On Automorphisms of Chain Rings

Abstract

Suppose R is a finite chain ring with invaraints $p,n,r,k,k^{\prime },m.$ Suppose G is also the subset of all $\phi \in$ Aut$\left(R\right),$ the automorphism group of $R,$ such that $\phi \left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }},$ where $\pi$ is a generator of the maximal ideal of $R.$ It was found that G is a group that is, in some sense, the set of all symmetries of $\left\{{\pi }^{k^{\prime }}\right\}.$ The main purpose of this article is to describe the structure of $G.$ The subgroup G helps us understand the structure of Aut$\left(R\right)$ in the general case which in turn provides immediate results in classifying chain rings.

1. Introduction

An associative ring which admits a unique chain of right (left) ideals is called a chain ring. A local ring is a chain if and only if its unique maximal ideal is principal. Finite chain rings have increasingly appeared in several areas [1,2,3,4,5,6,7]. One of the most significant areas is coding theory in which chain rings are used to create more compact error-correcting codes, see for instance [8,9,10,11]. In this paper, we aim to describe the construction of automorphisms of chain rings in order to better understand them and widen their applications.

A natural class that well represents the application of chain rings is the class of Galois rings which are formed by extending ${\mathbb{Z}}_{p^n}$ using specific basic irreducible polynomials of particular degrees. Every Galois ring has the construction ${\mathbb{Z}}_{p^n}\left[x\right]/\left(g\left(x\right)\right),$ where r is the degree of $g\left(x\right),$ and they are denoted by $GR(p^n,r)$. It is known that these rings are uniquely determined by the invariants $p,n,r,$ and moreover, their automorphism groups are cyclic with rank of r (see Krull [12]). Suppose p is a prime number and R is a chain ring with radical $J\left(R\right)$ of nilpotency index $m.$ Suppose also $F=R/J\left(R\right),$ residue field of $R,$ and of order $p^r.$ There is a positive integer n such that the additive order of 1 is $p^n;$ the characteristic of R is $p^n.$ Let $R_0$ be the coefficient subring of $R,$ then $R_0$ can be represented as

$$R_0=GR(p^n,r)\cong {\mathbb{Z}}_{p^n}\left[a\right],$$

where the multiplicative order of a is $o\left(a\right)=p^r-1.$ Furthermore, if $\pi \in J\left(R\right)\setminus J^2\left(R\right)$ and $\sigma \in Aut\left(R_0\right),$ then $J\left(R\right)=\pi R$ and $\pi u=\sigma \left(u\right)\pi ,$ where $u\in R_0$. It was proved that R and $R_0$ characterize $\sigma$ and it is called the associated automorphism. We call the pair $(\pi ,\sigma )$ the distinguished basis for R with respect $R_0.$ Assume $p\in J^k\left(R\right)\setminus J^{k+1}\left(R\right),$ then

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

(1)

(as $R_0-$module). By Equation (1), ${\pi }^k=p{\sum }_{i=0}^{k-1}{u}_i{\pi }^i,$ where $u_i\in R_0$ and $u_0$ is a unit, which means that $\pi$ is a zero of an Eisenstein polynomial,

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

(2)

Another representation of R is

$$R=R_0[x,\sigma ]/(g\left(x\right),x^m).$$

(3)

If there exists an element $\beta$ in the cyclic subgroup generated by a, denoted $<a>$, such that ${\pi }^k=p\beta$, then R is referred to as a very pure chain ring. If $n>1$, then the order of $\sigma$, denoted by $k^{\prime }$, divides k and ${\sigma }^k$ is the identity map, i.e., ${\sigma }^k=\mathrm{id}$. The integers $p,n,r,k,k^{\prime }$ and m are called the invariants of $R.$ To summarize, p is the characteristic of the residue field $F=R/J,$ $p^n$ is the characteristic of $R,$ $p^r$ is the order of $F,$ k is the degree of Eisenstein polynomial, $k^{\prime }$ is the order of the associated automorphism $\sigma$ and m is the index of nilpotency of $J.$ The symmetry of invariants of various chain rings injects more choice and flexibility into the theory of ring construction. For more details on finite chain rings and automorphism groups, we refer the interested reader to [13,14,15,16,17,18].

The purpose of this article is to characterize the structure of the automorphisms of R that fix ${\pi }^{k^{\prime }}$. The decomposition of $Aut\left(R\right)$ was investigated when $n=1$ by Alkhamees in [19]. The case when $n>1,$ a special class of automorphisms was considered in [20]. Moreover, Alabiad and Alkhamees [21] investigated $Aut\left(R\right),$ where R is very pure; for instance $p\nmid k$. The result of this article not only generalizes that in [20] but also opens the door to provide a full description of the $Aut\left(R\right)$ structure as well. Moreover, the method used in this study differs from that given in [21]. We suppose that R is a chain ring not necessarily pure or commutative and $n>1,$ in contrast to that in [21], where the ring R was pure and commutative.

If G is a subgroup of $Aut\left(R\right)$ which contains all automorphisms ${\phi }_{\zeta }^{\eta },$

$${\phi }_{\zeta }^{\eta }(\sum u_i{\pi }^i)=\sum \eta \left(u_i\right){\left(\zeta \pi \right)}^i,$$

(4)

such that ${\phi }_{\zeta }^{\eta }\left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }},$ where $\zeta \in U\left(R_1\right)$ and $\eta \in Aut\left(R_0\right).$ We first established results concerning norm function and properties of ${\phi }_{\zeta }^{\eta }.$ In Theorem 2, we describe the structure of G, and we also give its order.

