Finite Local Rings of Length 4

Abstract

This paper presents a comprehensive characterization of finite local rings of length 4 and with residue field ${\mathbb{F}}_{p^m},$ where p is a prime number. Such rings have an order of $p^{4m}$ elements. The current paper provides the structure and classification, up to isomorphism, of local rings consisting of $p^{4m}$ elements. We also give the exact number of non-isomorphic classes of these rings with fixed invariants $p,n,m,k.$ In particular, we have listed all finite local rings of 4-length and of order $p^8$ and $256.$

1. Introduction

Finite rings of relatively low orders and their classification, up to isomorphism, have drawn a lot of attention from both new and old researchers; see [1,2] for some recent references. If p is a prime number, then the additive group $(R,+)$ of a finite ring with identity R splits as the direct sum of its p-primary components $R_p$, which are pairwise orthogonal ideals. The rings $R_p$ are thus the direct summand of $R.$ Therefore, it suffices to deal with rings of prime-power order for the purpose of characterizing finite rings; henceforth, we will refer to R as having this form in our work. All rings are taken, in this paper, to be finite, commutative, and to have an identity element. Furthermore, all modules over these rings are considered to be finitely generated. As customary, the Galois field of order $p^m$ is denoted by ${\mathbb{F}}_{p^m},$ where m is a positive integer. A ring R is referred to as local if it has a single maximal ideal, denoted by J, which represents the Jacobson radical of $R.$ There is a special class of local rings known as principal ideal rings (PIRs) in which all the ideals are single-element generated. In particular, when J is principal, then R is given a name of a chain ring. Because a chain ring is a principal local ring, PIRs can benefit from many of the results obtained for chain rings [3,4]. Finite local rings are particularly well-suited for coding since two classical theorems by MacWilliams, the Extension Theorem, and the MacWilliams Identities, apply to these rings as they do to finite fields [5,6,7,8]. Local rings also enable crucial connections between the ring and its complex characters. To look for additional details, please refer to [9,10,11,12].

Local rings of length $l=1$ and $l=2$ have their J principal, and thus they are chain rings. It is well-known that there is only one local ring with $l=1,$ i.e., $p^m$ elements, up to isomorphism: the Galois field ${\mathbb{F}}_{p^m}.$ The local rings of length $l=2,$ i.e., of order $p^{2m}$ are $GR(p^2,m)$ and ${\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^2⟩}}.$ When $l=3,$ the non-isomorphic classes of local rings with $p^{3m}$ elements are as follows: if p is odd, the local rings are $GR(p^3,m),$ ${\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^3⟩}},$ ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p,pu⟩}}$ and ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p\beta ,pu⟩}},$ where $\alpha$ is a primitive element of ${\mathbb{F}}_p;$ if $p=2,$ there are three such rings which are $GR(8,m),$ ${\displaystyle \frac{{\mathbb{F}}_{2^m}\left[u\right]}{⟨u^3⟩}}$ and ${\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2-2,2u⟩}}$ (see [2]). The rings ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{{⟨u,v⟩}^2}},$ ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2,pu⟩}},$ ${\displaystyle \frac{{\mathbb{F}}_{2^m}[u,v]}{{⟨u,v⟩}^2}}$, and ${\displaystyle \frac{GR(8,m)\left[u\right]}{⟨u^2,2u⟩}}$ are non-chain local rings, because their maximal ideals J are not principal, and the other rings are chain rings.

Let A be an R-module over a finite local ring R with residue field ${\mathbb{F}}_{p^m}$, and let l denote the length of A. The formula $\left|A\right|=|{\mathbb{F}}_{p^m}{|}^l$, which will be introduced in Section 2, indicates that if R is a local ring with $p^{4m}$ elements, then the length must be $l=4.$ However, there are many such rings with length 4 for different values of the nilpotency index t of $J,$ where $1\le t\le l=4.$ In the literature, a certain class of such rings was considered in studying constacyclic codes [13]. The ring ${\displaystyle \frac{{\mathbb{F}}_p[u,v]}{{⟨u,v⟩}^2}},$ with $p^4$ elements, was introduced in [3], where linear codes were investigated over it. More recently, authors of [2] studied linear codes over local rings of order $p^4.$ Finding the family of local non-chain rings with $\left|A\right|=p^{4m}$ would be more interesting if $m>1$ and $p>2.$ This paper aims to accomplish two goals. First, we must finalize the construction of local rings of length four. Next, we classify and enumerate these rings up to isomorphism. This includes a listing of all local non-chain and chain rings with $p^8$ elements in Table 1, with particular emphasis on the rings with 16 and 256 elements.

Table 1. Local rings of orders $p^8,$ 16, and $256$.

Rings of Order ${\mathit{p}}^8,$ $\mathit{p}\ne 2$	Rings of Order 16	Rings of Order 256
${\displaystyle \frac{{\mathbb{F}}_{p^2}[u,v]}{⟨u^3,v^2,uv⟩}}$	${\displaystyle \frac{{\mathbb{F}}_2[u,v]}{⟨u^3,v^2,uv⟩}}$	${\displaystyle \frac{{\mathbb{F}}_4[u,v]}{⟨u^3,v^2,uv⟩}}$
${\displaystyle \frac{{\mathbb{F}}_{p^2}[u,v]}{⟨u^2-v^2,uv⟩}}$	${\displaystyle \frac{{\mathbb{F}}_2[u,v]}{⟨u^2-v^2,uv⟩}}$	${\displaystyle \frac{{\mathbb{F}}_4[u,v]}{⟨u^2-v^2,uv⟩}}$
${\displaystyle \frac{{\mathbb{F}}_{p^2}[u,v]}{⟨u^2-\alpha v^2,uv⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^2-2⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^2-2⟩}}$
${\displaystyle \frac{GR(p^2,2)\left[u\right]}{⟨u^2-p\alpha ⟩}}$	${\displaystyle \frac{{\mathbb{F}}_2[u,v]}{⟨u^2,v^2⟩}}$	${\displaystyle \frac{{\mathbb{F}}_{2^m}[u,v]}{⟨u^2,v^2⟩}}$
${\displaystyle \frac{{\mathbb{F}}_{p^2}[u,v,w]}{{⟨u,v,w⟩}^2}}$	${\displaystyle \frac{{\mathbb{F}}_2[u,v,w]}{{⟨u,v,w⟩}^2}}$	${\displaystyle \frac{{\mathbb{F}}_{2^2}[u,v,w]}{{⟨u,v,w⟩}^2}}$
${\displaystyle \frac{GR(p^2,2)\left[u\right]}{⟨u^2-p⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^2-2-2u⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^2-2-2u⟩}}$
${\displaystyle \frac{GR(p^2,2)\left[u\right]}{⟨u^2⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^2⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^2⟩}}$
${\displaystyle \frac{GR(p^2,2)\left[u\right]}{⟨u^3-p\beta ,pu⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^2-2u⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^2-2u⟩}}$
${\displaystyle \frac{GR(p^2,2)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,1)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,2)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}$
${\displaystyle \frac{GR(p^2,2)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,1)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,2)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}$
${\displaystyle \frac{GR(p^2,2)[u,v]}{⟨u^2-p\alpha ,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,1)[u,v]}{⟨u^2-p\beta ,v^2,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,2)[u,v]}{⟨u^2-p\beta ,v^2,uv,pu,pv⟩}}$
${\displaystyle \frac{GR(p^2,2)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,1)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,2)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}$
${\displaystyle \frac{GR(p^2,2)[u,v]}{⟨u^2-p,v^2-p\alpha ,uv,pu,pv⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^3-2,2u⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^3-2,2u⟩}}$
${\displaystyle \frac{GR(p^2,2)\left[u\right]}{⟨u^3-p,pu⟩}}$	${\displaystyle \frac{GR(4,1)[u,v]}{⟨u^2,v^2,uv-2,2u,2v⟩}}$	${\displaystyle \frac{GR(4,2)[u,v]}{⟨u^2,v^2,uv-2,2u,2v⟩}}$
${\displaystyle \frac{GR(p^3,2)\left[u\right]}{⟨u^2-p^2\beta ,pu⟩}}$	${\displaystyle \frac{GR(4,1)\left[u\right]}{⟨u^3,2u⟩}}$	${\displaystyle \frac{GR(4,2)\left[u\right]}{⟨u^3,2u⟩}}$
${\displaystyle \frac{GR(p^3,2)\left[u\right]}{⟨u^2-p^2,pu⟩}}$	${\displaystyle \frac{GR(8,1)\left[u\right]}{⟨u^2-4,2u⟩}}$	${\displaystyle \frac{GR(8,2)\left[u\right]}{⟨u^2-4,2u⟩}}$
${\displaystyle \frac{GR(p^3,2)\left[u\right]}{⟨u^2,pu⟩}}$	${\displaystyle \frac{GR(8,1)\left[u\right]}{⟨u^2,2u⟩}}$	${\displaystyle \frac{GR(8,2)\left[u\right]}{⟨u^2,2u⟩}}$
${\displaystyle \frac{{\mathbb{F}}_{p^2}\left[u\right]}{⟨u^4⟩}}$	${\displaystyle \frac{{\mathbb{F}}_2\left[u\right]}{⟨u^4⟩}}$	${\displaystyle \frac{{\mathbb{F}}_4\left[u\right]}{⟨u^4⟩}}$

