On J-Diagrams for the One Groups of Finite Chain Rings

Abstract

Let R be a finite commutative chain ring with invariants $p,n,r,k,m.$ The purpose of this article is to study j-diagrams for the one group $H=1+J\left(R\right)$ of $R,$ where $J\left(R\right)=\left(\pi \right)$ is Jacobson radical of $R.$ In particular, we prove the existence and uniqueness of j-diagrams for such one group. These j-diagrams help us to solve several problems related to chain rings such as the structure of their unit groups and a group of all symmetries of $\left\{{\pi }^{k^{\prime }}\right\},$ where $k^{\prime }\mid k$. The invariants $p,n,r,k,m$ and the Eisenstein polynomial by which R is constructed over its Galois subring determine fully the j-diagram for $H.$

1. Introduction

Suppose that $P_m=\{1,3,\dots ,m\},$ the function $j:P_m\to P_m$ is said to be admissible if $s<j\left(s\right)$ and if $j\left(s\right)=j\left(i\right),$ then $s=i.$ Admissible functions have been used as a significant tool to determine the structure of abelian p-groups which have certain types of j-diagram series [1,2,3]. Moreover, j-diagrams are used in classifying chain rings and in determining their groups of automorphisms [4]. Motivated by the important role of j-diagrams in group and ring theory, this article is aimed to investigate the existence and uniqueness of such j-diagrams. We focus our attention on j-diagrams for finite abelian p-groups, and particularly groups of units of finite commutative chain rings. Chain rings are associative rings that have a lattice of ideals that creates a unique chain. A finite ring R can easily be shown to be a chain ring if and only if its (Jacobson) radical $J\left(R\right)=J$ is principal and $\overline{R}=R/J$ is a field of order $p^r,$ p is prime. Every finite chain ring has five positive integers $p,n,r,k,m$ named the invariants. These rings occur in several applications, for details see [1,5,6,7,8,9,10,11,12]. For instance, they have widely appeared in coding theory [13,14,15,16,17]. However, the class of Galois rings is a distinguished class of finite chain rings, and every Galois ring is represented as:

$$GR(p^n,r)={\mathbb{Z}}_{p^n}\left[x\right]/\left(f\left(x\right)\right),$$

(1)

where $f\left(x\right)$ is a monic irreducible polynomial of degree $r.$

Finite chain rings are constructed in at least two different ways. Suppose that R is a finite chain ring that has the invariants $p,n,r,k,m.$ First, R can be viewed as an Eisenstein extension of $GR(p^n,r)$

$$R=GR(p^n,r)\left[x\right]/(g\left(x\right),x^m),$$

(2)

where $g\left(x\right)$ is an Eisenstein polynomial over $GR(p^n,r),$ i.e.,

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

(3)

where $s_0$ is a unit of $GR(p^n,r)$. Another way to construct R involves ${\mathbb{Q}}_p,$ the field of p-adic numbers. Every chain ring R is a quotient ring of the integers ring of a certain finite extension of ${\mathbb{Q}}_p,$ for more details see [6] and the references therein. The symmetry of invariants of various chain rings injects more choice and flexibility into the theory of ring construction.

The group of units (multiplicative group) $U\left(R\right)$ of R is defined by $U\left(R\right)=R\backslash J$, i.e., the set of all non-nilpotent elements of $R.$ From Ayoub (1972) [2], $U\left(R\right)\cong U\otimes H,$ where $U\cong {(R/J)}^{*}$ is cyclic of order $p^r-1$ and $H=1+J$ which is a p-group. Thus, the structure problem of $U\left(R\right)$ is reduced to that of $H.$ After Ayoub, we call H the one group. If $p-1$ does not divide $k,$ the structure of H is given by Ayoub [2] based on the results in [3]. However, the case when $(p-1)\mid k,$ the full structure of H is given by Alabiad and Alkhamees [1]. In this paper, we aim to study the existence and uniqueness of j-diagrams for the one group H of $R.$

In Section 2, we introduce the concept of j-diagrams, some notations and examples. In Section 3, we study the existence and uniqueness of complete and incomplete j-diagrams for the series $H=H_1>H_2>H_3>\cdots >H_m=1$ of the one group $H,$ for any finite commutative chain ring R with invariants $p,n,r,k,m,$ where $H_s=1+J^s.$ Moreover, among other results, we find an explanation of j-diagrams from a ring-theoretic point of view, see Theorem 5.

2. Preliminaries

Unless otherwise mentioned, all considered groups are multiplicative abelian groups, p denotes a fixed prime. See [2,3,18] for the details of this section.

Definition 1. 

If $P_m=\{1,2,\cdots ,m\}$, $j:P_m\to P_m$ is called an admissible function if j satisfies the following conditions:

$$\begin{array}{ccccccccccc}& \left(\mathit{i}\right)s<j\left(s\right),s\ne m;\hfill & & & & & & & & & \\ & \left(\mathit{i}\mathit{i}\right)Ifj\left(i\right)=j\left(s\right)\ne m,\mathrm{then}i=s.\hfill & & & & & & & & \end{array}$$

The following example is direct.

Example 1. 

