Recapturing the Structure of Group of Units of Any Finite Commutative Chain Ring

Abstract

A finite ring with an identity whose lattice of ideals forms a unique chain is called a finite chain ring. Let R be a commutative chain ring with invariants $p,n,r,k,m.$ It is known that R is an Eisenstein extension of degree k of a Galois ring $S=GR(p^n,r).$ If $p-1$ does not divide k, the structure of the unit group $U\left(R\right)$ is known. The case $(p-1)\mid k$ was partially considered by M. Luis (1991) by providing counterexamples demonstrated that the results of Ayoub failed to capture the direct decomposition of $U\left(R\right).$ In this article, we manage to determine the structure of $U\left(R\right)$ when $(p-1)\mid k$ by fixing Ayoub’s approach. We also sharpen our results by introducing a system of generators for the unit group and enumerating the generators of the same order.

1. Introduction

In this article, we consider finite commuative chain rings, although some results still correct under more general situation. Ayoub called these rings (cf. [1]) primary homogeneous rings. Such rings arise in various places; in Coding Theory (cf. for example [2,3]); in Geometry (cf. [4]). It is well-known that every finite chain ring R is an Eisenstein extension of some degree k over a Galois ring of the form $S=GR(p^n,r).$ There are positive integers $p,n,r,k$ and m associated with R, called invariants of $R.$ Our main aim in this study is to obtain the structure of the multiplicative group $U\left(R\right)$ of $R.$ In (1972) (cf. [1]), Ayoub obtained various results based on ideas (cf. [5]) regarding j-diagrams for abelian p-groups.

One major result claims that the factorization of an abelian $p-$group with incomplete j-diagram can be completely obtained by the mentioned diagram. This idea was then used to find the structure of $U\left(R\right)$ when R is not necessarily finite (Theorem 3, Section 4 [1]). However, later, it turned out that incomplete j-diagrams failed to determine the exact decomposition of multiplicative groups of some examples introduced by Luis [6]. On the other hand, Hou [7] by different approach gave the structure of $U\left(R\right)$ in a special case; $p\nmid k.$ This sudy demonstrates that the result (Theorem 3, [1]) is still valid in some cases, for instance the case studied in [7]. Additionally, we enhance the idea that incomplete j-diagrams are generally not enough to obtain the decomposition of bounded p-groups. That is, there is a relation amongst the generators which plays essential role in determining the structure. That relation depends not only on the related incomplete j-diagram, but also on the Eisenstein polynomial by which R is constructed over its coefficient subring $GR(p^n,r).$ However, with this relation being taken into consideration, we manage to make incomplete j-diagrams succeed in recapturing the structure of multiplcative groups of finite chain rings. In addition, a set of linearly independent generators for $U\left(R\right)$ is provided and, further, the number of such generators is computed for each possible order. Finally, we give the correct version of ([5], p. 458).

2. Preliminaries and Notations

In what follows, R will denote a finite commutative chain ring of characteristic $p^n$ and nonzero radical N (the case when $N=0$, R is a field) with index of nilpotency $m.$ The residue field $F=R/N$ is a finite field of order $p^r.$ We now state some facts and introduce notations that we shall use throughout. For the details, we refer the reader to [8] for finite chain rings and [1,5] for j-diagrams.

The ring R contains a subring (coefficient subring) S of the form $S=GR(p^n,r)\cong Z_{p^n}\left[a\right],$ where a is an element of S of multiplicative order $p^r-1.$ If $\pi \in N\backslash N^2,$ it is easy to see that $N=\left(\pi \right)$, since $N/N^2$ is of 1 dimension over $F.$ In this case, R is expressed as:

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

(1)

(as $S-$module) where k is the greatest integer $i\le m$ such that $p\in N^i.$ It follows that

$${\pi }^k=p\sum _{i=0}^{k-1}{s}_i{\pi }^i,$$

(2)

where $s_i\in S.$ This means that $\pi$ is a root of an Eisenstein polynomial over S of the form:

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

(3)

The positive integers $p,n,r,k$ and m are called invariants of $R.$ Furthermore, $m=(n-1)k+t,$ for some $1\le t\le k.$ Let $H_s=1+N^s,$ $s\in P_m=\{1,2,\dots ,m\}$ and consider the following filtering:

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

(4)

joined with a function j defined as:

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

(5)

where $k=(p-1)u+v,$ $0\le v<p-1.$ Note that if Char $\left(R\right)=p,$ i.e., $n=1,$ put $k=m.$ The series (4) with j defined in (5) and the p-th power homomorphisms ${\eta }_s$ from $H_s/H_{s+1}$ into $H_{j\left(s\right)}/H_{j\left(s\right)+1}$ form what we call j-diagram. However, we refer to the series (4) when we mention j-diagram. If there exists $m\ne s^{\prime }\in P_m$, such that ${\eta }_{s^{\prime }}$ is not an isomorphism, we then call the series (4) incomplete j-diagram at $s^{\prime }$ and complete j-diagram otherwise (all ${\eta }_s$ are isomorphisms).