The structure of this study is as follows: Section 2 provides a review of fundamental concepts regarding finite local rings and their modules. In Section 3, we finalize our analysis of all local rings containing $p^{4m}$ elements, where p is a prime. Specifically, we focus on rings with $p^8$, 16, and 256 elements. Moreover, the family of finite local chains and non-chain rings of length 4 is fully characterized.

2. Preliminaries

This section provides the essential information and notations for our analysis. Here, J denotes the radical of a finite local ring R. For outcomes mentioned, we refer readers to [5,14,15,16,17,18].

In this discussion, we consider R and T as rings, and A as an R-module. If $R\subseteq T$, then T is called an extension of $R.$ The ideal $IT$ is referred to as the expansion of I to $T,$ where I is an ideal of $R.$ For any A over $R,$ ann_R$\left(A\right)=\{r:ra=0,a\in A\}.$ Let ${\Omega }_R\left(A\right)$ denote the set of all R-submodules of A. If $A=R$, we denote the set of ideals of R by $\Omega \left(R\right)$. Recall that in a partially ordered set, or lattice, any two elements have both a supremum and an infimum. Thus, the set ${\Omega }_R\left(A\right)$ forms a lattice with respect to set-theoretic inclusion ⊆. Furthermore, for any ideal I of R contained in the annihilator of A, the quotient module ${\displaystyle \frac{R}{I}}$ on A has a natural structure, ensuring that the lattices of R-submodules and ${\displaystyle \frac{R}{I}}$-submodules of A are identical. A series of R-submodules of A is defined as a strictly increasing sequence of R-submodules; that is,

$$A=A_0\supset A_1\supset \cdots \supset A_{l-1}\supset A_l=0.$$

(1)

A chain of this type is said to have a length equal to l, or its number of joints. The chain is a composition series if each R-module ${\displaystyle \frac{A_i}{A_{i+1}}}$ lacks a non-zero simple module, or non-zero proper submodule. The length of a composition series for $A,$ denoted by $l_R\left(A\right)$, and it is ∞ if A does not have a finite composition series, as per the Jordan–Holder–Schreier Theorem [15]. It should be observed that such series have no relevance on the length. A ${\mathbb{F}}_{p^m}$-module’s length likewise equals the ${\mathbb{F}}_{p^m}$-vector space’s dimension. Recall in particular that a ring R with a single maximum ideal J is referred to as local, and the ${\displaystyle \frac{R}{J}}$ represents its residue field ${\mathbb{F}}_{p^m},$ which is isomorphic to a field $GF\left(p^m\right).$ In the light of Nakayamas’ Lemma ([17]), when R is a finite local ring, then R acquires an ideals chain, and also there is an integer $t\ge 1,$ called the nilpotency index of $J,$ that is $J^t=0$ and $J^{t-1}\ne 0,$ such that

$$R\supset J\supset J^2\supset \cdots \supset J^{t-1}\supset J^t=0.$$

This implies that $t\le l_R\left(R\right),$ and if A admits a composition series over $R,$ $A=A_0\supset A_1\supset \cdots \supset A_{l-1}\supset A_l=0,$ then ${\displaystyle \frac{A_i}{A_{i+1}}}\cong {\mathbb{F}}_{p^m}$ as vector spaces over ${\mathbb{F}}_{p^m}.$ Therefore,

$$\mid A\mid ={p^m}^{l_R\left(A\right)}.$$

As direct consequences, when R is a finite local ring, then there are l and m such that the order of J is $p^{(l-1)m}$ with $J^l=0,$ and $\left|R\right|=p^{lm}.$ The characteristic of R is expressed as $p^n,$ with $1\le n\le l.$ This implies that R possesses $R_0=GR(p^n,m)$ as a subring called Galois ring with $p,n,m.$ There is k such that (as $R_0$-module)

$$\begin{array}{cc}\hfill & R=R_0+u_1{R}_0+u_2{R}_0+\cdots +u_k{R}_0,\hfill \\ \hfill & J=p{R}_0+u_1{R}_0+u_2{R}_0+\cdots +u_k{R}_0,\hfill \end{array}$$

(2)

where $u_i\in J$ [19]. These generators $u_1,u_2,\cdots ,u_k$ of J will be crucial, as we see later, in the construction of such rings with parameters $p,n,m,l,k.$

With $p,n,m,l,k$, let R be a finite local ring. Furthermore, let A be a R-module. A subset S of A spans A if and only if its image $\overline{S}$ in ${\displaystyle \frac{A}{JA}}$ generates ${\displaystyle \frac{A}{JA}}.$ A set of generators of A obtained from lifting a basis of the ${\displaystyle \frac{R}{J}}$-vector space ${\displaystyle \frac{A}{JA}}$ is called a minimal generating set for A over R and ${\nu }_R\left(A\right)$ denotes the number of such generators. Note that

$${\nu }_R\left(A\right)={\dim }_{{\mathbb{F}}_{p^m}}\left({\displaystyle \frac{A}{JA}}\right)=l_R\left({\displaystyle \frac{A}{JA}}\right).$$

Recall that, in the context of set-theoretic inclusion, a ring R is defined as a chain ring if the lattice of its ideals forms a chain. The ring R is considered a chain if and only if R is a local ring and its maximal ideal J is principal; this condition is equivalent to $t=l$, where t denotes the nilpotency index of J. In this scenario, we have $u_i=u^i$ for $i=1,2,\dots ,k-1$. Additionally, we have $J=uR$. It can also be noted that $p\in J^k$, meaning $u^k\in ⟨p⟩$, and then

$$u^k=p(r_0+r_{1}u+r_2{u}^2+...+r_{k-1}{u}^{k-1}),$$

where $r_0$ is a unit. This means u is a root of $g\left(x\right),$

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

(3)

The numbers p, n, m, l, and k are associated with $R.$

$$R\cong {\displaystyle \frac{R_0\left[x\right]}{⟨g\left(x\right),p^{n-1}{x}^{t^{\prime }}⟩}},$$

(4)

where $t^{\prime }=l-(n-1)k.$ An element $f\in R\left[x\right]$ is called basic irreducible if $\overline{f}$ is irreducible over ${\mathbb{F}}_{p^m}.$ If $g\in R\left[x\right]$ is a basic irreducible polynomial and let $m=\mathrm{deg}\left(\overline{g}\right).$ There exists a unit $\beta$ in $R\left[x\right]$ and a monic polynomial f in $R\left[x\right]$ satisfying $f=\beta g$ (see [15]). When $J=⟨p⟩,$ we obtain $n=l$ and if $\alpha$ is a root of $g\left(x\right)\in {\mathbb{Z}}_{p^n}\left[x\right]$ with degree $m,$ we obtain

$$R=R_0=GR(p^n,m)\cong {\mathbb{Z}}_{p^n}\left[\alpha \right]\cong {\displaystyle \frac{{\mathbb{Z}}_{p^n}\left[x\right]}{⟨g\left(x\right)⟩}}.$$

This extension is Galois.

