0-1}$ contains all matrices of the form $\left[b_{iq}\right],$ where $b_{iq}=\sigma \left(a_{iq}\right)a_{iq}^{-1}.$ Define the mapping g as $g\left({\varphi }_{\lambda }^{id}\right)=\left[b_{iq}\right]\delta ,$ where $\lambda =\sigma \left(\alpha \right){\alpha }^{-1}{\eta }_{\sigma }\left(w\right)w^{-1}\in ker{N}_{\sigma }$ and $\delta =\sigma \left(\alpha \right){\alpha }^{-1}.$ Then, it is not hard to show that g is an isomorphism from $Au{t}_{R_0}\left(R\right)$ into $D_1⋊<{\varphi }_{z_1}^{id}>,$ where ${\varphi }_{z_1}^{id}\in G_0$ of order $(e,p^r-1/p^{r_0}-1).$

Example 1.

Let R be a commutative chain ring with invariants $p,n,r,k,m$ and with ${\pi }^{p-1}=-p$ (Example 2, [16]). Note that $-1=a^{(p^r-1)/2},$ i.e., $s=(p^r-1)/2.$ Moreover, since $(p,k)=1,$ then,

$$\mid G_1\mid =p^{r(p-2)}.$$

Also $c=2,$ and then, $\tau \left(c\right)=1.$ Hence, $\mid G_0\mid =re.$ Now, if $(p^r-1,k)=1,$ then $e=1,$ which implies,

$$Aut\left(R\right)\cong E_{k-1}⋊Aut\left(R_0\right).$$

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In addition, $Au{t}_{R_0}\left(R\right)\cong E_{k-1}.$

Remark 12.

In the case when $n=1$ ([14]), then obviously ${\varphi }_z^{\mu }={\varphi }_z{\alpha }_{\mu },$ for any $1\le z\le p^r-1$ and $\mu \in Aut\left(F\right)$ ($R_0=F$). Since ${\pi }^k=0,$ then ${\varphi }_z$ and ${\alpha }_{\mu }$ are automorphisms of $R.$ This means,

$$G_0=F^{*}⋊Aut\left(F\right).$$

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In addition, $G_1\cong E_{k_1-1}$ in general. Since $R_2$ is a subring, Corollary 6 concludes that,

$$Aut\left(R\right)\cong Inn\left(R\right)⋊E_{k_1-1}⋊F^{*}⋊Aut\left(F\right).$$

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Example 2.

Consider a finite chain ring R with invariants $p,n,r,k,k^{\prime },m$ and associated with ${\pi }^k=p.$ Then, it is clear that ${\varphi }_1^{\mu }$ is an automorphism for every $\mu \in Aut\left(R_0\right).$ This means, $G_1=<\chi >⋊Aut\left(R_0\right),$ where $\chi ={\varphi }_{z_0}^{id}.$ If $(p,k)=1$ and $m>k+1$, then,

$$Aut\left(R\right)\cong Inn\left(R\right)⋊E_{k_1-1}⋊<\chi >⋊Aut\left(R_0\right).$$

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