We denote, by $U\left(R\right)$, the units group of R, and so one can verify that

$$U\left(R\right)=<a>\otimes H,$$

(6)

where $<a>$ is a cyclic group of order $p^r-1$ and $H=1+\pi R$ is the p-Sylow subgroup of $U\left(R\right).$ The structure problem of $U\left(R\right)$ is then reduced to that of $H.$ After Ayoub [1] we call H the One Group of $U\left(R\right).$ If $N_m\left(e\right)$ is the number of basis elements in H whose orders $p^e,$ then H is decomposed as:

$$H\cong {\otimes }_e{\left(C_{p^e}\right)}^{N_m\left(e\right)},$$

(7)

where $C_{p^e}$ is a cyclic group of order $p^e.$ We define p-rank $N_m$ of H as

$$p^{N_m}=[H:H^p].$$

(8)

If $\pi \in$ N$\backslash N^2$ fixed, then from Eisenstein polynomial (3)

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

(9)

where $\beta \in <a>$ and $h\in H.$ Now, let ${\left\{{\alpha }_i\right\}}_{1\le i\le r}$ be a representatives system in R for a basis of F over its prime field ( ${\mathbb{Z}}_p$). Set

$$w_{is}=1+{\alpha }_i{\pi }^s(1\le i\le r\mathrm{and}s\notin R\left(j\right)),$$

(*)

where $R\left(j\right)$ is the range of $j.$ For each $s\in P_m,$ let $U_s$ be a subgroup of H generated by ${\left\{w_{is}\right\}}_{1\le i\le r}.$ Hence one can easily show that $H_s=U_s\times H_{s+1}.$ Now, if (4) is a complete $j-$diagram, then, from (cf. [1]), the system $\left\{w_{is}\right\}$ (*) forms a basis for $H.$

Proposition 1

(Theorem 3, [1]). Assume that (4) is a complete j-diagram. Then,

$$H={\otimes }_{s\notin R\left(j\right)}{U}_s.$$

(10)

The purpose of the present paper is to establish a basis for H when (4) is an incomplete j-diagram at $u.$ Moreover, $N_m\left(e\right)$ and $N_m$ are determined in both cases (complete and incomplete) in Theorems 3 and 4.

Unless otherwise mentioned, all of the symbols stated above will retain their meanings throughout and, additionally, if $s\in P_m,$ ${\nu }_m\left(s\right)$ is the least positive number fulfilling $j^{{\nu }_m\left(s\right)}\left(s\right)=m.$

3. The Determination of the Structure of $\mathit{H}$

Throughout this section, $k=k_1{p}^l,$ where $(k_1,p)=1$ and $l\ge 0.$ The following proposition is useful and it will be needed later.

Proposition 2.

If $m>k+u,$ then the following conditions are equivalent:

(i) The series (4) is an incomplete j-diagram at $u.$

(ii) The polynomial $x^{p-1}+p$ has a root in $R.$

(iii) $p-1\mid k$ and $\beta \in {F^{*}}^{p-1}.$

Proof. 

Assume that (i) holds, then ker ${\eta }_u\ne 1$ since ${\eta }_u$ is onto. Then there is $1+\alpha {\pi }^u\in$ ker ${\eta }_u$, such that ${(1+\alpha {\pi }^u)}^p=1+{\alpha }^p{\pi }^{up}-\beta \alpha {\pi }^{u+k}\xi =1\mathrm{mod}{H}_{up+1},$ where $\xi \in H.$ This only happens when $pu=u+k$ and ${\alpha }^p-\beta \alpha =0$ mod $p.$ Hence, if $(p-1)\mid k,$ ker ${\eta }_u$ is isomorhpic to that of the homomorphism, $f:F\to F$ defined by: $f\left(\alpha \right)={\alpha }^p-\beta \alpha .$ However, ker $f\ne 1$ if and only if $x^p-\beta x$ has nonzero solutions in $F.$ The remaining hypotheses follow immediately. □

Definition 1.

We call R incomplete (complete) chain ring if one of the equivalent conditions in Proposition 2 holds (otherwise).

Corollary 1.

If R is an incomplete chain ring, then ker ${\eta }_u$ is of rank $p.$

Remark 1.