Let $j:P_m\to P_m$ be an admissible function and let $1\le m^{\prime }\le m$. Define

(i) $j_1:P_{m^{\prime }}\to P_{m^{\prime }},$ given by $j_1\left(i\right)=j(m-m^{\prime }+i)-(m-m^{\prime });$

(ii) $j_2:P_{m^{\prime }}\to P_{m^{\prime }},$ given by:

$$j_2\left(i\right)=\left\{\begin{array}{cc}j\left(i\right),\hfill & \mathrm{if}j\left(i\right)<m^{\prime },\hfill \\ m^{\prime },\hfill & \mathrm{if}j\left(i\right)\ge m^{\prime }.\hfill \end{array}\right.$$

Then, $j_1$ and $j_2$ are admissible functions.

Definition 2.  

Let p be a fixed prime and A an abelian p-group. Then, the series

$$A=A_1>A_2>\cdots >A_m=1,$$

(4)

is called a complete $j-$diagram for A of length m with respect to p if $j:P_m\to P_m$ is admissible and

(i) 

If $j\left(s\right)=m,A_s^p=1.$

(ii) 

If $j\left(s\right)\ne m,{\eta }_s:A_s/A_{s+1}⟶A_{j\left(s\right)}/A_{j\left(s\right)+1},$ given by:

$${\eta }_s\left(x{A}_{s+1}\right)=x^p{A}_{j\left(s\right)+1},$$

(5)

defines an isomorphism.

In case when ${\eta }_s$, for some $s\in P_m,$ is only homomorphism, the series is called incomplete $j-$diagram at $i=s.$

Example 2. 

Let $A=<a>,$ where $o\left(a\right)=p^{m-1}.$ Then,

$$A=<a>=A_1>A_2=<a^p>\cdots >A_m=1,$$

is a $j-$diagram of length m if j defined by:

$$j\left(i\right)=\left\{\begin{array}{cc}i+1,\hfill & \mathrm{if}1\le i<m,\hfill \\ m,\hfill & \mathrm{if}i=m.\hfill \end{array}\right.$$

Example 3. 

Let A be an elementary abelian $p-$group and let

$$A=A_1>A_2\cdots >A_m=1,$$

(6)

be a chain of subgroups of A. Then if $j\left(i\right)=m,1\le i\le m,$ the chain (6) is a $j-$diagram of length m since $A_i^p=1,1\le i\le m.$ Thus, clearly a group need not have a unique j-diagram.

Notations and Terminologies

1.

$\nu \left(s\right)$ is the smallest positive integer such that $j^{\nu \left(s\right)}\left(s\right)=m.$

2.

o$\left(x\right)$ means the multiplicative order of $x.$

3.

$R\left(j\right)$ denotes the range of $j.$

4.

$\beta$ is always associated with Eisenstine polynomial $g\left(x\right)$, i.e., ${\pi }^k=p\beta h.$

5.

If $a\in A_i\backslash A_{i+1},$ we denote $wt\left(a\right)=i.$

6.

$rank\left(A\right)$ is the smallest number of generators of $A.$

7.

$dim\left(A\right)$ is the dimension of A as a vector space over ${\mathbb{Z}}_p.$

3. The J-Diagrams for One Group H

In what follows, R is a finite commutative chain ring with invariants $p,n,r,k,m.$ We focus on the following series of $H=1+J$ ( $J=J\left(R\right)$),

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

(7)

where $H_s=1+J^s.$

Definition 3. 

We call R a complete (incomplete) chain ring if H has complete (incomplete at $s^{*}$) $j-$diagram, where $s^{*}=\lfloor \frac{k}{p-1}\rfloor$.

Lemma 1. 

If x is a unit in $R.$ Then, $x=y$ mod $H_i$ if and only if $x=y$ mod $J^i.$

Proof. 

This is true because

$$y-x\in J^i\iff x^{-1}y-1\in J^i\iff x^{-1}y\in H_i.$$

□

The following lemma is easy to prove.

Lemma 2. 

Let $1\le s\le m.$

(1) 

The map ${\gamma }_s:H_s/H_{s+1}⟶J^s/J^{s+1}$ defined by:

$${\gamma }_s(1+x)H_{s+1}=x+J^{s+1}$$

(8)

is an isomorphism.

(2) 

Let ${\delta }_s:J^s/J^{s+1}⟶J^{s+k}/J^{s+k+1}$ defined by:

$${\delta }_s(x+J^{s+1})=px+J^{s+k+1}.$$

(9)

Then, ${\delta }_s$ is an isomorphism.

Remark 1. 

Note that $J^s/J^{s+1}$ is an elementary p-group.

Theorem 1. 

Let R be complete, and let $j\left(i\right)\ne m$ for some $i.$

(i) 

If $k+i<pi,$ then $j\left(i\right)=k+i.$

(ii) 

If $k+i=pi$ and $j(i+1)<m,$ then $j\left(i\right)=k+i.$

(iii) 

If $pi<k+i,$ then $j\left(i\right)=pi.$

Proof. 