Let $T={\displaystyle \frac{R\left[x\right]}{⟨g\left(x\right)⟩}}=\{r_0+r_{1}x+\cdots +r_s{x}^s:r_i\in R\},$ where g is basic irreducible and $s=\mathrm{deg}\left(g\right).$ This ring is an extension of $R,$ and moreover, it is local with maximal ideal $JT$ and residue field ${\mathbb{F}}_{p^{ms}},$ ([15]). Let $\alpha$ be as ${\mathbb{F}}_{p^m}^{*}=⟨\alpha ⟩,$ we set

$$\Gamma \left(m\right)=\left(\alpha \right)\cup \left\{0\right\}=\{0,1,\alpha ,{\alpha }^2,\cdots ,{\alpha }^{p^m-2}\}.$$

This set is called a set of representatives of ${\mathbb{F}}_{p^m}$ in R and the set $\Gamma \left(ms\right)\subseteq T$ is a set of representatives of ${\mathbb{F}}_{p^{ms}}.$ If I is an ideal of R, then $l_R\left(I\right)=l_T\left(IT\right),$ and $\left(\mathrm{ann}\left(I\right)\right)T={\mathrm{ann}}_T\left(IT\right).$ This follows that

$$l\left(\mathrm{ann}\left(J\right)\right)=l_T\left(an{n}_T\left(JT\right)\right),$$

$$v\left(J\right)=l\left({\displaystyle \frac{J}{J^2}}\right)=l_T\left({\displaystyle \frac{JT}{J^{2}T}}\right)=v_T\left(JT\right).$$

When $R={\mathbb{Z}}_{p^n},$ then $T=GR(p^n,s).$ The ring $T,$ in general, is not a Galois extension of $R,$ see [4].

All notations mentioned, in this section, will be in force throughout the manuscript.

3. Local Rings of Length 4

From now on, the ring R is a local ring with parameters $p,n,m,4,$ and k. By varying the values of n and k, we shall prove in this section all possible structures of R and their classes.

As we have $\mid R\mid =p^{4m},$ the all possible cases for R are

$$R=\left\{\begin{array}{cc}{\mathbb{F}}_{p^m}+u_1{\mathbb{F}}_{p^m}+u_2{\mathbb{F}}_{p^m}+u_3{\mathbb{F}}_{p^m},\hfill & \mathrm{if}k=3,\hfill \\ GR(p^2,m)+uGR(p^2,m)+vGR(p^2,m),\hfill & \mathrm{if}k=2,\hfill \\ GR(p^3,m)+uGR(p^3,m),\hfill & \mathrm{if}k=1,\hfill \\ GR(p^4,m)\hfill & \mathrm{if}k=0.\hfill \end{array}\right.$$

(5)

The group of units of R, denoted $U\left(R\right)$, plays a significant role in the construction of R. We have

$$U\left(R\right)=⟨\alpha ⟩\times H,$$

where $H=1+J$. The structure of $U\left(GR(p^n,m)\right)$ has been completely described in [20].

Suppose that $1\le i\le k$, and for each i, let $t_i$ be the additive order of $u_i$, such that $p^{t_i}{u}_i=0$. Since $\left|GR(p^n,m)\right|=p^{nm}$, we have

$$4=l=n+t_1+t_2+\cdots +t_k.$$

(6)

We observe that the shape and algebraic structure of R are entirely determined by the parameters $n,m,l,k$. Next, we establish a very helpful relation among the parameters of a local ring, namely $n,l,k$, which will reveal the nature of R.

Lemma 1.

Let R be a local ring with $p,n,m,l,k.$ Then,

$${\displaystyle \frac{l}{n}}-1\le k\le l-n.$$

(7)

Proof. 

Since $l=n+t_1+t_2+\cdots +t_k$ and $1\le t_i\le n$ for each $i.$ Then, $k\le t_1+t_2+\cdots +t_k.$ This follows that $k+n\le l,$ and so $k\le l-n.$ On the other hand, since there are k generators of R over $GR(p^n,m),$ and any element a of $GR(p^n,m)$ is written by

$$a=a_1+p{a}_1+\cdots +p^{n-1}{a}_n,$$

where $a_i\in \Gamma \left(m\right).$ Moreover, by Equation (5), we have $n+kn$ of such generators over ${\mathbb{F}}_{p^m}.$ Thus, l can not exceed $n+kn=(k+1)n,$ i.e., $l\le (k+1)n$ which implies that ${\displaystyle \frac{l}{n}}-1\le k.$ Therefore, ${\displaystyle \frac{l}{n}}-1\le k\le l-n.$ □

From the last equation, $1\le n\le 4,$ and consequently $0\le k\le 3=l-1.$ When $k=0,$ then $R=R_0=GR(p^n,m).$ Recall that t is the index of nilpotency of $J.$ Thus, $t\le 4,$ and $t=4$ occurs only when R is chain. When $t=1,$ then $J=0,$ and hence R is a field, i.e., $l=1.$ Therefore, in our work, we have always $2\le t\le 4.$ Also, notice that $\nu \left(J\right)\le k+1,$ and $\nu \left(J^2\right)\le {\left(\nu \left(J\right)\right)}^2.$ In general, we have

$$\begin{array}{cc}\nu \left(J\right)+\nu \left(J^2\right)+\cdots +\nu \left(J^{t-1}\right)=k,\hfill & \mathrm{if}n=1;\hfill \\ n+nk\ge \nu \left(J\right)+\nu \left(J^2\right)+\cdots +\nu \left(J^{t-1}\right)\ge k+1,\hfill & \mathrm{if}n>1.\hfill \end{array}$$

The following statement is straightforward.

Proposition 1.

Let R be a finite local ring with $p,n,m,l,k,$ and let $GR(p^n,m)$ be its coefficient subring. Then,

$$R\cong {\displaystyle \frac{GR(p^n,m)[u_1,u_2,\cdots ,u_k]}{I}},$$

where I is an appropriate ideal with respect to $u_i.$ Moreover, its maximal ideal is $J={\displaystyle \frac{⟨p,u_1,u_2,\cdots ,u_k⟩}{I}}.$

In order to establish a relation among $t,\nu \left(J\right)$, and $k,$ which are crucial for demonstrating the primary findings of this section, the following results are crucial.

Proposition 2.

If R is a finite local ring with $t=2$. Then,

$$k=\left\{\begin{array}{cc}\nu \left(J\right)=l-1,\hfill & \mathrm{if}n=1,\hfill \\ l-2,\hfill & \mathrm{if}n=2.\hfill \end{array}\right.$$

Furthermore, when $n=2,$ $\nu \left(J\right)=l-1.$

Proof. 

Assume that $t=2.$ Since $\nu \left(J\right)+\nu \left(J^2\right)+\cdots =k,$ this implies that $\nu \left(J\right)=k$ because $J^2=0.$ Moreover, since $n\le t,$ then $n=1$ or $n=2.$ The case when $n=1,$ we obtain

$$R=J^0\supset J=⟨u_1,u_2,\cdots ,u_k⟩\supset ⟨u_2,\cdots ,u_k⟩\supset \cdots \supset ⟨u_k⟩\supset ⟨0⟩.$$

As $⟨u_i,u_{i+1},\cdots ,u_k⟩/⟨u_{i+1},u_{i+2},\cdots ,u_k⟩\cong {\mathbb{F}}_{p^m},$ then this chain is clearly composition in R of length $l=k+1.$ Thus, $k=l-1.$ When $n=2,$ consider the following series

$$R=J^0\supset J=⟨p,u_1,u_2,\cdots ,u_k⟩\supset ⟨p,u_2,\cdots ,u_k⟩\supset \cdots \supset ⟨p,u_k⟩\supset ⟨p⟩\supset ⟨0⟩.$$

Also, we note that $⟨p,u_i,u_{i+1},\cdots ,u_k⟩/⟨p,u_{i+1},u_{i+2},\cdots ,u_k⟩\cong {\mathbb{F}}_{p^m},$ and hence the chain is composition with $l=k+2,$ and $k=l-2.$ The last assertion is direct. □

Remark 1.