2. Preliminaries

We present some necessary preliminaries and state symbols used subsequently. In the remainder of this paper, we suppose R is a finite chain ring which has $p,n,r,k,k^{\prime },m$ as its invariants with distinguished pair $(\pi ,\sigma ).$ Let $R_1$ be the centralizer of $R_0,C\left(R_0\right),$ in $R,$ then [21],

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

(5)

where $k_1=\frac{k}{k^{\prime }}$ and $m_1=\lfloor \frac{m}{k^{\prime }}\rfloor +1$. By (5), the radical of $R_1,$ $J\left(R_1\right)={\pi }^{k^{\prime }}{R}_1.$ The ring $R_1$ is chain with invariants $p,n,r,k_1,m_1.$

Now, suppose $Z\left(R\right)$ is the center of R and $S=R_0^{\sigma }$ is the fixed subring by $\sigma .$ Then,

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

(6)

where $\Omega =0$ if $k^{\prime }$ does not divide $m-1$ and $\Omega =J^{m_1-1}\left(R_1\right)=J^{m-1}\left(R\right)$ otherwise. Additionally, it is evident to show that $S=GR(p^n,s)=$ ${\mathbb{Z}}_{p^n}\left[b\right],$ $b\in <a>$ with $o\left(b\right)=p^s-1,$ and $s=r/k^{\prime }.$

Let

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

(7)

It follows that $R_2=Z\left(R\right)$ when $k^{\prime }$ does not divide $m-1;$ which means that $Z\left(R\right)$ is a chain with invariants p, n, s, $k_1$, $m_1.$ If $k^{\prime }\mid (m-1),$ then clearly $Z\left(R\right)=R_2+\Omega$ which means it is not a chain subring of $R.$ Nevertheless, the quotient $Z\left(R\right)/\Omega$ is a chain ring with invariants p, n, s, $k_1$, $m_1-1.$ Observe that ${\pi }^k\in Z\left(R\right)$, i.e., ${\pi }^k$ can be written as

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

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 $u_{m-1}\in <a>$. In addition, we denote $T_R$ all pairs ($\pi ,\sigma$).

Proposition 1

([21]). Suppose R is a finite chain ring, then

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

(8)

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

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

Suppose that $k^{\prime }=1;$R is commutative. Let $U\left(R\right)$ denote the unit group of $R,$ then $U\left(R\right)=<a>\otimes H,$ where $o\left(a\right)=p^r-1$ and $H=1+J\left(R\right)$ is the p-Sylow subgroup of $U\left(R\right)$ [22]. Let $H_s\left(R\right)=1+J^s\left(R\right),$ $s\in P_m=\{1,2,\cdots ,m\}.$

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 one-to-one maps:

$$\begin{array}{cc}& {\phi }_{\lambda }^{\mu }(\sum u_i{\pi }^i)=\sum \mu \left(u_i\right){\left(\lambda \pi \right)}^i,\hfill \\ & {\psi }_{\xi }(\sum u_i{\pi }^i)=\xi \sum u_i{\pi }^i{\xi }^{-1},\hfill \\ & {\alpha }_{\sigma }(\sum u_i{\pi }^i)=\sum \sigma \left(u_i\right){\pi }^i.\hfill \end{array}$$

Remark 1.

Note that ${\phi }_{\lambda }^{\mu },$ ${\psi }_{\xi }$ and ${\alpha }_{\sigma }$ are not necessarily automorphisms when $n>1,$ see ([21], Proposition 6).

Definition 1.

Assume that $T_2$ is a subset of $T_1$ and $T_1,T_2$ are commutative chain rings such that $T_1$ is cyclic Galois over $T_2.$ Let $G=<\chi >$ be the Galois group, define $N_{\chi }\left(y\right):U\left(T_1\right)\to U\left(T_2\right)$ as:

$$\begin{array}{cc}& N_{\chi }\left(y\right)=\prod _{\eta \in G}\eta \left(y\right),\hfill \\ & tr\left(y\right)=\sum _{\eta \in G}\eta \left(y\right),\hfill \end{array}$$

where $N_{\chi }$ and $tr$ are called the norm and trace function, respectively.

Remark 2.

Let $T_1=R_1/J^{m_1-1}\left(R_1\right)$ and $T_2=Z\left(R\right)/\Omega .$ In light of Lemma 1 and Proposition 2 [21],

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

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

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

(9)

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

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

(10)

3. The Main Results

In this section, we investigate a subgroup G of $Aut\left(R\right)$ which contains all automorphisms of R of the form ${\phi }_{\zeta }^{\eta }$ such that ${\phi }_{\zeta }^{\eta }\left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }},$ where $\zeta \in U\left(R_1\right)$ and $\eta \in Aut\left(R_0\right).$

Lemma 1.

(i) The pair $(\theta ,\eta )$ is distinguished if and only if $\theta =u\pi$ with u is a unit of $R_1$.

(ii) 

If $u\in U\left(R\right)$ such that $u^{-1}{R}_{0}u=R_0,$ then $u\in Z\left(R_0\right).$

Proof. 