In the light of Proposition 2 and its corollary, $v=0$ and $k_1=(p-1)s_0,$ i.e., $u=s_0{p}^l.$ Furthermore, if η is the composition of ${\eta }_{s_0},\dots ,{\eta }_u$ then ker η is of rank $p.$ Let

$$\xi =1+{\alpha }_1{\pi }^{s_0}$$

(11)

be a generator of ker $\eta .$

Assume that f is as mentioned in the proof of Proposition 2 and $\lambda =l+1$. Construct the following system:

$$\left\{\begin{array}{c}{w}_{is}=1+{\alpha }_i{\pi }^s,\mathrm{for}1\le i\le r\mathrm{and}s\notin R\left(j\right),\hfill \\ \gamma =1+{\alpha }_0{\pi }^{up},\hfill \\ \mathrm{where}{\alpha }_1^{p^{\lambda }}-\beta {\alpha }_1^{p^{\lambda -1}}=0\mathrm{and}{\alpha }_0\notin \mathrm{Im}f.\hfill \end{array}\right.$$

(**)

The following proposition is stated without proof, since it involves the same ideas of that in the complete case.

Proposition 3.

The set $\{w_{is},\gamma \}$ ($**$) represents a system of generators for $H.$

Put

$$B=\{(i,s):1\le i\le r\mathrm{and}s\notin R\left(j\right)\}\backslash \left\{(1,s_0)\right\}.$$

(12)

Observe that $\xi =w_{1s_0}\in$ ker $\eta$ and then

$${\xi }^{p^{\lambda }}=\prod _{(i,s)\in B}{w}_{is}^{a_{is}}·{\gamma }^{a_0}=y^{p^{\mu }},$$

(13)

where $a_{is}$ and $a_0$ are positive integers that are divisible by $p,$ i.e., $\mu \ge 1.$ Note that $\mu$ will keep its meaning throughout the paper, as described here. Let $H_0$ be a subgroup of H that is generated by ${\{\gamma ,w_{is}\}}_{(i,s)\in B}.$ Thus, by (cf. [1]), $H_0$ has a direct decomposition,

$$H_0={\otimes }_{(i,s)\in B}<w_{is}>\otimes <\gamma >.$$

(14)

Theorem 1.

Let R be an incomplete chain ring and $\lambda \le \mu ,$ then

$$H\cong H_0\otimes D,$$

(15)

where D is a cyclic group of order $p^{\lambda }.$

Proof. 

By Equation (13), it is clear that $H_0\cap <\xi >=<{\xi }^{p^{\lambda }}>$ and also $H_0$ is a p-pure subgroup of $H;$ every element is not of a smaller p-height in $H_0$ than in $H.$ Thus, by the results from (cf. [9]), $H_0$ is a direct summand of $H.$ This finishes the proof. □

Corollary 2.

If $\zeta =\xi y^{p^{\mu -\lambda }},$ then ${\{\gamma ,\zeta ,w_{is}\}}_{(i,s)\in B}$ is a basis for $H.$

Corollary 3.

The system ${\{\gamma ,\xi ,w_{is}\}}_{(i,s)\in B}$ is a basis for H if and only if $o\left(\xi \right)=\lambda$ (the order of ξ).

Remark 2.

(1) 

It is worthy to mention that Theorem 1 coincides with Ayoub’s results (Theorem 3 Section 4, [1]). That means in such case, incomplete j-diagrams succeed in retrieving the structure of $H.$

(2) 

If $p\nmid k,$ then obviously $\lambda =1\le \mu$ and, thus, the result that is given in [7] is just a particular case of Theorem 1.

From now on, we assume $\lambda >\mu .$ Consider Equation (13) and let

$$\left\{\begin{array}{c}A=\{(i,s)\in B\}\cup \left\{(i^{\prime },s^{\prime })\right\},\hfill \\ w_{i^{\prime }{s}^{\prime }}=\gamma \mathrm{and}{a}_{i^{\prime }{s}^{\prime }}^{\prime }=a_0.\hfill \end{array}\right.$$

(16)

Lemma 1.

If R is an incomplete chain ring and $\lambda >\mu ,$ then there is a positive integer d and p-pure subgroups $G_i$ of H, such that $G_{i+1}\subseteq G_i,$ $1\le i\le d-1.$

Proof. 