In fact, the converse of Proposition 2 is also true. To show that, suppose $k=\nu \left(J\right)=l-1.$ By Equation (7), it follows $\nu \left(J^2\right)=0,$ and thus $n=1$ and $J^2$ must be $0,$ i.e., $t=2.$

Corollary 1.

Assume R is of length 4 and if $t=2.$ Then, $k=\nu \left(J\right)=3.$

Proof. 

The result follows by substituting $l=4$ in Proposition 2. □

Remark 2.

If $t=2,$ $R=GR(p^n,m)+uGR(p^n,m)+vGR(p^n,m)+wGR(p^n,m)$ and $J=⟨u,v,w⟩,$ the generators must satisfy the relations: $u^2=v^2=w^2=uv=uw=vw=0.$

Lemma 2.

Suppose R is a ring with $l=4$ and $t=3.$ Then,

$$k=\left\{\begin{array}{c}3,\mathrm{if}n=1,\hfill \\ \left\{\begin{array}{cc}2,\hfill & \mathrm{if}pu\ne 0,p\in J^2,\hfill \\ 1,\hfill & \mathrm{if}p\in J\setminus J^2,pu=0,\hfill \end{array}\right.(n=2)\hfill \\ 1,\mathrm{if}n=3.\hfill \end{array}\right.$$

In particular, $\nu \left(J\right)=2.$

Proof. 

First, note that $J^2\ne 0$ and $J^3=0.$ In light of Equation (7), $\nu \left(J\right)+\nu \left(J^2\right)\ge k+1.$ Also, we know that $\nu \left(J^2\right)\le {\left(\nu \left(J\right)\right)}^2.$ Thus, when $n=1$ we must have $\nu \left(J\right)=2$ and $\nu \left(J^2\right)=1.$ However, $\nu \left(J\right)=2$ and $J^2=⟨p⟩$ in case of $n=2.$ Clearly, $k=1$ when $n=3$ or $p\in J\setminus J^2.$ □

Remark 3.

Based on the previous lemma, we note that $J=⟨u,v⟩$ and $J^2=⟨w⟩$ when $n=1;$ when $n=2$ and $p\in J^2,$ $J=⟨u,v⟩$ and $J^2=⟨p⟩;$ we have $J=⟨u,p⟩$ and $⟨p,pu⟩,$ otherwise. The last case is what we call it singleton local rings [1].

Remark 4.

When $t=3$ and $J=⟨p,u,v⟩,$ then u and v hold the equations $u^3=v^3=u{v}^2=u^{2}v=0$ because $J^3=0.$

Proposition 3.

If $t=3$ and $J=⟨p,u,v⟩.$ Then, only one of the following is satisfied

(a)

If $n=1,$

$$\left\{\begin{array}{c}\left(i\right)u^3=v^2=uv=0;\hfill \\ \left(ii\right)u^2-\beta v^2=0,uv=0;\hfill \\ \left(iii\right)u^2-{\beta }_{1}uv=0,v^2-{\beta }_{2}uv=0\mathrm{and}uv\ne 0\mathrm{provided}\mathrm{that}{\beta }_1\ne {\beta }_2^{-1}.\hfill \end{array}\right.$$

(b)

If $n=2,$

$$\left\{\begin{array}{c}\left(i\right)u^2=v^2=uv=pu=pv=0;\hfill \\ \left(ii\right)u^2-p\beta =0,v^2=uv=pv=pu=0;\hfill \\ \left(iii\right)\left\{\begin{array}{cc}{u}^2-p{\beta }_1=0,v^2-p{\beta }_2=0\mathrm{and}uv=pu=pv=0,\hfill & \mathrm{if}p\ne 2,\hfill \\ u^2-p{\beta }_1=0,v^2-p{\beta }_2=0\mathrm{and}pu=pv=0,\hfill & \mathrm{if}p=2\hfill \end{array}\right.\hfill \end{array}\right.$$

(c)

If $n=2$ and $p\in J\setminus J^2$ or $n=3.$

(i)

If $n=2$ and $p\in J\setminus J^2,$ $u^2=0$ when $p\ne 2$ and $u^2=2u,$ otherwise.

(ii)

If $n=3,$ then $u^2=p^2\gamma$ when $p\ne 2$ and $u^2=\gamma 4+2u,$ otherwise.

where $\beta \in {\Gamma }^{*}\left(m\right)$ and ${\beta }_1,{\beta }_2$ are in $\Gamma \left(m\right).$

Proof. 

Case (a). As $t=3,$ then $k=1,2$, or 3 by Lemma 2. First, assume $n=1,$ in this case, $R={\mathbb{F}}_{p^m}+u{\mathbb{F}}_{p^m}+v{\mathbb{F}}_{p^m}+w{\mathbb{F}}_{p^m}$ and $J=u{\mathbb{F}}_{p^m}+v{\mathbb{F}}_{p^m}+w{\mathbb{F}}_{p^m}.$ As $J^2=⟨w⟩\subseteq$ ann$\left(J\right)$ since $\nu \left(J^2\right)=1.$ Suppose that $v\in$ann$\left(u\right),$ then if ann$\left(u\right)=J^2,$

$$⟨u⟩=\mathrm{ann}\left(\mathrm{ann}\left(u\right)\right)=\mathrm{ann}\left(J^2\right)=J.$$

Which is impossible since $u\ne v.$ Now, if $v\notin$ann$\left(u\right)=J,$ then

$$⟨u⟩=\mathrm{ann}\left(\mathrm{ann}\left(u\right)\right)=\mathrm{ann}\left(J\right)\supseteq J^2=⟨v^2⟩.$$

This is again a contradiction since $v^2\notin ⟨u⟩.$ Hence, we obtain $J^2\subset \mathrm{ann}\left(u\right).$ When $v\notin$ann$\left(u\right),$ then one can easily deduce that ann$\left(u\right)=J^2=⟨uv⟩.$ Moreover, $\mathrm{ann}\left(v\right)\subset J$ and $\mathrm{ann}\left(u\right)\ne \mathrm{ann}\left(v\right).$ That being said, then $\mathrm{ann}\left(u\right)=⟨u^2,v⟩,$ $\mathrm{ann}\left(u\right)=⟨v⟩$, or $\mathrm{ann}\left(u\right)\supseteq ⟨uv⟩,$ and $\mathrm{ann}\left(v\right)=⟨v^2,u⟩,$ $\mathrm{ann}\left(v\right)⟨u⟩$ or $\mathrm{ann}\left(v\right)\supseteq ⟨uv⟩.$ Let us investigate the three cases: (i) Suppose $\mathrm{ann}\left(u\right)=⟨u^2,v⟩.$ Then, $\mathrm{ann}\left(v\right)=⟨v^2,u⟩.$ This implies $v^2=0,$ $uv=0$, and $u^3=0$ or $u^2=uv=v^3=0$. Thus, (i) is obtained. (ii) We have ann$\left(u\right)=⟨v⟩$ if and only if $\mathrm{ann}\left(v\right)=\mathrm{ann}\left(\mathrm{ann}\right(u\left)\right)=⟨u⟩.$ This shows that $uv=0.$ As $u\notin ⟨v⟩=\mathrm{ann}\left(u\right)$ and $v\notin ⟨u⟩=\mathrm{ann}\left(v\right),$ then $u^2\ne$ and $v^2\ne 0.$ Which leads to $J^2=⟨w⟩=⟨u^2,v^2⟩,$ and thus $w=u^2$ or $w=v^2.$ In either case, $u^2=\beta v^2$ for some $\beta \in {\Gamma }^{*}\left(m\right).$ (iii) The second case is when ann$\left(u\right)\supseteq ⟨uv⟩$ if and only if ann$\left(v\right)\supseteq ⟨uv⟩.$ From this, we have $J^2=⟨uv⟩$ and