(i) If $k^{\prime }=1,$ then R is commutative, and thus the result follows directly. Now, assume that $k^{\prime }>1$ and let $(\theta ,\eta )\in T_R.$ Then, $\theta ={\sum }_{i=0}^{k-1}{u}_i{\pi }^i,$ where $u_i\in R_0$ such that $u_0\in \left(p\right).$ If $u_1\in J,$ we have $\theta \in J^2,$ which leads to contradiction. This means $u_1$ is a unit of $R_0.$ For any $u\in R_0,$ $\theta u=\eta \left(u\right)\theta ,$ where $\eta$ is the associated automorphism of the pair $(\theta ,\eta ).$ So,

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

for $0\le i\le k-1.$ Since $u_1$ is a unit, then $\eta \left(u\right)-\sigma \left(u\right)\in J^{m-1}$ for every $u\in R_0,$ and thus $\sigma =\eta .$ If for some $i,$ $u_i{\pi }^i\ne 0,$ then $\sigma \left(u\right)-{\sigma }^i\left(u\right)\in J$ which leads to $\sigma ={\sigma }^i,$ $i>0$ and $k^{\prime }\mid (i-1).$ Thus, $\theta ={\sum }_j{u}_j{\pi }^{k^{\prime }j}\pi ,$ where $u_j\in R_0.$

(ii) Let $w\in R,$ there is $w^{\prime }\in R_0$ such that $wu=u{w}^{\prime }.$ Now, let $u={\sum }_{i=0}^{k-1}{u}_i{\pi }^i,$ where $u_i\in R_0.$ It follows that

$$\sum _{i=0}^{k-1}(w-{\sigma }^i\left(w^{\prime }\right))u_i{\pi }^i=0,(w-{\sigma }^i\left(w^{\prime }\right))u_i{\pi }^i=0,$$

for $1\le i\le k-1.$ As $u_0$ is a unit, then $w=w^{\prime }.$ If for some $i>0,$ $u_i{\pi }^i\ne 0,$ $\sigma \left(u\right)-{\sigma }^i\left(u\right)\in J,$ and hence $k^{\prime }\mid i.$ Thus, $u\in R_1=Z\left(R_0\right).$ □

Lemma 2.

$R_2$ is a subring of R if and only if ${\pi }^{k^{\prime }(k_1+1)}\in R_2.$

Proof. 

Suppose that $R_2$ is a subring of $R,$ then it is enough to show that ${\pi }^k={\pi }^{k_1{k}^{\prime }}\in R_2.$ If $k^{\prime }$ does not divide $m-1,$ then $R_2=Z\left(R\right)$ which means that ${\pi }^k$ is in $R_2$ since ${\pi }^k\in Z\left(R\right).$ Now let $k^{\prime }\mid m-1,$ we consider two cases.

Case (a) When $k^{\prime }=m-1,$ then in this case, $h=h_1$ from Equation (8). Because $h_1\in R_2,$ hence easily ${\pi }^k\in R_2.$

Case (b) If $k^{\prime }<m-1.$ Assume that $p{\alpha }_0{\pi }^{m-k-1}=0,$ thus by Equation (8) ${\pi }^k=p\beta h=p\beta h_1$ which is in $R_2.$ Now suppose that $p{\alpha }_0{\pi }^{m-k-1}\ne 0,$ then $p{\alpha }_0{\pi }^{m-k-1}=w{\pi }^{m-1}$ for some unit $w\in R_0$ and $w{\pi }^{m-1}{\sum }_{j=0}^{k_1}{b}_j{\pi }^{k^{\prime }j},$ where $b_j\in S.$ This means that there exists $0\le j\le k_1$ such that $b_j{\pi }^{k^{\prime }j}R=J^{m-1}$ and for any $i\ne j,$ we have $b_i{\pi }^{k^{\prime }i}=0.$ Now $b_j={\pi }^{n^{\prime }}v$ for some $0\le n^{\prime }<n$ and a unit element v of $S.$ Hence, $p{\alpha }_0{\pi }^{m-k-1}=b_j{\pi }^{k^{\prime }j}=w{\pi }^{n^{\prime }}v{\pi }^{k^{\prime }j}.$ This implies that $m-1=k{n}^{\prime }+k^{\prime }j.$ Since $k^{\prime }j\le k-1,$ we get $n^{\prime }=n-1$ and $m=k(n-1)+(k^{\prime }j+1).$ Now since $h=h_1+h_2\pi ,$ where $h_2\in R,$ then $h^{-1}=h_1^{-1}+h_2^{\prime }\pi$ for some $h_2^{\prime }\in R.$ Thus, $p={\pi }^k(h_1^{-1}+h_2^{\prime }\pi ),$ and therefore $p{\alpha }_0{\pi }^{m-k-1}=b_j{\pi }^{k^{\prime }j}$ gives $h_1^{-1}{\alpha }_0{\pi }^{m-1}=h_1^{-n^{\prime }}v{\pi }^{m-1}$ and ${\alpha }_0-h_1^{-n^{\prime }+1}v\in ann\left(J^{m-1}\right).$ Hence, $p({\alpha }_0-h_1^{-n^{\prime }+1}v){\pi }^{m-k-1}=0$ and this leads to ${\pi }^k\in R_2.$

Conversely, if ${\pi }^{k^{\prime }(k_1+1)}\in R_2,$ then clearly R is closed under multiplication, and thus is a subring of R. □

Remark 3.

If ${\pi }^k=0,$ then clearly $R_2$ is a subring of R.

Proposition 2.

The following statements are equivalent:

(i) 

$R_2$ is a subring.

(ii) 

β and h can be chosen in $R_2.$

Proof. 

If $R_2$ is a subring of R, ${\pi }^k\in R_2.$ In this case, we have

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

(11)

It is obvious that ${\alpha }_{\sigma }$ is an automorphism of R with ${\alpha }_{\sigma }\left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }}.$ Which means that $\sigma \left(\beta \right)=\beta$ and $p{\alpha }_{\sigma }\left(h\right)=ph,$ and thus $\beta$ and h are in $R_2.$ Conversely, if $\beta$ and h are in $R_2,$ then clearly ${\pi }^{k^{\prime }}\in R_2.$ Thus, $R_2$ is a subring. □

Remark 4.