For $(i,s)\in A,$ let $a_{is}=v_p\left(a_{is}^{\prime }\right),$ where $v_p$ is the p-adic valuation. Subsequently, it is clear that $\mu \le a_{is}$ for all $(i,s)\in A.$ Choose $(i_1,s_1)\in A$ such that $a_{i_1{s}_1}=\mu .$ If there are more than one of s satisfying $a_{is}=\mu ,$ choose the smallest one. Let $h_{is}$ be defined as $a_{is}^{\prime }=h_{is}{p}^{a_{is}},$ $(h_{is},p)=1.$ If $B_1=A\backslash \left\{(i_1,s_1)\right\},$ $h_1=h_{i_1{s}_1}$ and $w_1=w_{i_1{s}_1}^{h_1}$, then Equation (13) can be rewritten as:

$$w_1^{p^{\mu }}={\left({\xi }^{p^{\lambda -\mu }}\tau \right)}^{p^{f_1}},$$

(17)

where $\tau ={\prod }_{(i,s)\in B_1}{w}_{is}^{{\delta }_{is}}, {\delta }_{is}=-h_{is}{p}^{a_{is}-{\mu }_1}$ and $f_1=\min \{a_{is}:(i,s)\in B_1\}.$ Define $G_1$ to be the subgroup of H that is generated by ${\{w_{is},\xi \}}_{(i,s)\in B_1}.$ Thus, from (17), $G_1\cap \left\{w_1\right\}=<w_1^{p^{\mu }}>.$ Because $\mu \le f_1,$ then the p-height of $w_1$ in $G_1$ is not lesser than that in H and thus $G_1$ is a p-pure subgroup of $H.$ Now, if we raise Equation (17) to $p^{\nu \left(s_1\right)-\mu },$ we obtain a new relation among the generators of $G_1,$

$${\displaystyle {\xi }^{p^{\lambda +\nu \left(s_1\right)-\mu }}=\prod _{(i,s)\in B_1^{\prime }}{w_{is}}^{b_{is}^{\prime }},}$$

(18)

where $b_{is}^{\prime }=c_{is}{p}^{a_{is}+\nu \left(s_1\right)-\mu },$ and $B_1^{\prime }\subseteq B_1$, because some generators might disappear due to raising (17) to a power of $p.$ Note that since the generators on the right side of Equation (18) are linearly independent, thus the left side cannot vanish. Again, there is $(i_2,s_2)$ such that $a_{i_2{s}_2}=a_2\ge \mu$ is the smallest $a_{is}$ ($(i,s)\in B_1^{\prime }$) and then

$${\mu }_2=b_{i_2{s}_2}=v_p\left(b_{i_2{s}_2}^{\prime }\right)=a_2+\nu \left(s_1\right)-\mu$$

(19)

is the smallest $b_{is}=v_p\left(b_{is}^{\prime }\right),$ $(i,s)\in B_1^{\prime }.$ Assume that $w_2=w_{i_2{s}_2}^{h_{i_2{s}_2}},$ then we have a similar situation to that of $w_1.$ Let $B_2=B_1^{\prime }\backslash \left\{(i_2,s_2)\right\}$ and $G_2$ be a subgroup of $G_1$ generated by $\xi$ and all $w_{is}$ with $(i,s)\in B_2.$ By a similar arguement, $G_2$ is a p-pure subgroup of $G_1$ and, thus, transitively (Lemma 26.1, [9]) $G_2$ is a p-pure subgroup of $H.$ Therefore, after a finite number d of similar processes, we obtain:

$$\left\{\begin{array}{c}I=\{(i_1,s_1),\dots ,(i_d,s_d)\}\subseteq A;\hfill \\ J=\{w_1,\dots ,w_d\},w_q=w_{i_q{s}_q}^{h_{i_q{s}_q}}\mathrm{and}(i_q,s_q)\in I;\hfill \\ G_d\subseteq G_{d-1}\subseteq \dots \subseteq G_1\subseteq H,G_i=G_{i+1}\times <w_i>;\hfill \\ {\mu }_1=\mu \le {\mu }_2\le \dots \le {\mu }_d\mathrm{subjected}\mathrm{to}\left\{\begin{array}{c}{w}_i^{p^{{\mu }_i}}=y_i^{p^{f_i}},\hfill \\ y_i\in G_i\mathrm{and}{\mu }_i\le f_i;\hfill \end{array}\right.\hfill \\ {\xi }^{p^{{\lambda }_0}}=1,{\lambda }_0=\lambda +{\sum }_{i=1}^d(\nu \left(s_i\right)-{\mu }_i).\hfill \end{array}\right.$$

(20)

 □

Remark 3.