$$u^2-{\beta }_{1}uv=0,v^2-{\beta }_{2}uv=0.$$

This implies that $uv\ne 0.$ Indeed, if $uv=0,$ then $J^2$ will be 0 which is not possible. In this case, $w=uv.$ As $u^2$ and $v^2$ are in $J^2=⟨uv⟩,$ then the required relations are attained. Note that if ${\beta }_1{\beta }_2=1,$ then after straightforward computations

$$uv=(1-{\beta }_1{\beta }_2)uv.$$

So, $uv=0,$ which is impossible in this case. Thus, ${\beta }_1\ne {\beta }_2^{-1}.$ Which means we return to previous cases. Case (b). Next, we proceed to prove assuming that $n=2.$ As we know from Lemma 2 that $k=2$ if $J^2=⟨p⟩,$ and hence

$$R=GR(p^2,m)+uGR(p^2,m)+vGR(p^2,m).$$

This leads to that $u^2=p{\beta }_1$ and $v^2=p{\beta }_2,$ where ${\beta }_1$ and ${\beta }_2$ are elements of $\Gamma \left(m\right).$ Now, consider the choices for ${\beta }_1$ and ${\beta }_2.$ If ${\beta }_1={\beta }_2=0,$ then we attain (i). Secondly, if ${\beta }_1=\beta \ne 0$ and ${\beta }_2=0,$ we obtain (ii). Lastly, if p is odd and when ${\beta }_1\ne 0$ and ${\beta }_2\ne 0,$ we have (iii). On the other hand, when $p=2,$ we have an extra case which is when $uv=2$ and the other multiplications are $0.$ Note that the case when ${\beta }_2=\beta \ne 0$ and ${\beta }_1=0$ is equivalent to (ii) by replacing u and v with each other. Finally, we prove Case (c). If we assume that $n=3$ or $n=2$ and $p\in J\setminus J^2,$ by Lemma 2, $k=1.$ Thus, $R=GR(p^n,m)+uGR(p^n,m).$ When $n=3,$ then clearly we have $u^2={\beta }_1{p}^2+apu,$ where ${\beta }_1\in \Gamma \left(m\right)$ and $a\in R.$ While if $n=2,$ then $u^2={\beta }_{2}up,$ where ${\beta }_2\in \Gamma \left(m\right).$ Note that the case when $p\ne 2,$ $u^2={\beta }_1{p}^2+apu$ can be reduced to $u^2=p^2\gamma ,$ where $\gamma \in \Gamma \left(m\right)$ and $u^2={\beta }_{2}pu$ to $u^2=0.$ □

In the general setting of ring theory, we have the following result.

Proposition 4.

Let $t=l,$ then R is a chain ring with $p,n,m,t$ and $k=t-1.$

Proof. 

Because $J^l=0$ and $J^{l-1}\ne 0$, the chain

$$R=J^0\supset J\supset J^2\supset \cdots \supset J^{l-1}\supset J^l=0$$

is a composition. Therefore, we have $\nu \left(J^i\right)=1$ for $1\le i\le l-1$. Moreover, $\nu \left(J\right)=1$ as $k\le l-1$. This implies that J is principal, and hence R is a chain ring with $\nu \left(J^i\right)=1$ for all i. Therefore, it follows that $k=l-1.$ □

Prior to moving further with the proof, we recall, in particular, that $\mid A\mid ={\left(p^m\right)}^{l_R\left(A\right)}$ if R is a local ring with residue field ${\mathbb{F}}_{p^m}$ and A is an R-module of finite length $l_R\left(A\right).$ With R having a length of $4,$ in particular, $\mid R\mid =p^{4m}$, and since J’s nilpotency index is $2\le t\le 4,$ so char$\left(R\right)\in \{p,p^2,p^3,p^4\}.$ Therefore, the following are the only possible values for t and $n.$ For convenience, we divide and address each of them in a proposition.

(i)

If $t=2,$ then we consider

(a)

When $n=1.$

(b)

When $n=2.$

(ii)

If $t=3,$ then there are three possibilities

(a)

When $n=1.$

(b)

When $n=2.$

(c)

When $n=3.$

(iii)

If $t=4,$ then we have the following subcases

(a)

When $n=1.$

(b)

When $n=2.$

(c)

When $n=3.$

(d)

When $n=4.$

Observe that when $t=3,$n could be 3 since $p^3\in J^3=0.$ In the sake of simplicity, we intend to set $\alpha$ as a primitive element of ${\mathbb{F}}_{p^m}$ and

$$A=\{{\alpha }^{2i+1}:1\le i\le {\displaystyle \frac{p^m-1}{2}}-1\}=\{\alpha \in {\Gamma }^{*}\left(m\right):\alpha \notin {\left({\Gamma }^{*}\left(m\right)\right)}^2\};$$

$$B=\{{\alpha }^{2i}:1\le i\le {\displaystyle \frac{p^m-1}{2}}-1\}=\{\alpha \in {\Gamma }^{*}\left(m\right):\alpha \in {\left({\Gamma }^{*}\left(m\right)\right)}^2\}.$$

Proposition 5.

Let R be a local ring with $l=4$ and $t=2.$ Then, R is uniquely isomorphic to

$${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v,w]}{{⟨u,v,w⟩}^2}}\mathrm{or}{\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}.$$

Proof. 

Since $t=2,$ thus if $n=1,$ then $\nu \left(J\right)=k=3$ from Proposition 2. Let us start with assumption $n=1.$ Thus, all multiplications among the generators in J are $u^2=v^2=w^2=uv=u=vw=0.$ As $R={\mathbb{F}}_{p^m}+u{\mathbb{F}}_{p^m}+v{\mathbb{F}}_{p^m}+w{\mathbb{F}}_{p^m},$ therefore,

$$R\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v,w]}{{⟨u,v,w⟩}^2}}.$$

Clearly, this ring is unique under isomorphisms and with parameters $p,n=1,m,4,3.$ When $n=2,$ then $\nu \left(J\right)=2,$ and also $k=2,$ i.e., $J=⟨p,u,v⟩.$ By similar reasoning, we obtain one ring which has the structure,

$$R\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}.$$

□

Proposition 6.

Assume that R is a local ring of 4-length with $t=3$ and $n=1.$ Then, R is isomorphic to only one of the following rings:

 (1) 

${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^3,v^2,uv⟩}}.$

 (2) 

${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}.$

 (3) 

${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\alpha v^2,uv⟩}}.$

 (4) 

When $p=2,$ ${\displaystyle \frac{{\mathbb{F}}_{2^m}[u,v]}{⟨u^2,v^2⟩}}.$

Proof. 

The proof will depend essentially on Lemma 2 and Proposition 3. Thus, we divide the proof according to relations in (i), (ii), and (iii). Since $n=1,$ then ${\mathbb{F}}_{p^m}\subset R,$ and moreover, $t=3,$ thus

$$R={\mathbb{F}}_{p^m}+u{\mathbb{F}}_{p^m}+v{\mathbb{F}}_{p^m}+w{\mathbb{F}}_{p^m}.$$

Furthermore, the maximal ideal $J=⟨u,v,w⟩$ is equal

$$\begin{array}{cc}& J=u{\mathbb{F}}_{p^m}+v{\mathbb{F}}_{p^m}+J^2,\hfill \\ & J^2=w{\mathbb{F}}_{p^m}.\hfill \end{array}$$

The minimal generating elements u and v over ${\mathbb{F}}_{p^m}$ satisfy (i), (ii), and (iii) in Proposition 3 (a). By the same proposition, (i) suggests that ann$\left(u\right)=⟨u^2⟩,$ and so $J^2=⟨u^2⟩=⟨w⟩,$ and hence,

$$R_{\left(i\right)}\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^3,v^2,uv⟩}}.$$