If there is an extension of σ to an automorphism ψ of R fixing ${\pi }^{k^{\prime }},$ then

$$p\beta h={\pi }^k=\psi \left({\pi }^k\right)=p\sigma \left(\beta \right)\psi \left(h\right).$$

(12)

This means, $\sigma \left(\beta \right)=\beta$ and $\psi \left(h\right)=h$ mod ${\pi }^{m-k}.$ Now, if $m-1=k,$ then $h=1.$ Also, if $m-1>k,$ then $\psi \left(u_{t_1}\right)=\sigma \left(u_{t_1}\right)=u_{t_1},$ where $t-1=t_1{k}^{\prime }.$ So, in both cases, ${\pi }^k\in R_2,$ and therefore, $R_2$ is a subring. The converse is also true.

Corollary 1.

$R_2$ is a subring of R if and only if the map ${\alpha }_{\sigma }$ is an automorphism of R, i.e., ${\alpha }_{\sigma }\in$ Aut$\left(R\right).$

Proposition 3.

(i) $J\left(R_2\right)=\left({\pi }^{k^{\prime }}\right)$ with index of nilpotency $m_1.$

(ii) 

$$R_2\cong S\left[x\right]/(g\left(x\right),x^{m_1}),$$

where $g\left(x\right)=x^{k_1}-ph\left(x\right)$ and $h\left(x\right)\in S\left[x\right]$ of degree less than $k_1.$

(iii) 

$$R_1\cong R_0\left[x\right]/(g\left(x\right),x^{m_1}),$$

where $g\left(x\right)=x^{k_1}-ph\left(x\right)$ and $h\left(x\right)\in R_0\left[x\right]$ of degree less than $k_1.$ The automorphisms gets extended to an automorphism $\eta \in$ Aut$\left(R_1\right)$ leaving ${\pi }^{k^{\prime }}$ fixed such that $R_1^{\eta }=R_2.$

(iv) 

$R,$ as an $R_2$-algebra, has $R_1$ as its coefficient subring.

Proof. 

(i) Let $u={\sum }_{j=0}^{k_1}{u}_j{\pi }^{k^{\prime }j},$ where $u_j\in S.$ Then, u is a non-unit if and only if $u_0=pv$ for some $v\in S.$ Moreover, $p={\pi }^k{w}^{-1}$ with $w\in R_2$ gives $p={\pi }^{k^{\prime }}{w}^{\prime },$ where $w^{\prime }\in R_2.$ It follows that $J\left(R_2\right)={\pi }^{k^{\prime }}{R}_2.$ Since $m-1<k^{\prime }{m}_1,$ then the index of nilpotency of ${\pi }^{k^{\prime }}$ is $m_1.$

(ii) The ring $S\left[x\right]/(x^{k_1}-ph\left(x\right))$ is a chain ring with radical of nilpotency index $n{k}_1\ge m_1.$ Consider the epimorphism $\phi$ such that $\phi :S\left[x\right]/(x^{k_1}-ph\left(x\right))\to R_2$ defined by: $\phi \left(f\left(x\right)\right)=f\left({\pi }^{k^{\prime }}\right).$ Now, the radical of $S\left[x\right]/(x^{k_1}-ph\left(x\right))$ is $\left(x\right)$ with index of nilpotency $m_1$ and $J\left(R_2\right)=\left(\phi \left(x\right)\right).$ This implies that $\phi$ induces an isomorphism

$$\overline{\phi }:S\left[x\right]/(x^{k_1}-ph\left(x\right),x^{m_1})\to R_2.$$

(iii) Similarly to (ii), consider the epimorphism $\phi :R_0\left[x\right]/(x^{k_1}-ph\left(x\right))\to R_1$ and by the same argument $\phi$ induces an isomorphism from $R_0\left[x\right]/(x^{k_1}-ph\left(x\right),x^{m_1})$ into $R_1$ defined by: $\phi \left(f\left(x\right)\right)=f\left({\pi }^{k^{\prime }}\right).$ Let $\delta$ be an extension of $\sigma$ to $R_0\left[x\right]$ with $\delta \left(x\right)=x.$ Then, clearly

$$\delta \left((x^{k_1}-ph\left(x\right),x^{m_1})\right)=(x^{k_1}-ph\left(x\right),x^{m_1}),$$

which induces the required automorphism $\eta$ of $R_1.$ The last part follows easily.

(iv) Since $J\left(R_1\right)={\pi }^{k^{\prime }}{R}_1=H\left(R_2\right)R_1,$ then $R_1$ is an unramified $R_2$-algebra. Obviously, $R=R_1+J,$ and $R_1\cap J=J\left(R_1\right).$ □

Theorem 1.

Let $R_2$ be a subring of R and η be the extension of σ to $R_1$ fixing ${\pi }^{k^{\prime }}.$

(i) 

As an $R_2$-algebra, then $R_1$ is a coefficient subring of R and $(\pi ,\eta )$ is a distinguished pair of R with respect to $R_1$.

(ii) 

If ${\pi }^{{}^{\prime }}={\pi }^{k^{\prime }},$

$$R\cong R_1[x,\eta ]/(x^{k^{\prime }}-{\pi }^{{}^{\prime }},x^m).$$

(iii) 

There is an extension φ of the associated automorphism σ to the ring R that leaves π fixed.

Proof. 

(i) For any $u\in R_1,$ by the definition of $\eta ,$ $\pi u=\eta \left(u\right)\pi .$ Also, by Lemma 1, $(\pi ,\eta )$ is a distinguished pair of R with respect to $R_1.$

(ii) This is direct.

(iii) Let $\phi :R_1[x,\eta ]\to R_1[x,\eta ]$ defined by:

$$\phi \left(\sum _i{u}_i{x}^i\right)=\sum _i\eta \left(u_i\right)x^i.$$

Then, $\phi$ is the desired extension of $\eta$ which fixes the ideal $(x^{k^{\prime }}-{\pi }^{{}^{\prime }},x^m).$ Thus, (iii) follows. □

Example 1.