Based on notations that were introduced in Lemma 1, put $\mathsf{\Omega }=\{s_1,\dots ,s_d\}$ and let

$$G_d=\left\{\begin{array}{cc}\prod _{s\in \Lambda }{U}_s\times \prod _{s\in {\mathsf{\Omega }}^{\prime }}{U}_s^{*}\times <\xi >,\hfill & \mathrm{if}(i^{\prime },s^{\prime })\in I,\hfill \\ \prod _{s\in \Lambda }{U}_s\times \prod _{s\in {\mathsf{\Omega }}^{\prime }}{U}_s^{*}\times <\xi >\times <\gamma >,\hfill & \mathrm{if}(i^{\prime },s^{\prime })\notin I,\hfill \end{array}\right.$$

(21)

where $\Lambda =\{P_m\backslash R\left(j\right)\}\backslash {\mathsf{\Omega }}^{\prime }$ and ${\mathsf{\Omega }}^{\prime }=\mathsf{\Omega }\cup \left\{s_0\right\}.$

Theorem 2.

Let R be an incomplete finite chain ring with invariants $p,n,r,k,m$ such that $\lambda >\mu .$ Then,

$$H={\otimes }_{s\in \Lambda }{U}_s{\otimes }_{i=1}^d(U_{s_i}^{*}\otimes D_i)\otimes D\otimes C,$$

(22)

(i) 

$U_s$ is a direct product of r cyclic groups of order $p^{\nu \left(s\right)}.$

(ii) 

$U_{s_i}^{*}$ is a direct product of $r-1$ cyclic groups of order $p^{\nu \left(s_i\right)},$ and $D_i$ is a cyclic group of order $p^{{\mu }_i}.$

(iii) 

D is a cyclic group of order ${\lambda }_0=\lambda +{\sum }_{i=1}^d(\nu \left(s_i\right)-{\mu }_i).$

(iv) 

C is a cyclic group of order $p^{\nu \left(u\right)-1}$ if $(i^{\prime },s^{\prime })\notin I$ and $C=<1>$ otherwise.

Proof. 

Because, for each $1\le i\le d-1,$ $G_{i+1}$ is a p-pure subgroup of $G_i$ (Lemma 1) and also a bounded p-group, then (Theorem 27.5, [9]) $G_{i+1}$ is a direct summand of $G_i;$ there is a subgroup $L\subseteq G_i$, such that $G_i=G_{i+1}\otimes L.$ Thus, based on (20) and Remark 3,

$$L\cong G_i/G_{i+1}=G_{i+1}\times <w_i>/G_{i+1}=<w_i>/G_{i+1}\cap <w_i>.$$

By the argument of the proof of Lemma 1, $G_{i+1}\cap <w_i>=<w_i^{p^{{\mu }_i}}>.$ Hence, $G_i=G_{i+1}\otimes D_i,$ where $D_i$ is a cyclic group of order $p^{{\mu }_i}.$ Therefore, the decomposition (22) is just a result of successive application to the above argument. Note that $D=<\xi >.$ □

Corollary 4.

With the same assumption in Theorem 2, $N_m=cr+1.$

Corollary 5.

$\left\{w_{is}\right\}((i,s)\in B\backslash I),$$\left\{w_i{y}_i^{-p^{f_i-{\mu }_i}}\right\}$ ($1\le i\le d$), ξ and η if $(i^{\prime },s^{\prime })\notin I$ form a basis of $H.$

The following example demonstrates the failure of Ayoub’s results (Theorem 3, [1]); non-isomorphic abelian p-groups may have the same incomplete j-diagrams. Furthermore, it shows how Eisenstein polynomials play a key role in factorizing $H.$

Example 1.

Let R be a finite chain ring with invariants $2,3,1,2,6$ and ${\pi }^2=2.$ Forward computation leads to

$${\xi }^{2^2}=1+{\pi }^5={(1+{\pi }^3)}^2,$$

(23)

where $\xi =1+\pi .$ Note that $H_0$ here is not p-pure subgroup of $H_R.$ Since $3\notin R\left(j\right),$ $(1,3)\in I,$ $w_1\ne \gamma$ and $\mu =1,$ thus from Theorem 2

$$H_R=C_{2^3}\otimes C_2\otimes C_2.$$

(24)

Now, if we consider T is a finite chain ring with same invariants $2,3,1,2,6$ (same incomplete j-diagram) and different Eisenstein polynomial $x^2-2(1+x).$ Let $\xi =1+\theta ,$ where $\theta =1+a$ (root of $x^2-2(1+x)$) and a is a root of $x^2-3.$ Then, ${\xi }^{2^2}=1$ and, hence, from Theorem 1,

$$H_T=C_{2^2}\otimes C_2\otimes C_{2^2}.$$

(25)

Next, we state the correct version of (cf. p. 458, [1]).

Corollary 6.