Such a ring is uniquely determined by the relations $u^3=v^2=uv=0.$ While if ann$\left(u\right)=⟨v⟩,$ which means $J^2=⟨v^2⟩=⟨u^2⟩,$ then by the same proposition (ii), we obtain $u^2=\beta v^2$ and $uv=0.$ Hence,

$$R_{\left(ii\right)}\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\beta v^2,uv⟩}},$$

where $\beta \in {\Gamma }^{*}\left(m\right).$ One can see that $\beta \ne 0$ because if it equals $0,$ we would return to (i). Now, if $p\ne 2$ and $\beta \in B,$ then we replace $\sqrt{\beta }v$ by $v,$ and thus clearly ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\beta v^2,uv⟩}}\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}.$ In addition, when p is odd, and $\beta \in A,$ then ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\beta v^2,uv⟩}}≆{\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}.$ Indeed, one can check easily that ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\beta v^2,uv⟩}}\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}$ if and only if $p=2.$ Let the two rings be isomorphic. Then, there are ${\gamma }_1,{\gamma }_2,{\gamma }_3,{\gamma }_4\in \Gamma \left(m\right)$ such that $u_1={\gamma }_{1}u+{\gamma }_{2}v$ and $v_1={\gamma }_{3}u+{\gamma }_{4}v$ are the minimal generators of $J.$ These generators must satisfy the relations of u and $v,$ and this gives equations of the form ${\beta }_1^2+{\beta }_2^2=\alpha ({\beta }_3^2+{\beta }_4^2),$ ${\beta }_3^2+{\beta }_4^2\ne 0$ and ${\beta }_1{\beta }_3+{\beta }_2{\beta }_4=0$ if and only if $p=2.$ When $J^2=⟨uv⟩,$ in such a case, ann$\left(u\right)=⟨u-\beta v⟩,$ where $\beta \in {\Gamma }^{*}\left(m\right).$ The generators, therefore, hold the relations (iii) of Proposition 3. This means that $u^2-{\beta }_{1}uv=0$ $v^2-{\beta }_{2}uv=0,$ for some ${\beta }_1$ and ${\beta }_2$ in ${\Gamma }^{*}\left(m\right)$ with condition that ${\beta }_1{\beta }_2\ne 1.$ This implies that R is

$$R_{\left(iii\right)}={\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-{\beta }_{1}uv,v^2-{\beta }_{2}uv⟩}}.$$

For shortness, put $T={\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-{\beta }_{1}uv,v^2-{\beta }_{2}uv⟩}}.$ Observe that T is isomorphic to ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}$ when ${\beta }_1,{\beta }_2\in B,$ with an isomorphism of form $u⟼\sqrt{{\beta }_2}u-\sqrt{{\beta }_1}v$ and $v⟼\sqrt{{\beta }_2}u+\sqrt{{\beta }_1}v.$ Moreover, if $\beta \in A,$ then $T\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\beta v^2,uv⟩}}$ if either ${\beta }_1$ or ${\beta }_2$ is in $A,$ and the desired isomorphism is the one takes $u⟼\sqrt{{\beta }_1^{-1}}u$ and $v⟼\sqrt{{\beta }_1^{-1}}u-v,$ assuming ${\beta }_1\in {\Gamma }^{*}\left(m\right).$ The same argument might be applied when ${\beta }_2\in {\Gamma }^{*}\left(m\right).$ In the case ${\beta }_1,{\beta }_2\in A,$ we have $T\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}$ since one can check that ${\beta }_1{\beta }_2^{-1}\in B,$ and the correspondence can be chosen to take $u⟼u-\sqrt{{\beta }_2^{-1}{\beta }_1}v$ and $v⟼u+\sqrt{{\beta }_2^{-1}{\beta }_1}v.$ Furthermore, when $p\ne 2$ and ${\beta }_1={\beta }_2=0,$ then $T={\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2,v^2⟩}}$ is isomorphic to either ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}},$ when $\sqrt{-1}\in {\Gamma }^{*}\left(m\right)$ by $u⟼u+\sqrt{-1}v$ and $v⟼u-\sqrt{-1}v$ or $T\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\alpha v^2,uv⟩}}$ when $\sqrt{-1}\notin {\Gamma }^{*}\left(m\right)$ via $u⟼{\alpha }^{i+1}u$ and $v⟼u-{\alpha }^{i+1}v,$ where ${\alpha }^{i+1}$ satisfies the equations $x^2+\alpha =0,$ for some $1\le i\le p^m-2,$ and in fact, $4i+3=p^m.$ Note that ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-{\beta }_{1}uv,v^2-{\beta }_{2}uv⟩}}\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-{\beta }_{2}uv,v^2-{\beta }_{1}uv⟩}}$ by the isomorphism $u⟼v$ and $v⟼u.$ Finally, for $p=2,$ since all elements of the maximal ideal of ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2,v^2⟩}}$ have square zero and the elements u and v${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}$ such that $u^2=v^2\ne 0,$ therefore, ${\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2,v^2⟩}}≆{\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}}.$ □

Proposition 7.

Let R be a local ring of length 4 with $t=3$ and $n=2.$ Then, R is one of the following rings:

 (1) 

$R_1={\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2⟩}}.$

 (2) 

When $p=2,$ $R_2={\displaystyle \frac{GR(2^2,m)[u,v]}{⟨u^2-2u⟩}}.$

 (3) 

$R_3={\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3,pu⟩}}.$

 (4) 

$R_4={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}.$

 (5) 

$R_5={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}.$

 (6) 

$R_6={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2,uv,pu,pv⟩}}.$

 (7) 

$R_7={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}.$

 (8) 

$R_8={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p\beta ,uv,pu,pv⟩}},$ where $\beta \in A.$

 (9) 

When $p=2,$ $R_9={\displaystyle \frac{GR(2^2,m)[u,v]}{⟨u^2,v^2,uv-2,2u,2v⟩}}.$

Proof. 

Throughout the proof, assume $n=2.$ We have

$$\begin{array}{cc}& R=GR(p^2,m)+uGR(p^2,m)+vGR(p^2,m).\hfill \end{array}$$

Let $J=⟨u,v⟩$ such that $u,v$ satisfy the equations in Proposition 3 (b). As $t=3,$ then we can not have $p,u,v\in J\setminus J^2,$ i.e., $\nu \left(J\right)=2$ and k must be 2 or $1.$ Therefore, we are left with exactly two cases: the first is when, $p\notin J^2,$ i.e., $k=1$ and so, we must have $pu\ne 0;$ the second case is when $J^2=⟨p⟩.$ Also, we obtain ann$\left(J\right)=J^2=⟨pu⟩$ or ann$\left(J\right)=J^2=⟨p⟩.$ Case a. When $p\notin J^2,$ which results in $pu\ne 0,$ and then $k=1.$ Thus,

$$\begin{array}{cc}& R=GR(p^2,m)+uGR(p^2,m),\hfill \\ & J=pGR(p^2,m)+uGR(p^2,m)+J^2,\hfill \\ & J^2=puGR(p^2,m).\hfill \end{array}$$

In such a case, the generators of J are $p,u.$ Note that $\nu \left(J\right)=2,$ i.e., the minimal generating set of J contains only $p,u$ over ${\mathbb{F}}_{p^m}.$ As $J^2=⟨pu⟩$ and $t=3,$ thus we have a unique relation among the elements $p,u,pu$ which is $u^2=pu\beta ,$ where $\beta \in \Gamma \left(m\right).$ When $p\ne 2,$ the relation can be reduced to $u^2=0,$ either $\beta =0$ or by completing squares otherwise. Thus, when $p\ne 2$ and $pu\ne 0,$ we obtain $R\cong {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2⟩}}.$ This ring uniquely satisfies these conditions when $p\ne 2.$ If $p=2,$ then we obtain $R\cong {\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2⟩}}$ or $R\cong {\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2-2u⟩}}$ according to $\beta =0$ or $\beta \ne 0,$ respectively. While if $pu=0,$ thus $k=2,$ and $J^2=⟨u^2⟩.$ Consequently. R has the form ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3,pu⟩}}$ which is unique under those specifications. Case b. We assume that $J^2=⟨p⟩.$ Therefore, $u,v$ are generating $J/J^2$ and $\nu \left(J\right)=2.$ In this case,

$$\begin{array}{cc}& R=GR(p^2,m)+uGR(p^2,m)+vGR(p^2,m),\hfill \\ & J=uGR(p^2,m)+vGR(p^2,m)+J^2,\hfill \\ & J^2=pGR(p^2,m).\hfill \end{array}$$