Suppose that $R=R_0\left[\pi \right],$ where $R_0=GR(3^5,4)$ and π is a root of the Eisenstein polynomial $g\left(x\right)=x^4-3.$ Let $\sigma \in Aut\left(R_0\right)$ such that $\sigma ={\rho }^2,$ where ρ is the Frobenius map of $R_0,$

$$R=R_0[x,\sigma ]/(g\left(x\right),x^{20}).$$

Thus, R is a chain ring with invariants $p=3,n=5,r=4,k=4,k^{\prime }=2,m=20$ and R is can be written as

$$R={\oplus }_{i=0}^3{R}_0{\pi }^i=R_0\oplus R_0\pi \oplus R_0{\pi }^2\oplus R_0{\pi }^3.$$

Also $R_1$ is a commutative chain ring with invariants $p=3,n=5,r=4,k_1=2,m_1=10$ of the form

$$R_1={\oplus }_{i=0}^1{R}_0{\pi }^{2i}=R_0\oplus R_0{\pi }^2.$$

Let η be an extension of σ to $R_1,$, i.e.,

$$\eta \left(R_1\right)=\eta \left({\oplus }_{i=0}^1{R}_0{\pi }^{2i}\right)=\sigma \left(R_0\right)\oplus \sigma \left(R_0\right){\pi }^2.$$

Now $R_2$ is a subring which is a commutative chain ring with invariants $p=3,n=5,s=2,$ $k_1=2,m_1=10$ of the form

$$R_2={\oplus }_{i=0}^{1}S{\pi }^{2i}=S\oplus S{\pi }^2,$$

where $S=GR(3^5,2)=R_0^{\sigma }.$ Now as $R_2-$algebra, we get

$$R=R_1\oplus R_1\pi .$$

Which means that $R_1$ is the coefficient subring of R and $(\pi ,\eta )$ is a distinguished pair of R with respect to $R_1$. Next, we have

$$R=R_1[x,\eta ]/(x^2-{\pi }^2,x^{20}).$$

The extension of η is defined by:

$$\phi \left(\sum _i{u}_i{x}^i\right)=\sum _i\eta \left(u_i\right)x^i$$

which fixes the ideal $(x^2-{\pi }^2,x^{20}).$

Lemma 3.

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,

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

(13)

where $T_1$ and $T_2$ are the rings defined in Remark 2.

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 [21] and Remark 2, 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 2. □

Proposition 4.

Let ${\psi }_s:H_s\left(T_1\right)\to H_s\left(T_1\right)$ defined by: ${\psi }_s\left(w\right)={\alpha }_{\sigma }\left(w\right)w^{-1}.$ Then,

$$ker{\psi }_s\cap Im{\psi }_s=1.$$

In particular, $Im{\psi }_{m_1-1}=\{1+\alpha {\pi }^{m_1-1}:\alpha \in ker\overline{tr}\}.$

Proof. 

It is clear that $Im{\psi }_s=ker{N}_{\sigma }\cap H_s\left(T_1\right)=ker{N}_{\sigma }^s=H_s\left(ker{N}_{\sigma }\right).$ Further, $ker{\psi }_s=H_s\left(T_2\right).$ Thus, by Lemma 3, $ker{\psi }_s\cap Im{\psi }_s=1.$ To prove the last assertion, let $y\in Im{\psi }_{m_1-1}.$ Then, $y\in ker{N}_{\sigma }\cap H_{m_1-1}\left(T_1\right)$ and, hence, $y=1+\delta {\pi }^{m_1-1}={\alpha }_{\sigma }\left(\zeta \right){\zeta }^{-1},$ where $\delta \in <a>$ and $\zeta \in U\left(T_1\right).$ If $\zeta =\alpha w,$ where $\alpha \in <a>$ and $w=1+\pi ϵ\in H\left(T_1\right)$ for some $ϵ\in T_1,$ then

$$\alpha (1+\delta {\pi }^{m_1-1})(1+\pi ϵ)=\sigma \left(\alpha \right)(1+\pi {\alpha }_{\sigma }\left(ϵ\right)).$$

This leads to

$$\begin{array}{cc}& \sigma \left(\alpha \right)=\alpha ,\hfill \\ & 1+\delta {\pi }^{m_1-1}+\pi ϵ=1+\pi {\alpha }_{\sigma }\left(ϵ\right).\hfill \end{array}$$

Now, consider two cases. (i) If $m_1=2,$ then $\delta +ϵ={\alpha }_{\sigma }\left(ϵ\right);$ hence,

$$\delta ={\alpha }_{\sigma }\left({\delta }^{\prime }\right)-{\delta }^{\prime },$$

(14)

where ${\delta }^{\prime }\in <a>.$ Thus, we get the result. (ii) When $m_1>2,$

$$\delta {\pi }^{m_1-1}+\pi ϵ=\pi {\alpha }_{\sigma }\left(ϵ\right).$$

(15)

Thus,

$$\pi (\delta {\pi }^{m_1-2}+ϵ-{\alpha }_{\sigma }\left(ϵ\right))=0,$$

(16)

which means that $\delta {\pi }^{m_1-2}+ϵ-{\alpha }_{\sigma }\left(ϵ\right)\in {\pi }^{m_1-1}{T}_1.$ This implies that $ϵ-{\alpha }_{\sigma }\left(ϵ\right)\in {\pi }^{m_1-2}{T}_1;$ thus, $ϵ={ϵ}_0+{\pi }^{m_1-2}{ϵ}_1,$ where ${ϵ}_0\in T_2$ and ${ϵ}_1\in T_1.$ Note that from Equation (16),

$$\pi {\alpha }_{\sigma }\left(ϵ\right)=\pi {ϵ}_0+{\pi }^{m_1-1}{\alpha }_{\sigma }\left({ϵ}_1\right)=\delta {\pi }^{m_1-1}+\pi {ϵ}_0+{\pi }^{m_1-1}{ϵ}_1.$$

(17)

Hence, $\delta {\pi }^{m_1-1}\in {\pi }^{m_1}({\alpha }_{\sigma }\left({ϵ}_1\right)-{ϵ}_1)$ which completes the proof since

$$ker\overline{tr}=\{\eta \left(\alpha \right)-\alpha :\eta \in G,\alpha \in F\}.$$

□

From now on, we assume that $R_2$ is a subring of $R.$

Lemma 4.