Let R and T be two incomplete chain rings with same invariants $p,n,r,k,m$ and same Ω and ${\left\{{\mu }_i\right\}}_{1\le i\le d}.$ Then, $H_R\cong H_T.$

In general, Lemma 1 gives an algorithm for calculating $\mathsf{\Omega }$ and ${\mu }_i^{\prime }$s. The following example illustrates that algorithm.

Example 2.

Let R be an incomplete chain ring with invariants $p,n,r,k,m$ and ${\pi }^k=p\beta h,$ where $h\in H\left(R_0\right)$ and $R_0\subset R$ is an Eisenstein extension over S by $x^{p-1}+p,$ i.e., $R_0=S\left[{\pi }_0\right],$ ${\pi }_0$ is a root of $x^{p-1}+p$ (Proposition 2). Clearly, $R_0$ is an incomplete chain ring with $p,n,r,p-1,m_1,$ where $m_1=(n-1)(p-1)+t_1$ and $t_1=\lceil \frac{t}{k_1{p}^l}\rceil .$ Consider the (admissible) function $j_0$ with

$$H_1\left(R_0\right)>H_2\left(R_0\right)>\dots >H_{m_1}\left(R_0\right)=1.$$

A similar equation of that in (13),

$${\xi }_0^p=\prod _{(i,s)\in A}{\omega }_{is}^{a_{is}}=y^p,$$

(26)

where ${\xi }_0=1+{\pi }_0.$ Let $R_1=R_0\left[{\pi }_1\right]$ be an Eisenstein extension over $R_0$ by $x^p-{\pi }_0{h}_1,$ where $h_1^{p-1}=h.$ $R_1$ is an incomplete chain ring with invariants $p,n,r,p,m_2$, such that $m_2=p{m}_1.$ Now, put ${\pi }_0={\pi }_1^p{h}_1^{-1}$ in (26) and after forward computations we get

$${\xi }_1^{p^2}=x_{s_1}^p\dot{y},$$

(27)

where ${\xi }_1=1+{\pi }_1,$ $y\in H_s{\left(R_1\right)}^p$ and $s>s_1=p(p-1)+1.$ If $j\left(s_1\right)<m_1,$ then based on Lemma 1 ${\mathsf{\Omega }}_1=\left\{s_1\right\}.$ Continuing in this way and after l steps ($k=s_0(p-1)p^l$), R is an Eisenstein extension over $R_l$ by $x^{s_0}-{\pi }_l\alpha ,$ where ${\alpha }^{(p-1)p^l}=\beta .$ Thus, as a result $\mathsf{\Omega }=\{s_0{s}_1,\dots ,s_0{s}_l\},$ where $s_i=(p-1)(p^i+p^{i-1}+\dots +1)+1$ for $1\le i\le l.$

4. Enumeration of Generators of the Same Order

In this section, we compute p-rank $N_m$ and $N_m\left(e\right)$ for the decomposition that is given in (15) and (22). However, first we determine $N_m$ and $N_m\left(e\right)$ for the complete case (10). If we denote $c=\mid P_m\backslash R\left(j\right)\mid ,$ then using the j-diagram it is not hard to prove the following:

$$c=\left\{\begin{array}{cc}m-\lfloor \frac{m}{p}\rfloor ,\hfill & \mathrm{if}m<k+u,\hfill \\ k,\hfill & \mathrm{otherwise},\hfill \end{array}\right.$$

(28)

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

$$m_0=\left\{\begin{array}{cc}k+t,\hfill & \mathrm{if}t<u,\hfill \\ t,\hfill & \mathrm{if}t\ge u.\hfill \end{array}\right.\left(n>1\right)$$

(29)

Note that, if $n=1,$ $m=m_0=k.$ By the j-diagram, one can easily prove that ${\nu }_m\left(s\right)={\nu }_{m_0}\left(s\right)+n-2$ when $t<u,$ and ${\nu }_m\left(s\right)={\nu }_{m_0}\left(s\right)+n-1$ when $t\ge u.$ Thus, for $e\ge 1,$ $N_m(e+(n-2))=N_{m_0}\left(e\right)$ if $t\le u$ and $N_m(e+(n-1))=N_{m_0}\left(e\right)$ otherwise. Hence, it suffices to obtain $N_{m_0}\left(e\right).$

Theorem 3.

Assume that R is a complete chain ring, then $N_m=cr$ and if $e\ge 1,$

$$N_{m_0}\left(e\right)=(\lfloor \frac{m_0-1}{p^{e-1}}\rfloor -2\lfloor \frac{m_0-1}{p^e}\rfloor +\lfloor \frac{m_0-1}{p^{e+1}}\rfloor )r.$$

(30)

Proof. 