Consider $1+x\in H_i\backslash H_{i+1}.$ Then, ${(1+x)}^p\in H_{j\left(i\right)}\backslash H_{j\left(i\right)+1}.$ Moreover, ${(1+x)}^p=1+py+x^p$ with $wt\left(y\right)=wt\left(x\right)=i.$ Suppose $k+i<pi,$ then $k+i=wt\left(py\right)<pi.$ As $wt\left(x^p\right)=\min \{m,pi\}>k+i,$ we get ${(1+x)}^p\in H_{k+i}\backslash H_{k+i+1}.$ Hence, $j\left(i\right)=k+i$ and this ends (i). For (ii), assume that $k+i=pi$ and $j\left(i\right)<m.$ Then, $k+(i+1)<p(i+1).$ Now $1+py+x^p\in H_{k+i}.$ So $j\left(i\right)\ge k+i.$ However, $k+i+1<p(i+1).$ If $j(i+1)<m,$ then $j(i+1)=k+i+1>j\left(i\right)$ gives $j\left(i\right)=k+i.$ Similarly, one can prove (iii). □

Lemma 3. 

Let R be complete.

(i) 

Let e be the smallest positive integer such that $j\left(e\right)=m.$ Then either $k+e=pe$ or $k+e\ge m,$ $pe\ge m.$

(ii) 

Any homomorphic image of R is complete.

Proof. (i)

As $j\left(e\right)=m,$ by definition $H_e^p=1.$ For any $x\in J,$ ${(1+x)}^p=1+py+x^p$ for some $y\in R$ with $wt\left(y\right)=wt\left(x\right).$ Consider any $0\ne x\in J^e,$ then $1={(1+x)}^p=1+py+x^p,$ $py+x^p=0.$ If $py=0,$ then $x^p=0,k+e\ge m,pe\ge m.$ Suppose that $py\ne 0.$ Then, $py=-x^p$ gives $k+wt\left(x\right)=pwt\left(x\right).$ (ii) Let I be a non-zero ideal of $R,$ $I=J^s$ for some $1\le s\le m.$ For $T=R/I,$ $J\left(T\right)=J/J^s,$ $H_i\left(T\right)=1+J^i\left(T\right)$ and $H_i\left(T\right)/H_{i+1}\left(T\right)\cong H_i/H_{i+1}$ for $1\le i\le s.$ Thus, whenever $i<s$ and $j\left(i\right)<s,$

$$H_i\left(T\right)/H_{i+1}\left(T\right)\cong H_{j\left(i\right)}\left(T\right)/H_{j\left(i\right)+1}\left(T\right).$$

For some $i<s$ suppose that $j\left(i\right)\ge s.$ Then, $H_{j\left(i\right)}\subseteq H_s.$ However, either $H_i^p=1$ or $H_i/H_{i+1}\cong H_{j\left(i\right)}/H_{j\left(i\right)+1}$ under the mapping $(1+x)H_{i+1}\to {(1+x)}^p{H}_{j\left(i\right)+1}.$ Hence, $H_i^p\subseteq 1+J^s.$ This shows that $H_i^P\left(T\right)=1,$ and thus T is complete with the admissible function $j^{\prime }$ defined on $P_s$ as follows: $j^{\prime }\left(i\right)=j\left(i\right)$ if $j\left(i\right)<s,$ otherwise $j^{\prime }\left(i\right)=s.$ □

Lemma 4. 

Let R be complete, and let $(p-1)\mid k,$ i.e., $k+s=ps<m.$ If $j\left(s\right)>k+s,$ then the followings hold

(i) 

$m=k+s+1.$

(ii) 

$\mid \overline{R}\mid =p.$

(iii) 

There exists a unit $v\in R$ such that $p{x}_0=x_0^{p}v$ and $1+v\in J,$ where $wt\left(x_0\right)=s.$

Conversely, if there exists $x_0\in J$ with $wt\left(x_0\right)=s,$ $k+s=ps<m,$ and if R satisfies (i), (ii) and (iii), then there exists an admissible function j on $P_m$ for which R is complete.

Proof. 

Now $k+s+1<p(s+1).$ If $k+s+1<m,$ by Theorem 1, $j(s+1)=k+s+1.$ As $j\left(s\right)<j(s+1),$ we get $j\left(s\right)=k+s.$ This is a contradiction. Hence, $m=k+s+1$ and $j\left(s\right)=m.$ For any $x\in R,$ the binomial expansion gives ${(1+x)}^p=1+px+x^p+pz$ for some $z\in R$ with $wt\left(z\right)\ge \min \{m,2wt(x\left)\right\}.$ Consider any unit $u\in R.$ As $p{x}_0^2=0,$ then,

$${(1+u{x}_0)}^p=1+pu{x}_0+u^p{x}_0^p=1.$$

So, $pu{x}_0+u^p{x}_0^p=0,$ and in particular $p{x}_0+x_0^p=0.$ It follows that $p(u-u^p)x_0=0,$ and hence $u-u^p\in J.$ This proves $\mid \overline{R}\mid =p.$ The hypothesis gives that $p{x}_0=x_0^{p}v$ for some unit $v\in R.$ Then, $p{x}_0+x_0^p=0$ gives that $1+v\in J.$

For the converse, define j such that $j\left(i\right)=pi$ for $i<s,$ and $j\left(i\right)=m$ for $i\ge s.$ Then for any unit $u\in R,$ ${(1+u{x}_0)}^p=1+pu{x}_0+u^p{x}_0^p.$ As $u-u^p\in J,1+u\in J,$

$$pu{x}_0+u^p{x}_0^p=p(u-u^p)x_0+u^p{x}_0^p(1+u)=0.$$

Hence, ${(1+u{x}_0)}^p=1.$ By using this, and the argument in [2], it can be easily verified that j is an admissible function and that R is complete. □

Theorem 2. 