Based on Proposition 3 (b) (i), $R\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}}.$ This ring is unique up to isomorphism. Since, $u^2,v^2,uv$ are in $J^2=⟨p⟩,$ thus

$$\begin{array}{c}\hfill u^2={\beta }_{1}p,v^2={\beta }_{2}p,uv={\beta }_{3}p,pu=pv=p^2=0,\end{array}$$

where ${\beta }_1,{\beta }_2,{\beta }_3\in \Gamma \left(m\right).$ Suppose that ${\beta }_1\ne 0$ and ${\beta }_2={\beta }_3=0.$ If ${\beta }_1\in B,$ then ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-{\beta }_{1}p,v^2,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}$ by $u⟼\sqrt{{\beta }_1^{-1}}u$ and $v⟼v.$ While if ${\beta }_1\in A,$ then ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}}$ if and only if $p=2.$ The clarification is similar to that in proof of Proposition 6. Let us assume now ${\beta }_1\ne$ and ${\beta }_2\ne 0.$ In case of ${\beta }_1,{\beta }_2\in B,$ thus ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2-p{\beta }_2,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}$ using the correspondence $u⟼\sqrt{{\beta }_1^{-1}}u$ and $v⟼\sqrt{{\beta }_2^{-1}}v.$ If only one of ${\beta }_1,$ ${\beta }_2$ is in $A,$ hence ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-{\beta }_{1}p,v^2-{\beta }_2,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p{\beta }_2,uv,pu,pv⟩}}$ using the correspondence $u⟼\sqrt{{\beta }_1^{-1}}u$ and $v⟼v.$ If both ${\beta }_1\ne {\beta }_2$ are in $A,$ then ${\beta }_1={\alpha }^{2i+1}$ and ${\beta }_2={\alpha }^{2j+1}$ for some i and $j.$ The map that takes $u⟼{\alpha }^{i+j+1}v$ and $v⟼{\alpha }^{j}u$ is an isomorphism, and thus ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2-p{\beta }_2,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2-p,uv,pu,pv⟩}}.$ In case of ${\beta }_1={\beta }_2,$ and without loss of generality, we let ${\beta }_1={\beta }_2=\alpha .$ Asuume $u⟼\alpha ({\gamma }_{1}u+{\gamma }_{2}v)$ and $u⟼(-{\gamma }_{1}u+{\gamma }_{2}v),$ where ${\gamma }_1^2+{\gamma }_2^2=\alpha ,$ then ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2-p\alpha ,uv,pu,pv⟩}}\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2-p,uv,pu,pv⟩}}.$ We finish the proof assuming that ${\beta }_3\ne 0.$ Denote $T={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2-p{\beta }_2,uv-p{\beta }_3,pu,pv⟩}}.$ First note that the ring T with $({\beta }_1,{\beta }_2,{\beta }_3)=(0,{\beta }_2,{\beta }_3)$ and T with $({\beta }_1,{\beta }_2,{\beta }_3)=(\beta ,0,{\beta }_3)$ are isomorphic by the map $u⟼v$ and $v⟼u.$ So, we suppose that ${\beta }_1\ne 0,$ then note that $u^2={\beta }_1{\beta }_3^{-1}uv,$ and hence $u(u-{\beta }_1{\beta }_{3}v)=0.$ If we let $u⟼u$ and $v⟼u-{\beta }_1{\beta }_{3}v,$ thus $u^2=p{\beta }_1$ and $v^2=p\gamma ,$ where $\gamma ={\beta }_1(-1-{\beta }_1{\beta }_3^{-2}{\beta }_2).$ If $\gamma =0,$ then $T\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2,uv,pu,pv⟩}},$ when $\gamma \ne 0,$ $T\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p{\beta }_1,v^2-p\gamma ,uv,pu,pv⟩}}.$ These two rings were already discussed. Finally, we study the case when ${\beta }_1={\beta }_2=0,$ i.e., $T={\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv-p{\beta }_3,pu,pv⟩}}.$ It can be easily observed that $T\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv-p,pu,pv⟩}}$ using the correspondence $u⟼u$ and $v⟼{\beta }_{3}v.$ If $p\ne 2,$ then $(u^2-v^2)=(u-v)(u+v),$ and if we take $u⟼u-v$ and $v⟼u+v,$ thus $u^2=2p$ and $v^2=-2p.$ When $\sqrt{-1}\in {\Gamma }^{*}\left(m\right),$ then we have $u^2=2p$ and $v^2=2p$ by $v⟼\sqrt{-1}v.$ Thus, $T\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}.$ On the other hand, if $\sqrt{-1}\notin {\Gamma }^{*}\left(m\right),$ thus, by a similar discussion, $T\cong {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2-p,uv,pu,pv⟩}}.$ The case when $p=2,$ it is clear that ${\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv-p,pu,pv⟩}}≆{\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}}.$ □

Proposition 8.

Suppose $t=n=3.$ Then, R is isomorphic to only one of ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\beta ,pu⟩}},$ ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2,pu⟩}}$ or ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2,pu⟩}},$ where $\beta \in A.$

Proof. 

If we have $n=3=t$, we must have $k=1$ by Equation (7). Now, if $p\in J^2$, then we can not have $p^2\in J^4=0.$ Thus, we always have $p\in J\setminus J^2$. This implies that p and u are minimal generators of $J,$ i.e., $\nu \left(J\right)=2$.

$$\begin{array}{cc}& R=GR(p^3,m)+uGR(p^3,m),\hfill \\ & J=pGR(p^3,m)+uGR(p^3,m)+J^2,\hfill \\ & J^2=p^{2}GR(p^3,m).\hfill \end{array}$$

As $J^2=⟨p^2⟩=⟨pu⟩=⟨u^2⟩,$ then we have the following relations

$$\left\{\begin{array}{cc}{u}^2=p^2\beta ,\hfill & \mathrm{if}pu=0;\hfill \\ u^2=pu{\beta }_1,p^2=pu\gamma ,\hfill & \mathrm{if}pu\ne 0,\hfill \end{array}\right.$$

where $\beta ,{\beta }_1,\gamma \in \Gamma \left(m\right).$ Thus, we have $R\cong {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\beta ,pu⟩}}$ or $R\cong {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-pu{\beta }_1,p^2-pu\gamma ⟩}}.$ First, note that there are three different rings; ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2,pu⟩}},$ ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2,pu⟩}}$, and ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\beta ,pu⟩}}$ according to $\beta =0,\beta \in B$ and $\beta \in A,$ respectively. Also, ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\alpha ,pu⟩}}\cong {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2,pu⟩}}$ if and only if $p=2.$ Since $\gamma \ne 0$ since $n=3,$ then $p(p-\gamma u)=0.$ Let $u⟼p-\gamma u,$ then we have $pu=0$ and $u^2=p^2{\beta }^{*},$ where ${\beta }^{*}=-1+\gamma {\beta }_1.$ Thus, ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-pu{\beta }_1,p^2-pu\gamma ⟩}}$ is isomorphic to ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2{\beta }^{*},pu⟩}}$ which is isomorphic to only one of ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2,pu⟩}},$ ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2,pu⟩}}$ or ${\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\beta ,pu⟩}}$ depends on ${\beta }^{*}=0,$ i.e., $\gamma {\beta }_1=1,$ ${\beta }^{*}\in B$, or ${\beta }^{*}\in A,$ respectively. □

Proposition 9.

Assume $t=l=4,$ then R belongs to the family of the following rings:

$$\begin{array}{ccc}{\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^4⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3,pu⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3-p\beta ,pu⟩}},\\ {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p\alpha ⟩}},& {\displaystyle \frac{GR(2^2,m)\left[u\right]}{⟨u^2-2-2u⟩}},\\ GR(p^4,m)& \end{array},$$

where $\beta \in {\Gamma }^{*}\left(m\right).$

Proof. 