The first claim is easy. Note that $N_{m_0}\left(e\right)$ is exactly the cardinality of $A=\{s\notin R\left(j\right):v_{m_0}\left(s\right)=e\}.$ For each s, there is a positive integer $b_s$, such $s{p}^{b_s-1}\le u<s{p}^{b_s}$ and, hence,

$$j^b\left(s\right)=\left\{\begin{array}{cc}s{p}^b,\hfill & \mathrm{if}b\le b_s,\hfill \\ s{p}^{b_s}+(b-b_s)k,\hfill & \mathrm{if}b>b_s.\hfill \end{array}\right.$$

However, since $m_0=k+t$ or $m_0=t,$ then, if $s\in A$, either $b_s=e$ or $b_s=e-1.$ Consider two cases: (i) let $m_0\ge pu,$ then clearly $b_s=e-1$ for all $s\in A$ and, thus, by the definition of $b_s,$ $s{p}^{e-1}\le pu\le m_0.$ However, we know that $m_0\le s{p}^e$ and, hence, $\frac{m_0}{p^e}\le s<\frac{m_0}{p^{e-1}}.$ (ii) Assume $m_0<pu,$ then if $b_s=e,$ by definition of $b_s,$ $s{p}^{e-1}\le u<m_0.$ Also if $b_s=e-1,$ which means that $s{p}^{e-1}>u,$ hence

$$s{p}^e\ge j\left(s{p}^{e-1}\right)=s{p}^{e-1}+k=j^e\left(s\right)>m_0.$$

(31)

On the other hand, $s{p}^{e-1}<m_0.$ Thus, in this case $\frac{m_0}{p^e}<s<\frac{m_0}{p^{e-1}}.$ From (i) and (ii), we conclude that, for every $s\in A,$ $\frac{m_0}{p^e}\le s<\frac{m_0}{p^{e-1}}.$ That implies A is the set of all positive integers in the interval $(\lfloor \frac{m_0-1}{p^{e-1}}\rfloor ,\lfloor \frac{m_0-1}{p^e}\rfloor ]$ except those in $R\left(j\right),$ i.e., $s\equiv 0$ mod $p.$ Therefore, the proof follows. □

Corollary 7.

Based on the assumptions of Theorem 3,

$$\left\{\begin{array}{cc}{N}_m(n-2)=u-t+1-\lfloor \frac{u-t+1}{p}\rfloor ,\hfill & \mathrm{if}t\le u\mathrm{and}n>2,\hfill \\ N_m(n-1)=k+u-t+1-\lfloor \frac{k+u-t+1}{p}\rfloor ,\hfill & \mathrm{if}t>u.\hfill \end{array}\right.$$

We now compute $N_m\left(e\right)$ for the incomplete case. In this case, $m>k+u$, which means either $n=2$ and $t>u$ or $n>2.$ Put

$$m_0=\left\{\begin{array}{cc}\iota k+t,\hfill & \mathrm{if}u\le t,\hfill \\ (\iota +1)k+t,\hfill & \mathrm{if}u>t,\hfill \end{array}\right.n_0=\left\{\begin{array}{cc}n-1-\iota ,\hfill & \mathrm{if}t\ge u,\hfill \\ n-2-\iota ,\hfill & \mathrm{if}t<u,\hfill \end{array}\right.$$

(32)

where

$$\iota =\left\{\begin{array}{cc}1,\hfill & \mathrm{if}\mu \ge \lambda ,\hfill \\ \mu ,\hfill & \mathrm{if}\mu <\lambda .\hfill \end{array}\right.$$

Proposition 4.

Let $e\ge \iota$ and $\mu <n-1.$ Then with same notations mentioned above,

$$\left\{\begin{array}{c}{N}_m\left(\mu \right)=1+N_{m_0}(\mu -n_0),\hfill \\ N_m(\mu +n_0)=N_{m_0}\left(\mu \right)-1,\hfill \\ N_m(e+n_0)=N_{m_0}\left(e\right),\mathrm{if}e\ne \mu \mathrm{and}e\ne \mu +n_0.\hfill \end{array}\right.$$

(33)

Proof. 

By the j-diagram, one can check that $H_{m_0}$ is a subgroup of $H^{p^{\mu }}.$ Because there is an element in H with $p^{\mu }$ (Theorem 2), then clearly $N_m\left(\mu \right)=1+N_{m_0}(\mu -n_0)$ and $N_m(\mu +n_0)=N_{m_0}\left(\mu \right)-1.$ On the other hand, the proof of Lemma 1 implies ${\mu }_i\ge n-1$ for all $i\ge 2$ (since $\nu \left(s_i\right)\ge n-1$) and, hence, we only consider $N_{m_0}\left(e\right),$ $e\ge \mu .$ This immediately implies $N_m(e+n_0)=N_{m_0}\left(e\right)$ for all e other than $\mu$ and $\mu +n_0.$ □

Remark 4.