Let R be a finite commutative chain ring with invariants $p,n,r,k,m,$ ${\pi }^k=p\beta h,$ and for some $s\in P_m,$ $k+s=ps<m.$ Then there exists an admissible function j on $P_m$ such that $j\left(s\right)>k+s,$ and R is complete if and only if R satisfies the following conditions:

(i) 

$m=k+s+1.$

(ii) 

$\mid \overline{R}\mid =p.$

(iii) 

$\beta =-1$ in $\overline{R}.$

Proof. 

Let R be complete and $j\left(s\right)>k+s.$ Let $x_0\in J$ such that $wt\left(x_0\right)=s.$ By Lemma 4, (i) and (ii) hold, and there exists a unit $u\in R$ such that $p{x}_0=x_0^{p}u$ and $1+u\in J.$ Now $x_0={\pi }^{s}w$ for some unit $w.$ Then,

$$p{\pi }^{s}w={\pi }^{ps}{w}^{p}u={\pi }^{k+s}{w}^{p}u=p{\pi }^s{w}^{p}uy.$$

It follows that $w-w^{p}uy\in J^{m-k-s}=J.$ In $\overline{R},$ $\overline{u}=-1,$ by (ii), ${\overline{w}}^{p-1}=1,$ so $\overline{y}=-1$, i.e., $\beta =-1.$ Conversely, let R satisfy (i), (ii) and (iii). By (iii), $-y^{-1}\in H=H^{p-1}.$ Hence, $-y^{-1}=w^{p-1}$ for some $w\in H.$ Consider $x={\pi }^{s}w,u=-1.$ Then, $px=x^{p}u$ and $1+u=0\in J.$ By Lemma 4, the desired j exists. □

Theorem 3. 

Let R be a finite commutative chain ring with invariants $p,n,r,k,m,$ and for some $s\in P_m,$ $k+s=ps<m.$ Then there exists an admissible function j on $P_m$ such that $j\left(s\right)=k+s$ and R is complete if and only if $-\beta \notin {\overline{R}}^{p-1}.$

Proof. 

Suppose that R is complete and $j\left(s\right)=k+s.$ For any $0\ne y\in J,{(1+y)}^p=1+py+y^p+pz$ for some $z\in R$ with $wt\left(z\right)\ge \min \{m,2wt(y\left)\right\}.$ Fix an $x\in H_s\backslash H_{s+1}.$ As ${(1+x)}^p\in H_{k+s}\backslash H_{k+s+1},$ then we get $wt(px+x^p)=k+s.$ Moreover, $px=x^{p}u$ for some unit $u\in R.$ So $x^p(u+1)\in J^{k+s}\backslash J^{k+s+1},$ and thus $1+u$ is a unit. For any $c\in R,$ as ${(1+cx)}^p\in H_{k+s}\backslash H_{k+s+1},$ $pcx+c^p{x}^p=x^p(cu+c^p)$ has weight $k+s,$ so $u+c^{p-1}$ is a unit. Thus, in $\overline{R},$ $u\notin {\overline{R}}^{p-1}.$ Now, $x={\pi }^{s}w$ for some unit $w.$ Then, $x^{p}u={\pi }^{k+s}{w}^{p}u,$ and $px={\pi }^{k+s}w{y}^{-1},$ so $w^{p-1}-{\left(yu\right)}^{-1}\in J^{m-k-s}.$ Thus, $\overline{yu}\in {\overline{R}}^{p-1}.$ As $\overline{u}\notin {\overline{R}}^{p-1,}$ we get $\overline{y}\notin {\overline{R}}^{p-1}.$ Consequently, $-\beta \notin {\overline{R}}^{p-1}.$ Conversely, let $-\beta \notin {\overline{R}}^{p-1}.$ Consider $u=-y^{-1},$ then for any unit $c\in R,$ $c^{p-1}+u$ is a unit. It follows that for $x={\pi }^s,$ $px=x^{p}u,$ $pcx+c^p{x}^p=x^p(cu+c^p)$ has weight $k+s, {(1+cx)}^p\in H^{k+s}\backslash H^{k+s+1},$ and thus

$$H_s/H_{s+1}\cong H_{s+k}/H_{s+k+1}.$$

For $i<s,$ $pi<k+i,$ define $j\left(i\right)=pi,$ and for $i\ge s,$ define $j\left(i\right)=\min \{m,k+i\}$. By using ([2], Propositions 1 and 2), it follows that j is the desired admissible function. □

Theorem 4. 

Let R be a finite commutative chain ring with invariants $p,n,r,k,m$. If R is complete, then there exists only one admissible function j on $P_m.$

Proof. 