If $\eta \in Aut\left(R_0\right)$ and ${\phi }_1^{\eta }\left(y\right)=y.$ Then, $y={\sum }_{i=0}^{k-1}{u}_i{\pi }^i,$ where $u_i\in R_0^{\eta }.$

Proof. 

Let $y={\sum }_{i=0}^{k-1}{u}_i^{\prime }{\pi }^i,$ then ${\phi }_1^{\eta }\left(y\right)=y$ implies $u_i^{\prime }{\pi }^i=\eta \left(u_i^{\prime }\right){\pi }^i$ for all $1\le i\le k-1.$ Now, fix i and note that $u_i^{\prime }{\pi }^i=\eta \left(u_i^{\prime }\right){\pi }^i$ leads to $(u_i^{\prime }-\eta \left(u_i^{\prime }\right)){\pi }^i=0,$ and thus $(u_i^{\prime }-\eta \left(u_i^{\prime }\right))\in p{R}_0.$ Then, $u_i^{\prime }=a_1+a_2{p}^q,$ where $a_1\in R_0^{\eta }$ and $q\ge 1.$ If $a_2{p}^q{\pi }^i=0,$ then put $u_i=a_1,$ and however if $a_2{p}^q{\pi }^i\ne 0,$ repeat the same process with $a_2.$ This concludes the result. □

Proposition 5.

Assume that R is a finite chain ring and $\zeta \in U\left(R_1\right)$. If $\eta \in Aut\left(R_0\right),$ then

(i) 

${\phi }_{\zeta }^{\eta }$ is a ${\mathbb{Z}}_{p^n}-$automorphism of $R.$

(ii) 

For $u\in R_0,$ ${\phi }_{\zeta }^{\eta }\left(uy\right)=\eta \left(u\right){\phi }_{\zeta }^{\eta }\left(y\right)$ and ${\phi }_{\zeta }^{\eta }\left(yu\right)={\phi }_{\zeta }^{\eta }\left(y\right)\eta \left(u\right).$

(iii) 

If ${\phi }_{\zeta }^{\eta }$ is an automorphism of R fixing ${\pi }^{k^{\prime }},$ then

$$p\beta h=p{\beta }^{\prime }{h}^{\prime },$$

(18)

where ${\beta }^{\prime }\in R_0^{\eta }$ and $h^{\prime }\in R_0^{\eta }\left[{\pi }^{k^{\prime }}\right].$

Proof. 

(i) and (ii) are easy to check. For (iii), note that ${\phi }_{\zeta }^{\eta }\left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }}$ implies that $N_{\sigma }\left(w\right){\pi }^{k^{\prime }}={\pi }^{k^{\prime }}$ and, thus, $N_{\sigma }{\left(w\right)}^{k_1}{\pi }^k={\pi }^k.$ On the other hand, from Proposition 5 [21], we obtain

$$p\mu \left(\beta \right){\phi }_{\zeta }^{\eta }\left(h\right)=p\beta h.$$

(19)

Hence, $\beta =\eta \left(\beta \right),$, i.e., $\beta \in R_0^{\eta }$ and $p{h}^{\prime }=ph,$ where $h^{\prime }={\phi }_{\zeta }^{\eta }\left(h\right).$ Using Lemma 4, $h^{\prime }\in R_0^{\eta }\left[{\pi }^{k^{\prime }}\right].$ □

Remark 5.

If $n=1,$ then clearly ${\phi }_{\zeta }^{\eta }$ is an automorphism of $R.$

Proposition 6.

Let ${\eta }_1$ and ${\eta }_2$ be any two automorphisms of $R_0,$ and ${\zeta }_1$ and ${\zeta }_2\in U\left(R_1\right).$ Then,

(i) 

${\phi }_{{\zeta }_1}^{{\eta }_1}{\phi }_{{\zeta }_2}^{{\eta }_2}={\phi }_{\zeta }^{{\eta }_1{\eta }_2},$ where $\zeta ={\phi }_{{\zeta }_1}^{{\eta }_1}\left({\zeta }_2\right).$

(ii) 

Assume that ${\phi }_{{\zeta }_1}^{{\eta }_1}={\phi }_{{\zeta }_2}^{{\eta }_2}.$ Then, ${\eta }_1={\eta }_2,$ and moreover

$$\left\{\begin{array}{cc}{\zeta }_1={\zeta }_2,\hfill & \mathit{if}{k}^{\prime }\nmid (m-1)\hfill \\ {\zeta }_1={\zeta }_2\mathrm{mod}{H}_{m_1-1}\left(R_1\right),\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Proof. 