By Proposition 4, when $\mu <n-1$, it suffices to compute $N_{m_0}\left(e\right).$ However, in case of $\mu \ge n-1$ nothing can be said since $N_{m_0}\left(e\right)=N_m\left(e\right).$

Note that, in the case when $\lambda >\mu ,$ once we determine (13) and $N_{m_0}\left(e\right)$ for the part ${\otimes }_{s\in \Lambda }{U}_s{\otimes }_{i=1}^d{U}_{s_i}^{*},$ we then completely obtain $H.$ Denote $z_e=\mid {\mathsf{\Omega }}^{\prime }\cap P_e\mid ,$ where $P_e=\{w\in H_s:o\left(w\right)=e,s\notin R\left(j\right)\}.$

Theorem 4.

Let R be an incomplete chain ring, then

(i) $N_{m_0}=N_m=cr+1$

(ii)

$$N_{m_0}\left(e\right)=\left\{\begin{array}{cc}(2k-m_0+\lfloor \frac{m_0-k}{p}\rfloor )r+\delta ,\hfill & \mathrm{if}e=1,\hfill \\ (\lfloor \frac{m_0-ak-1}{p^{e-1-a}}\rfloor -2\lfloor \frac{m_0-ak-1}{p^{e-a}}\rfloor +\lfloor \frac{m_0-ak-1}{p^{e-a+1}}\rfloor )r+z,\hfill & \mathrm{if}e>\iota ,\hfill \end{array}\right.$$

where

$$\begin{array}{cc}\hfill \delta =\left\{\begin{array}{cc}2,\hfill & \mathrm{if}\lambda =1,\hfill \\ 1,\hfill & \mathrm{if}\lambda \ne 1,\hfill \end{array}\right.& z=\left\{\begin{array}{c}\left\{\begin{array}{cc}1,\hfill & \mathrm{if}e=\lambda \ge 2,\hfill \\ -1,\hfill & \mathrm{if}e=\lambda +1,\hfill \\ 0,\hfill & \mathrm{otherwise},\hfill \end{array}\right.\left(\iota =1\right)\hfill \\ -z_e,\mathrm{if}\iota =\mu ,\hfill \end{array}\right.\hfill \\ a=\left\{\begin{array}{cc}\iota -1,\hfill & \mathrm{if}t=u,\hfill \\ \iota ,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

(i) is obvious. For (ii), we first let $e=1.$ Note that $o\left(\gamma \right)=1$ and $o\left(\xi \right)=1$ if $\lambda =1,$ thus we have $\delta .$ Now, consider two cases: (a) when $u\le t,$ then $m_0=k+t$ and clearly ${\nu }_{m_0}\left(s\right)=1$ if and only if $s\ge t.$ Consequently, there are $k+u-t-\mid \{s\ge t:s\equiv 0\mathrm{mod}p\}\mid$ of such s and, thus

$$N_{m_0}\left(1\right)=(k+u-t-(u-\lfloor \frac{t}{p}\rfloor ))r+\delta =(2k-m_0+\lfloor \frac{m_0-k}{p}\rfloor )r+\delta .$$

(34)

(b) In the case when $u>t.$ Firstly, observe that ${\nu }_{m_0}\left(s\right)=1=e$ if and only if $s\ge k+t.$ By the same reasoning of Case (a), we obtain the results. Secondly, when $e>\iota ,$ in this case all s satisfying ${\nu }_{m_0}\left(s\right)>1$ occur in the interval $s<t$ if $u\le t$ or $s<k+t$ when $t<u.$ Therefore, we return to a similar situation of Theorem 3 with $m_0-ak$ and $e-a$ instead of $m_0$ and e, respectively. Observe that, if $\iota =1$ and $e=\lambda +1,$ then there exist $r-1$ generators of $U_{s_0}$ and then put $z=-1.$ When $e=\lambda ,$ we have to add $z=1$, since $o\left(\xi \right)=\lambda .$ On the other hand, if $\iota =\mu ,$ there are $z_e$ of $U_s$ each having $r-1$ elements of basis of the same order, so put $z=-z_e.$ □

In conclusion, this article considered the multiplicative groups of finite commutative chain rings. The case when $(p-1)\mid k$ was completed for every incomplete chain ring. Moreover, a set of linearly independent generators for H was established by connecting those generators to a basis of F over its prime field.

Recently, there has been increasing interest in using finite commutative chain rings in Coding Theory. Researchers may wish to further apply this knowledge about finite commutative chain rings to Coding Theory.