Suppose that $k=m.$ Then, $charR=p,$ ${(x+y)}^p=x^p+y^p.$ Using this it follows that any finite chain ring R of characteristic p is complete and the underlying admissible function j on $P_m$ is such that $j\left(i\right)=pi,$ whenever $pi<m,$ and $j\left(i\right)=m$ otherwise. Suppose $k<m,$ and $j,j^{\prime }$ are two different admissible functions on $P_m$ such that R is complete with respect to j as well as $j^{\prime }.$ It follows from the proof of Theorem 1 that if for some $i<m,$ $k+i\ne pi$ and $\min \{k+i,pi\}<m,$ then $j\left(i\right)=\min \{k+i.pi\}=j^{\prime }\left(i\right).$ If for some $i<m$ and $\min \{k+i.pi\}\ge m,$ then $j\left(i\right)=m=j^{\prime }\left(i\right).$ So, there exists an $s<m$ such that $k+s=ps<m$ and $j\left(s\right)\ne j^{\prime }\left(s\right).$ It follows from Lemma 4 that $\mid \overline{R}\mid =p,$ and we can take $j\left(s\right)=s+k,$ $j^{\prime }\left(s\right)=s+k+1=m.$ Let $J_1$ and $J_2$ be the restriction of j and $j^{\prime }$, respectively, to $L_s=\{i:s\le i\le m\}.$ Set $X=\{i:s\le i\le s+k\}.$ By applying (Theorem 1 (4) [3]), we get two sets of cyclic $p-$subgroups $\{U_i:1\le i\le m\},$ $\{U_i^{\prime }:1\le i\le m\}$ of H corresponding to j and $j^{\prime }$, respectively. By Proposition 6, $H_s={\oplus }_{i\in X}{U}_i,$ so $rank\left(H_s\right)=k.$ Furthermore, $H_s={\oplus }_{i\in Y}{U}_i^{\prime }$ gives $rank\left(H_s\right)=k+1.$ This is a contradiction, and thus proves the result. □

Remark 2. 

By a similar discussion, the above results hold if we assume that R is incomplete.

Remark 3. 

Consider a finite commutative chain ring R with $p,n,r,k,m$. Let ${\pi }^k=p\beta h$ be an Eisenstein polynomial of $R.$ By looking at the invariants $p,k$ and the element $\beta ,$ one knows whether a given R is complete or incomplete using Theorems 2 and 3. In any case, the form of the underlying admissible function j on $P_m$ is well defined by:

$$j\left(i\right)=\left\{\begin{array}{cc}\min \{pi,m\},\hfill & \mathrm{if}i\le s^{*},\hfill \\ \min \{i+k,m\},\hfill & \mathrm{if}i>s^{*},\hfill \end{array}\right.$$

(10)

where $k=(p-1)s^{*}+q,$ where $0\le q<p-1.$

Example 4. 

Let R be a chain ring with invariants $2,3,5,1,3$ and suppose that $j\left(1\right)=2,j\left(2\right)=3$ and $j\left(3\right)=3.$ Then, R is clearly a complete j-diagram with unique admissible function j. This means if there is another admissible function j’ such that R is also a complete j’-diagram, then $j=j^{\prime }.$ For the converse, note that if $j_1\left(1\right)=j_1\left(2\right)=j_1\left(3\right)=3$ which is an admissible function but R is not $j_1-$diagram. This means the existence of an admissible function j on $P_m$ is not enough to say R is j-diagram (either complete or not), see Definition 2. Thus, in general, the converse is not true.

Proposition 1. 

Let R be a finite commutative chain ring with $p,n,r,k,m.$ If $(p-1)\nmid k$ or $m\le k+s^{*},$ then R is complete.

Proof. 

Note that for $x\in U\left(R\right),$

$${(1+{\pi }^{s^{*}}x)}^p=\left\{\begin{array}{cc}1+p{\pi }^{s^{*}}u+{\pi }^{s^{*}p}{x}^p,\hfill & \mathrm{if}m>k+s^{*},\hfill \\ 1,\hfill & \mathrm{if}m\le k+s^{*},\hfill \end{array}\right.$$

(11)

where $u\in U\left(R\right).$ Thus, if $m\le k+s^{*},$ then clearly the series (7) is a complete j-diagram, and hence R is complete. Now, assume that $m>k+s^{*}.$ If $(p-1)\nmid k$, $q\ne 0,$ and hence $s^{*}p<k+s^{*}.$ It follows from Equation (11) that

$${(1+{\pi }^{s^{*}}x)}^p=1+{\pi }^{s^{*}p}{x}_1\mathrm{mod}{H}_{s^{*}p+1},$$

for some $x_1\in U\left(R\right).$ Furthermore, when $s>s^{*},$ ${\eta }_s={\gamma }_{s+k}^{-1}·{\delta }_s·{\gamma }_s,$ where ${\gamma }_s$ and ${\delta }_s$ are defined in Lemma 2 Thus, ${\eta }_s$ is an isomorphism. In case of $s\le s^{*},$ consider the map

$$\begin{array}{cc}& {\beta }_s:J^s/J^{s+1}⟶J^{sp}/J^{sp+1}\hfill \\ & x+J^{s+1}⟼x^p+J^{sp+1}\hfill \end{array}$$

One can prove easily that ${\beta }_s$ is well-defined, and moreover, is a monomorphism. For epimorphism, note that since $\overline{R}$ is a finite field, then ${\overline{R}}^p=\overline{R},$ a basic field. That is, if $y\in J^{sp}\backslash J^{sp+1},$ then $y={\pi }^{sp}{y}_0$ mod $J^{sp+1}$ where $y_1\in {\overline{R}}^{*}$. Then, there is $y_2$ such that $y_1=y_2^p$ and then $\beta \left(x^s{y}_2\right)=y$ mod $N^{ps+1}.$ Therefore, ${\beta }_s$ is an isomorphism and ${\eta }_s={\gamma }_{sp}^{-1}·{\beta }_s·{\gamma }_s,$ which means ${\eta }_s$ is an isomorphism. □

Corollary 1. 