(i) is obvious. (ii) First, observe that the restriction of ${\phi }_{{\zeta }_1}^{{\eta }_1}$ to $R_0$ is ${\eta }_1,$ then clearly ${\eta }_1={\eta }_2.$ Moreover, since ${\phi }_{{\zeta }_1}^{{\eta }_1}\left(\pi \right)={\phi }_{{\zeta }_2}^{{\eta }_2}\left(\pi \right),$ then ${\zeta }_1\pi ={\zeta }_2\pi ;$ thus, $({\zeta }_1-{\zeta }_2)\pi =0.$ This means ${\zeta }_1-{\zeta }_2\in <{\pi }^{m-1}>$. However, it is evident to see that

$$\left({\pi }^{m-1}\right)\cap R_1=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}{k}^{\prime }\nmid (m-1),\hfill \\ {\pi }^{m-1}{R}_1,\hfill & \mathrm{if}{k}^{\prime }\mid (m-1).\hfill \end{array}\right.$$

This completes the proof. (iii) Note that ${\phi }_{\alpha w}^{\eta }\left({\pi }^{k^{\prime }}\right)=N_{\sigma }\left(\alpha w\right){\pi }^{k^{\prime }},$ then ${\varphi }_{\alpha w}\in Aut\left(R\right)$ fixing ${\pi }^{k^{\prime }}$ if and only if $N_{\sigma }\left(w\right){\pi }^{k^{\prime }}={\pi }^{k^{\prime }},$ and hence if and only if $N_{\sigma }\left(w\right)\in H_{m_1-1}\left(R_2\right).$ □

Remark 6.

From Equation (18), we have $ph=p{h}^{\prime },$ where $h^{\prime }\in R_0^{\eta }\left[{\pi }^{k^{\prime }}\right].$ This means, if $h=1+{\sum }_{i=0}^{k_1-1}{u}_i{\pi }^{i{k}^{\prime }}$ and $h^{\prime }=1+{\sum }_{i=0}^{k_1-1}{u}_i^{\prime }{\pi }^{i{k}^{\prime }},$ where $u_i\in R_0$ and $u_i^{\prime }\in R_0^{\eta };$ thus, $p(u_i-u_i^{\prime }){\pi }^{i{k}^{\prime }}=0.$ For each $i,$ assume

$$an{n}_S\left(p{\pi }^{i{k}^{\prime }}\right)=p^{n_i}S.$$

(20)

Now, let A be the intersection of all Galois subrings $A_1$ of S such that $u_i\in A_1+p^{n_i}S.$ This means that A is the intersection of all these $A_1=S^{\eta },$ and then η is $A-$automorphism of $R_0.$ In contrast, suppose η is an $A-$automorphism of $R_0.$ Then, ${\phi }_1^{\eta }\in Aut\left(R\right)$ leaves ${\pi }^{k^{\prime }}$ fixed. Furthermore, for some $u\in S$

$$u_i=u_i^{\prime }+p^{n_i}u=\eta \left(u_i^{\prime }\right)+p^{n_i}u.$$

(21)

This implies $p(u_i-u_i^{\prime }){\pi }^{i{k}^{\prime }}=0$ and, hence, $ph=p{h}^{\prime },$ for some $h^{\prime }\in R_0^{\eta }\left[{\pi }^{k^{\prime }}\right].$ Denote L as the set of all $A-$automorhisms of $R_0.$

Lemma 5.

Let R be a finite chain ring with $p,n,r,k,k^{\prime },m$ and let

$$\begin{array}{cc}& Y=\{\eta \in Aut\left(R_0\right):{\phi }_{\zeta }^{\eta }\in Aut\left(R\right)\mathrm{fixes}{\pi }^{k^{\prime }}\mathrm{for}\mathrm{some}\zeta \in U\left(R_1\right)\},\hfill \\ & D=\{s\in S:\eta (s)=s\mathrm{for}\mathrm{every}\eta \mathrm{such}\mathrm{that}{\phi }_{\zeta }^{\eta }\in Y\}.\hfill \end{array}$$

Then, $A\subseteq D$ and, moreover, $Y=L.$

Proof. 

From Remark 6, it is obvious that $A\subseteq D$ and also $Y\subseteq L.$ Assume that $\eta \in L,$ then one can see that ${\phi }_1^{\eta }$ fixes ${\pi }^{k^{\prime }},$ and thus $\eta \in Y.$ Therefore, $Y=L.$ □

Theorem 2.

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

$$G\cong \left\{\begin{array}{cc}U\left(R_1\right)⋊Aut\left(R_0\right),\hfill & if{\pi }^{k^{\prime }}=0,\hfill \\ Q⋊L,\hfill & if{\pi }^{k^{\prime }}\ne 0,\hfill \end{array}\right.$$

(22)

where

$$Q=\left\{\begin{array}{cc}\left\{1\right\},\hfill & if{k}^{\prime }\nmid (m-1),\hfill \\ ker{N}_{\sigma }\times H_{m_1-1}\left(R_1\right)/H_{m_1-1}\left(R_1\right),\hfill & if{k}^{\prime }\mid (m-1).\hfill \end{array}\right.$$

Proof. 

If ${\pi }^{k^{\prime }}=0,$ then it is not hard to see that ${\phi }_{\zeta }^{\eta }\in G$ for every $\zeta \in U\left(R_1\right)$ and $\eta \in Aut\left(R_0\right).$ Thus, the result follows by Proposition 6 (i). Next, suppose ${\pi }^{k^{\prime }}\ne 0,$ then from Lemma 5, $Y=L.$ By Proposition 6, we have ${\phi }_{\zeta }^{\eta }\in G$ if and only if $N_{\sigma }\left(w\right)\in H_{m_1-1}\left(R_2\right),$ where $\zeta =\alpha w.$ The proof of Lemma 3 also implies that $N_{\sigma }\left(w\right)\in H_{m_1-1}\left(R_2\right)$ if and only if $w\in ker{N}_{\sigma }\times H_{m_1-1}\left(R_1\right).$ Furthermore, ${\phi }_{w_1}^{{\eta }_1}={\varphi }_{w_2}^{{\eta }_2}$ if and only if $w_1=w_2$ when $k^{\prime }\nmid (m-1)$ and $w_1=w_2\mathrm{mod}{H}_{m_1-1}\left(R_1\right),$ otherwise. Therefore, Proposition 6 (i) concludes that G is semi product of Q by $L.$ □

Example 2.