Since $t=4=l,$ then R is a chain ring, Proposition 4. In this case, $\nu \left(J\right)=1,$ let $J=⟨u⟩.$ Consider the chain

$$R=J^0\supset J\supset J^2\supset J^3\supset J^4=0.$$

We will prove by considering values for $n,$ where $1\le n\le 4.$ If $n=4,$ then $p\in J\setminus J^2,$ and hence $J=⟨u⟩=⟨p⟩.$ Thus,

$$R=J^0\supset ⟨p⟩\supset ⟨p^2⟩\supset ⟨p^3⟩\supset ⟨p^4⟩=0.$$

Therefore, R is a Galois ring of the form $GR(p^4,m),$ and such rings are uniquely determined by $p,n,m.$ Next, let $n=1.$ As $k=3=l-1$ and $u^4=0,$ then R is written as

$$R={\mathbb{F}}_{p^m}+u{\mathbb{F}}_{p^m}+u^2{\mathbb{F}}_{p^m}+u^3{\mathbb{F}}_{p^m}.$$

Therefore, $R\cong {\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^4⟩}},$ which is unique up to isomorphism. Nevertheless, when $n=3,$ then $GR(p^3,m)\subseteq R.$ Because $J=⟨u⟩,$ so $p\notin J\setminus J^2$ or $p^3\ne 0$ as $J^3\ne 0.$ Now if $p\in J^2,$ then $p^2\in J^4=0,$ and thus $p^2=0$ which is impossible since $n=3.$ We conclude that n can not take $3,$ i.e., no chain rings with $t=4$ and $n=3.$ If $n=2,$ we then study two possible cases. When $p\in J^2\setminus J^3,$ then

$$\begin{array}{cc}& R=GR(p^2,m)+uGR(p^2,m),\hfill \\ & J=uGR(p^2,m)+J^2,\hfill \\ & J^2=pGR(p^2,m)+J^3,\hfill \\ & J^3=u^{3}GR(p^2,m).\hfill \end{array}$$

Observe, $u^2\in J=⟨p,u⟩,$ and hence $u^2=p\beta +p{\beta }_{1}u$ because $pu\ne 0,$ where $\beta ,{\beta }_1\in \Gamma \left(m\right).$

$$\begin{array}{c}\hfill {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p\beta -p{\beta }_{1}u,p{u}^2⟩}}.\end{array}$$

Now, if ${\beta }_1\ne 0,$ then one can apply completing squares in case of $p\ne 2,$ and thus, we obtain $u^2=p\beta .$ Furthermore, when $\beta =0,$ so $u^2=0$ and R possesses the form ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2⟩}}.$ However, the case when $\beta \ne 0,$ the construction of R relays on whether $\beta \in A$ or $\beta \in B.$ Which means that R acquires the form of ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p,p{u}^2⟩}}$ or ${\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p,p{u}^2\alpha ⟩}},$ respectively. Thus, when $p\ne 2,$R is isomorphic to either of

$$\begin{array}{cc}{\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p,p{u}^2⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p\alpha ,p{u}^2⟩}}.\end{array}$$

Conversely, there are two formulas when $p=2$ are $u^2=2u$ and $u^2=2+2u$. The first relation, $u^2=2u$, is not a chain. Furthermore, when $\beta ={\beta }_1=0$, we have $u^2=0$, which leads us back to the previous case. Thus, for $p\ne 2$, we obtain exactly three copies of such rings, while there are four classes when $p=2$. Therefore, we list them as follows:

$$\begin{array}{cc}{\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2-2,2u^2⟩}},& {\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2-2-2u,2u^2⟩}}.\end{array}$$

(8)

The last assertion happens when $p\in J^3.$ In such a case,

$$\begin{array}{cc}\hfill & R=GR(p^2,m)+uGR(p^2,m)+u^{2}GR(p^2,m),\hfill \\ \hfill & J=uGR(p^2,m)+J^2,\hfill \\ \hfill & J^2=u^{2}GR(p^2,m)+J^3,\hfill \\ \hfill & J^3=pGR(p^2,m).\hfill \end{array}$$

(9)

As $pu\in J^4,$ then $pu=0,$ and thus there is a unique relation which is $u^3=p\beta ,$ where $\beta \in {\Gamma }^{*}\left(m\right).$ With respect to $\beta ,$ we give the structure of R as

$$R\cong {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3-p\beta ,pu⟩}}.$$

(10)

The number of rings of the this form N is $(p-1,3).$ This number is given by

$$N=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}p\not\equiv 1\left(\mathrm{mod}3\right),\hfill \\ 3,\hfill & \mathrm{if}p\equiv 1\left(\mathrm{mod}3\right).\hfill \end{array}\right.$$

In summary, there $N+1$ of rings of types in Equation (10). □

As the main outcomes of this section, we consolidate the previously established.

Theorem 1.

Suppose that R is a finite local ring with $p,n,m,l=4,k.$ Then, R is constructed as one of the following:

$$\begin{array}{ccc}{\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^3,v^2,uv⟩}},& {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-v^2,uv⟩}},& {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v]}{⟨u^2-\alpha v^2,uv⟩}},\\ {\displaystyle \frac{{\mathbb{F}}_{2^m}[u,v]}{⟨u^2,v^2⟩}},& {\displaystyle \frac{{\mathbb{F}}_{p^m}[u,v,w]}{{⟨u,v,w⟩}^2}}& {\displaystyle \frac{GR(2^2,m)\left[u\right]}{⟨u^2-2-2u⟩}},\\ {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2⟩}},& GR(p^4,m),& {\displaystyle \frac{GR(4,m)\left[u\right]}{⟨u^2-2u⟩}},\\ {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2,v^2,uv,pu,pv⟩}},& {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2,uv,pu,pv⟩}},& {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p\alpha ,v^2,uv,pu,pv⟩}},\\ {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p,uv,pu,pv⟩}},& {\displaystyle \frac{GR(p^2,m)[u,v]}{⟨u^2-p,v^2-p\alpha ,uv,pu,pv⟩}},& {\displaystyle \frac{GR(2^2,m)[u,v]}{⟨u^2,v^2,uv-2,2u,2v⟩}},\\ {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2\alpha a,pu⟩}},& {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2-p^2,pu⟩}},& {\displaystyle \frac{GR(p^3,m)\left[u\right]}{⟨u^2,pu⟩}},\\ {\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^4⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3,pu⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^3-p\beta ,pu⟩}},\\ {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p⟩}},& {\displaystyle \frac{GR(p^2,m)\left[u\right]}{⟨u^2-p\alpha ⟩}}.\end{array}$$

.

Corollary 2.

The number of non-isomorphic local rings of length 4 is

$$N=\left\{\begin{array}{c}18,\mathrm{if}p=2,\hfill \\ \\ \left\{\begin{array}{cc}19,\hfill & \mathrm{if}p\not\equiv 1\left(\mathrm{mod}3\right)\hfill \\ \\ 21,\hfill & \mathrm{if}p\equiv 1\left(\mathrm{mod}3\right).\hfill \end{array}\right.(p\ne 2)\hfill \end{array}\right.$$

Example 1.

Table 1 provides a complete classification of finite local rings of length 4 with orders $8,$ 16, and $256$.

4. Conclusions

The relevance of local rings characterized by parameters $p,n,m,l$, and k in coding theory has made them a significant focus of research. They are especially valuable for their applications in distance distributions and error correction processes. For instance, the ring ${\displaystyle \frac{{\mathbb{F}}_{p^m}\left[u\right]}{⟨u^4⟩}}$ can be used to generate new sequences with optimal Hamming correlation properties, making it an appropriate choice in coding theory. These sequences are especially beneficial in applications like spread spectrum communication, where they reduce interference and provide reliable multiple access for users sharing the same frequency bandwidth. By producing sequences with superior correlation characteristics, systems can improve performance in noisy environments and counteract various disruptions. This paper examined local rings of length $4,$ meaning rings for which $l=4.$ We successfully categorized all such rings, up to isomorphism, according to their invariants in Theorem 1. As a direct result, we present a comprehensive list of local rings of orders $p^8,$ $16,$ and 256 in Table 1.