Any finite commutative chain ring R with characteristic p is complete.

Proof. 

Since $n=1,$ then $k=m$ which means that $m<k+s^{*},$ and by Proposition 1, R is complete. □

Remark 4. 

By Proposition 1, when $q\ne 0$ $\left(\right(p-1)\nmid k)$ the j-diagram for H is independent of the Eisenstein polynomial ${\pi }^k=p\beta h.$ However, this is not true when $q=0$, i.e., $k+s^{*}=p{s}^{*}.$ Let $x\in U\left(R\right),$ by Equation (11),

$$\begin{array}{cc}\hfill {(1+{\pi }^{s^{*}}x)}^p& =1+p{\pi }^{s^{*}}x+{\pi }^{s^{*}p}{x}^p\mathrm{mod}{H}_{j\left(s^{*}\right)+1}\hfill \\ & =1+{\pi }^{s^{*}+k}({\left(\beta h\right)}^{-1}x+x^p)\mathrm{mod}{H}_{j\left(s^{*}\right)+1}.\hfill \end{array}$$

Thus, ${(1+{\pi }^{s^{*}}x)}^p=1$ mod $H_{j\left(s^{*}\right)+1}$ if and only if ${\left(\beta h\right)}^{-1}x+x^p=0$ mod $J$, i.e., $x^{p-1}+\beta =0$ in ${\overline{R}}^{*}.$

Proposition 2. 

If $m>k+s^{*},$ then ${\eta }_{s^{*}}$ is an isomorphism if and only if $-\beta \notin {\overline{R}}^{*}{}^{p-1}.$

Proof. 

If ${\eta }_{s^{*}}$ is an isomorphism, then ker ${\eta }_{s^{*}}=\left\{H_{s^{*}+1}\right\},$ which means that ${(1+{\pi }^{s^{*}}a)}^p\ne 1$ mod $H_{j\left(s^{*}\right)+1},$ for any $a\in {\overline{R}}^{*}.$ Hence, $x^{p-1}+\beta$ has no zeros in ${\overline{R}}^{*}$ and thus $-\beta \notin {\overline{R}}^{*}{}^{p-1}.$ The converse is direct by Theorem 3. □

The following theorem gives a characterization of incomplete chain rings.

Theorem 5. 

Suppose that R has invaraints $p,n,r,k,m$ with $m>k+s^{*}.$ Therefore, the subsequent hypotheses are equivalent:

(i) 

R is incomplete.

(ii) 

There is $\alpha \in R$ such that ${\alpha }^{p-1}+p=0.$

(iii) 

$p-1$ divides k and there exists $\alpha \in {\overline{R}}^{*}$ such that $-\beta ={\alpha }^{(p-1)}.$

Proof. 

Let (i) be satisfied, thus ker ${\eta }_{s^{*}}\ne 1$ because ${\eta }_{s^{*}}$ is surjective. In this case, there is $1+\alpha {\pi }^{s^{*}}$ in ker ${\eta }_{s^{*}}$ with

$${(1+\alpha {\pi }^{s^{*}})}^p=1+{\alpha }^p{\pi }^{s^{*}p}-\beta \alpha {\pi }^{s^{*}+k}\xi =1$$

$\mathrm{mod}{H}_{s^{*}p+1},$ where $\xi \in H.$ However, the above equation holds when $p{s}^{*}=s^{*}+k$ and ${\alpha }^p+\beta \alpha =0$ mod $\pi .$ Now, assume that $(p-1)\mid k,$ then ker ${\eta }_{s^{*}}\cong$ ker $f,$ where f is a homomorphism; $f:\overline{R}\to \overline{R}$ and $f\left(\alpha \right)={\alpha }^p+\beta \alpha .$ Moreover, ker $f=1$ if and only if $x^p+\beta x$ has only zero solution. Thus, ${\left({\beta }_1{h}_1\pi \right)}^{s^{*}}$ is a root of $x^{p-1}+p$ in $R,$ where ${\beta }_1={\alpha }^{-1}$ and $h_1^{p-1}=h.$ The remaining hypotheses follow immediately by Proposition 2 and Theorem 3. □

Corollary 2. 

If R is an incomplete chain ring, then ker ${\eta }_{s^{*}}$ is of rank $p.$

Proof. 

Since any element in ker ${\eta }_{s_1}$ is of the form $1+\alpha {\pi }^{s^{*}},$ where $\alpha$ is a zero of the polynomial $x^p+\beta x.$ Thus, the order of ker ${\eta }_{s^{*}}$ is exactly p since there are p distinct zeros of $x^p+\beta x$ in $\overline{R}.$ □

Lemma 5. 

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

(a) 

If $n>2$ or $n=2$ and $t>s^{*}.$ Then, $m>k+s^{*}.$

(b) 

If $n\le 2,$ $t\le s^{*}.$ Then, $m\le k+s^{*}.$

Proof. 