Let R be a chain ring of the form,

$$R=GR(3,4)[x,\sigma ]/\left(x^4\right),$$

where $GR(3,4)$ is a Galois ring of order $3^4$ and σ is defined as $\sigma ={\rho }^2\in Aut\left(GR(3,4)\right),$ρ is the Frobenius map of $GR(3,4).$ Note that $R_0=GR(3,4)=GF\left(3^4\right)=F.$ Also, we can write the invariants of R as $p=3,n=1,r=4,k=4,k^{\prime }=2,m=4.$ Let $R_1$ be the centralizer of $GR(3,4)$ in $R,$ then $R_1$ is a commutative chain ring with invariants $p=3,n=1,r=4,k_1=k/k^{\prime }=2,$ $m_1=2.$ Now we investigate the group of units of $R_1,$ $U\left(R_1\right).$ Consider the following series,

$$H_1\left(R_1\right)<H_2\left(R_1\right),$$

where $H_i\left(R_1\right)=1+J^i\left(R_1\right)$ with the admissible function $j,$ $j\left(1\right)=j\left(2\right)=2$ and the range of $j,$ $R\left(j\right)=\left\{1\right\}.$ Then by results from [23], we have

$$H\left(R_1\right)=C_3^4,$$

and thus

$$U\left(R_1\right)=C_{3^4-1}\otimes C_3^4.$$

Now from Theorem 2,

$$G=C_3^4⋊C_{3^4-1}⋊Aut\left(GR(3,4)\right).$$

But since $Aut\left(GR(3,4)\right)=C_4,$ and thus

$$G=C_3^4⋊C_{3^4-1}⋊C_4.$$

Remark 7.

The case when ${\pi }^{k^{\prime }}=0,$, i.e., $n=1.$ Then every automorphism ϕ in $Aut\left(R\right)$ satisfies $\varphi \left({\pi }^{k^{\prime }}\right)={\pi }^{k^{\prime }}=0.$ This means that $\varphi \in G,$ and hence

$$G=Aut\left(R\right).$$

Corollary 2.

Assume the hypotheses of Theorem 2. Then,

(i) 

If ${\pi }^{k^{\prime }}\ne 0,$

$$\mid G\mid =\left\{\begin{array}{cc}[\overline{R_0}:\overline{A}]\frac{(p^r-1)p^{(m_1-1)(r-s)+s}}{p^s-1},\hfill & if{k}^{\prime }\nmid (m-1),\hfill \\ \overline{R_0}:\overline{A}]\frac{(p^r-1)p^{(m_1-1)(r-s)+s-r}}{p^s-1},\hfill & if{k}^{\prime }\mid (m-1).\hfill \end{array}\right.$$

(ii) 

If ${\pi }^{k^{\prime }}=0,$

$$\mid G\mid =\mid H\left(R_1\right)\mid \times (p^r-1)\times r.$$

Proof. 

First, note that by Remark 6, $\mid L\mid =[\overline{R_0}:\overline{A}].$ Now, by Proposition 4,

$$\mid ker{N}_{\sigma }\cap H_{m_1-1}\left(R_1\right)\mid =\mid ker\overline{tr}\mid .$$

Thus,

$$\mid ker{N}_{\sigma }\cap H_{m_1-1}\left(R_1\right)\mid =p^{r-s}.$$

Moreover, $\mid ker{N}_{\sigma }\mid =\frac{\mid U\left(R_1\right)\mid }{\mid U\left(R_2\right)\mid }$ and $\mid H_{m_1-1}\left(R_1\right)\mid =\mid \overline{R}\mid =p^r.$ Hence,

$$\mid Q\mid =\left\{\begin{array}{cc}\frac{(p^r-1)p^{(m_1-1)(r-s)+s}}{p^s-1},\hfill & \mathrm{if}{k}^{\prime }\nmid (m-1),\hfill \\ \frac{(p^r-1)p^{(m_1-1)(r-s)+s-r}}{p^s-1},\hfill & \mathrm{if}{k}^{\prime }\mid (m-1).\hfill \end{array}\right.$$

Therefore, the result follows from Theorem 2. For the second part, the result is obvious since $U\left(R_1\right)=C_{p^r-1}\otimes H\left(R_1\right).$ □

Remark 8.

In fact, the structure of $H\left(R_1\right)$ is well-known in [23]. If $n=1,$ then the structure is given by

$$H\left(R_1\right)={\otimes }_{s\notin R\left(j\right)}{C}_{p^e}^{N_{k_1}\left(e\right)r},$$

where $N_{k_1}\left(e\right)=(\lfloor \frac{k_1-1}{p^{e-1}}\rfloor +2\lfloor \frac{k_1-1}{p^e}\rfloor +\lfloor \frac{k_1-1}{p^{e+1}}\rfloor )$ and $R\left(j\right)$ the range of the admissible function $j.$

Example 3.

Consider the chain ring R as in the Example 2, using the results of corollary, we have

$$\mid G\mid =3^4\times (3^4-1)\times 4.$$

Remark 9.

The results in Theorem 2 hold under more general situations. For instance, when R is not necessarily finite.

4. Conclusions

In this article, we consider a subgroup G of the automorphism group of a finite chain ring R which contains all automorphisms that leave ${\pi }^{k^{\prime }}$ invariant. Moreover, we describe the structure of $G.$ The results of this article are useful in studying $Aut\left(R\right)$ in the general case and thus in enumerating chain rings. As a future problem, the authors might suggest studying $Aut\left(R\right)$ of chain rings in general in order to expand their application in different regions.