Part (b) is obvious; note that if $n=1,$ then $m=k=t$. For part (a),

$$\begin{array}{cc}\hfill m& =(n-1)k+t=(n-1)((p-1)s^{*}+q)+t\hfill \\ & =(n-1)(p{s}^{*}-s^{*}+q)+t\hfill \\ & =(n-1)(p{s}^{*}+q))-(n-1)s^{*}+t\hfill \\ & =(n-1)(k+s^{*})+t-(n-1)s^{*}\hfill \\ & =(k+s^{*})+(n-2)k+t-(n-1)s^{*}\hfill \\ & =(k+s^{*})+(n-2)k++(n-2-n+1)s^{*}\hfill \\ & =(k+s^{*})+(n-2)k+t-s^{*}.\hfill \end{array}$$

However, $s^{*}<k$ and $n>2,$ then $(n-2)k-s^{*}>0,$ thus, $m>k+s^{*}.$ □

Proposition 3. 

If $s>k+s^{*},$ then $s\in R\left(j\right).$ Furthermore,

$$c_0=\mid P_m\backslash R\left(j\right)\mid =\left\{\begin{array}{cc}m-\lfloor \frac{m}{p}\rfloor ,\hfill & \mathrm{if}m<k+s^{*},\hfill \\ k,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

(12)

where $\lfloor x\rfloor$ means the greatest integer that is less than or equal to $x.$

Proof. 

Let$s=k+s^{*}+e,$ for some $e>0,$ then clearly $s=j(s^{*}+e),$ which means $s\in R\left(j\right).$ If $m\ge k+s^{*},$ then it is clear that $P_m\backslash R\left(j\right)=\{s\in P_m:p\nmid s,1\le s\le k+s^{*}\}.$ Thus,

$$c_0=k+s^{*}-\lfloor \frac{k+s^{*}}{p}\rfloor =\lfloor \frac{(p-1)s^{*}+q+s^{*}}{p}\rfloor =\lfloor \frac{p{s}^{*}+q}{p}\rfloor =k+s^{*}-s^{*}=k.$$

For the case $m<k+s^{*},$ $P_m\backslash R\left(j\right)=\{s\in P_m:p\nmid s,s<m\},$ and thus,

$$c_0=m-\lfloor \frac{m}{p}\rfloor .$$

□

Proposition 4. 

Assume the admissible function j satisfies: if $j\left(s\right)\ge p,$ then $s\in R\left(j\right)$ for all $s.$ Then, $H_s^{p^i}=H_{j^i\left(s\right)},$ in particular, $H^{p^i}=H_{j^i\left(1\right)}.$

Proof. 

The proof is conducted by induction on $i.$ First, let $i=1,$ and note that $H_s^p\subseteq H_{j\left(s\right)}.$ If $y\in H_{j\left(s\right)},$ then $y=u_{j\left(s\right)}{y}_1,$ where $u_{j\left(s\right)}\in U_{j\left(s\right)}$ and $y_1\in H_{j\left(s\right)+1}.$ Moreover, $u_{j\left(s\right)}=u_s^p$ for some $u_s\in U_s,$ and $y_1=u_{j\left(s\right)+1}{y}_2,$ where $u_{j\left(s\right)+1}\in U_{j\left(s\right)+1}$ and $y_1\in H_{j\left(s\right)+2}.$ Since

$$j\left(j\right(s)+2)\ge j\left(j\right(s)+1)\ge j\left(1\right)=p,$$

(13)

it follows that $j\left(s\right)+2,j\left(s\right)+1$ are elements of $R\left(j\right).$ This means $u_{j\left(s\right)+1}=u_{s_1}^p.$ As we proceed, we get $y=y_0^p,$ and thus $H_{j\left(s\right)}\subseteq H_s^p.$ Therefore, $H_{j\left(s\right)}=H_s^p.$ If $i>1,$ observe that $H_s^{p^i}={\left(H_s^{p^{i-1}}\right)}^p,$ and hence the conclusion is drawn from the induction step. □

Next, we give an important result; that is useful in capturing the structure of the subgroups $H_s$ of H via the following j-subdiagram:

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

Which in turn helps us to investigate the group of automorphisms of $R,$ for more details see Remark 4.2.10 in [4].

The following result for finite abelian groups can be easily proved.

Lemma 6. 

Let G be a finite direct product of cyclic groups, each of order $p^e$ for some $e\ge 1.$ Let $U^{\prime }$ be a subgroup of G which is a direct product of cyclic groups $B_i,$ each of order $p^{e^{\prime }},$ and for which $rank\left(U^{\prime }\right)=rank\left(G\right).$ Then, $G=A_1\otimes A_2\otimes \cdots \otimes A_s$ such that each $A_i$ is a cyclic group and $B_i=U^{\prime }\cap A_i.$ Moreover, for any $s\ge 1,$ $G^{p^{e-e^{\prime }-c}}=\{g\in G:g^{p^s}\in U^{\prime }\},$ where $c=\mathit{min}\{e-e^{\prime },s\}.$

Theorem 6. 

Let $A=A_1>A_2>\cdots >A_m=1$ be a complete $j-$diagram for an abelian p-group A, $0\le s<m$ and $L_{m-s}=\{i:m-s\le i\le m\}.$ Let $j^{\prime }=j{\mid }_{L_{m-s}}$ and $X_s=\{i\in L_{m-s}:i\notin R\left(j^{\prime }\right)\}.$ Then,

(a) 

$A_{m-s}={\otimes }_{i\in X_s}{U}_i.$

(b) 

$X_s=B\cup C$ satisfying the following conditions:

(i) 

$B\cap R\left(j\right)=\varphi ,$

(ii) 

There exists a subset D of $P_m\backslash R\left(j\right)$ disjoint form B and a one to one mapping $j_1:D\to C$ such that for any $i\in D,j_1\left(i\right)=j^{e_i}\left(i\right)$ for some $e_i\ge 1.$ Suppose $P^{\prime }=(P_m\backslash R\left(j\right))\backslash (D\cup B),$ $E={\otimes }_{i\in P^{\prime }}{U}_i$ and $F^{\prime }={\otimes }_{i\in (B\cup D)}{U}_i.$

(iii) 

$A=E\otimes F^{\prime },$ $A_{m-s}=\left({\otimes }_{i\in B}{U}_i\right)\otimes \left({\otimes }_{i\in D}{U_i}^{p^{e_i}}\right)\subseteq F$ and $rank\left(A_{m-s}\right)=rank\left(F^{\prime }\right).$

(iv) 

Let $c\ge 0$ and $P_1=\{i\in P^{\prime }:\nu \left(i\right)\le c\}.$ Then,

$$G=\{x\in A:x^{p^c}\in A_{m-s}\}=\left({\otimes }_{i\in P_1}{U}_i\right)\otimes \left({\otimes }_{i\in B}{U}_i\right)\otimes \left({\otimes }_{i\in D}{U}_i^{p^{e_i-\min \{e_i,c\}}}\right).$$

(14)

Proof. (a)

Put $K_i=A_{m+i-s-1},$ $1\le i\le s+1.$ Then, $A_{m-s}=K_1>K_2>\cdots >K_{s+1}=1$ is a complete $j^{*}-$diagram, where $j^{*}:P_{s+1}\to P_{s+1}$ is given by $j^{*}\left(i\right)=j(m-s-1-i)-(m-s-1).$ Note that $K_i=K_{i+1}\times U_{m-s-1+1}.$ By [Theorem 1 [3]],

$$A_{m-s}={\otimes }_{i\notin R\left(j^{*}\right)}{U}_{m-s-1+i}={\otimes }_{i\in X_s}{U}_i.$$

(15)

(b) Write $X_k=B\cup C$ with $B\subseteq P_m\backslash R\left(j\right)$ and $C\subseteq R\left(j\right).$ It is clear that $m\notin C.$ Suppose that all $U_i$ have the same rank. For each $i\in C,$ there exists a positive integer $e_i,$ and a unique $i^{\prime }$ such that $i=j^{e_i}\left(i^{\prime }\right).$ Thus, $U_i=U_{j^{e_i}\left(i^{\prime }\right)}=U_{i^{\prime }}^{p^{e_i}}\subseteq U_{i^{\prime }}.$ It follows that from the definition of $j,$ $D=\{i^{\prime }:i\in C\}$ is disjoint from B and there exists a bijection $j_1:D\to C,$ such that $j_1\left(i\right)=j^{e_i}\left(i\right).$ This proves (ii). Hence,

$$A_{m-s}={\otimes }_{i\in B}{U}_i{\otimes }_{i\in D}{p}^{e_i}{U}_i\subseteq F^{\prime }.$$

This proves (iii). Finally, consider $c\ge 0$ and $G=\{x\in A:x^{p^c}\in A_{m-s}\}.$ Observe that any $x\in A$ is in G if and only if each of its components in the decomposition $A=E\otimes F^{\prime }$ is in $G.$ For any $x\in E,$ $x^{p^c}\in A_{m-s}$ implies $x^{p^c}=1.$ For any $i\in P^{\prime },$ $U_i^{p^c}=1,$ whenever $\nu \left(i\right)\le c.$ So $E\cap G={\times }_{i\in P_1}{U}_i.$ Consider any $i\in D.$ Now, $U_i\cap G=\{x\in U_i:x^{p^c}\in U_i^{p^{e_i}}\}.$ As the order of $U_i$ is $p^{\nu \left(i\right)}$ and the order of $U_i^{p^{e_i}}$ is $p^{\nu \left(i\right)-e_i},$ then by Lemma 6,

$$U_i\cap G=U_i^{p^{e_i-\min \{e_i,c\}}}.$$

Thus,

$$G=\left({\otimes }_{i\in P_1}{U}_i\right)\otimes \left({\otimes }_{i\in B}{U}_i\right)\otimes \left({\otimes }_{i\in D}{U}_i^{p^{e_i-\min \{e_i,c\}}}\right).$$

□

4. Conclusions

In this article, we have investigated j-diagrams for one group of finite commutative chain rings. Under certain conditions concerning the invariants $p,n,r,k,m$ and Eisenstein polynomials, we proved the existence and uniqueness of such j-diagrams. These j-diagrams have been found helpful tools in investigating finite chain rings.