# Classification of Finite Rings of Orderp6 by Generators and Relations

Authors: P. Karimi Beiranvand; R. Beyranvand; M. Gholami
Journal: Journal of Mathematics
 (2013)

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## Abstract

For any finite abelian group(R,+), we define a binary operation or “multiplication” on R and give necessary and sufficient conditions on this multiplication for R to extend to a ring. Then we show when two rings made on the same group are isomorphic. In particular, it is shown that there are n+1 rings of order pn with characteristic pn, where p is a prime number. Also, all finite rings of order p6 are described by generators and relations. Finally, we give an algorithm for the computation of all finite rings based on their additive group.


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## Body

## 1. Introduction

The problem of determining and classifying up to isomorphism finite rings has received considerable attention, both old and new, see [1–4]. It is well known that if R is a finite ring (with identity), then the additive group of R splits as the direct of its p-primary components R(p), where p is a prime number, and these are pairwise orthogonal ideals. Thus R is the direct sum of the rings R(p). We provide a new proof of this fact in the paper.Now letp be a prime number. It is easy to see that there are two rings of order p, ℤp and the null ring of order p. In [5], Raghavendran (1969) proved that there exist eleven rings of order p2, only four of these rings have identity. Also in [6], Gilmer and Mott (1973) showed that there exist 4p+48 rings of order p3(p≠2), only twelve of these rings have identity, and 59 rings of order 8, only eleven of these rings have identity. For p4, a comprehensive list of noncommutative rings was first only drawn up in [7]. Commutative rings of order p4 have been characterized by Wilson [4]. Finally, Corbas and Williams (2000) in [1, 8], determined all rings of order p5. The rings of upper order are not still characterized.Throughout the paper, all rings are associative (not necessarily commutative or with identity). For a setX, |X| denotes the cardinal of X. For two integers a,b and positive integer n, we denote a≡nb in case a and b are congruent modulo n, and a∣b in case a divides b. Also, for two integers a and b, gcd(a,b) denotes the greatest common divisor of a and b. In [9], the authors introduced a multiplication (similar to Theorem 3) on a finitely generated S-module to extend to an S-algebra, where S is a commutative ring. Also a number of necessary and sufficient conditions for any S-module to extend to an S-algebra, where S is a commutative ring, is given by Behboodi et al. [9]. Here instead of modules over a commutative ring, we focus our attention to the ℤ-modules. In Section 2, by using the method in [9], we define a binary operation or “multiplication" on a finite abelian group and extend to a ring. In particular, we prove that there are n+1 rings (up to isomorphism) of order pn whose abelian group is cyclic. Also, all rings of order p6 whose abelian group is ℤp⊕ℤp5 are determined. In Section 3, an algorithm for the computation of all finite rings based on their additive group is given.


## 2. A Representation of Finite Rings and Fundamental Theorems

We begin the paper with the following well known fact and give a new proof of this fact.Theorem 1.
LetR be a ring of order p1n1p2n2⋯pknk, where the pi are distinct primes and the ni are positive integers. Then R is expressible, in a unique manner, as the direct sum of rings Ri, where |Ri|=pini,i=1,2,…,k.Proof.
We know that the additive groupR is expressible, in a unique manner, as the direct sum of groups Ri, where |Ri|=pini,i=1,2,…,k. We claim that Ri is an ideal of R, for i=1,…,k. Suppose that ai,bi∈Ri and aibi=x1+x2+⋯+xk where xj∈Rj, for 1≤j≤k. Then aibi-xi = x1 + ⋯ + xi-1 + xi+1 + ⋯ + xk. Since p1n1 ⋯ pi-1ni-1 pi+1ni+1 ⋯ pknk(x1+⋯+xi-1+xi+1+⋯+xk)=0, p1n1 ⋯ pi-1ni-1 pi+1ni+1 ⋯ pknk(aibi-xi)=0. If aibi-xi≠0, then we must have pi∣p1n1⋯pi-1ni-1pi+1ni+1⋯pknk, a contradiction. Thus aibi-xi=0 and so aibi∈Ri. Now suppose that ai∈Ri and aj∈Rj where i≠j and 1≤i,j≤k. We show that aiaj=0. Let aiaj=x1+x2+⋯+xk, where xt∈Rt, for 1≤t≤k. Then piniaiaj=(piniai)aj=0 implies that pini(x1+x2+⋯+xk)=0 and hence pinixt=0 for 1≤t≤k. Now for any t≠i, if xt≠0, then pt∣pini, a contradiction. Thus for all t≠i, xt=0. On the other hand, since pjnjaiaj=ai(pjnjaj)=0, by a similar argument, we conclude that for all t≠j,xt=0. Thus x1=x2=⋯=xk=0 and so for all 1≤i≤k, Ri is an ideal of R, and the proof is complete.Remark 2.
IfR is a finite ring, then its additive group is a finite abelian group and is thus a direct product of cyclic groups. Suppose these have generators a1,a2,…,an of orders m1,m2,…,mn. Then the ring structure is determined by the n2 products
(1)aiaj=∑k=1nwijkakwithwijk∈ℤ
and thus by the n3 structure constants wijk. As [6] we introduce a convenient notation, motivated by group theory, for giving the structure of a finite ring. A presentation for a finite ring R consists of a set of generators a1,a2,…,an of the additive group of R together with relations. The relations are of two types: (i)
miai=0 for i=1,…,n;(ii)
aiaj=∑k=1n‍wijkak with wijk∈ℤ for i,j=1,…,n.
If the ringR has the presentation above we write
(2)R=〈a1,…,an:miai=0,aiaj=∑k=1nwijkak,i,j=1,…,n〉.Theorem 3.
LetR be a finite ring with additive group ⊕i=1nAi, where Ai=〈ai〉 (1≤i≤n) are cyclic subgroups of orders mi, respectively. Assume wijk∈ℤ, λij=(mi,mj), where 1≤i,j,k≤n, and “·” is the following operation
(3)·:R×R→R(∑i=1n‍siai,∑j=1n‍tjaj)→∑k=1n(∑i,j=1n‍sitjwijk)ak.
Then (1)
“·” is well-defined if and only if
(*)mk∣λijwijk∀1≤i,j,k≤n;(2)
“·” is always distributive, in the case when “·” is well-defined;(3)
“·” is associative if and only if
(**)∑α=1nwijαwαlk≡mk∑α=1nwjlαwiαk∀1≤i,j,k,l≤n.Consequently,
(4)〈a1,…,an:miai=0,aiaj=∑k=1nwijkak,i,j=1,…,n〉
is a ring if and only if (*) and (**) above hold.Proof.
(1) (⇒). Assume “·” is well-defined and fix i,j where 1≤i,j≤n. Let x∈miℤ. Then by definition of “·”, we have
(5)∑k=1nxwijkak=(xai)·(aj)=(0)·(aj)=0.
Thus for each k=1,…,n, we deduce xwijkak=0thatis,miwijkℤ⊆mkℤ, for k=1,…,n. Similarly, mjwijkℤ⊆mkℤ, for k=1,…,n. Therefore,
(6)[miℤ+mjℤ]wijk⊆mkℤ∀1≤i,j,k≤n.
Now one can easily check that the last relation holds if and only if (*) holds.
(1) (⇐). Assume
(7)∑i=1nsiai=∑i=1nuiai,∑j=1ntjaj=∑j=1nvjaj,
where si,ui,tj,vj∈ℤ. Then siai=uiai and tjaj=vjaj for all i,j. Let xi=ui-si and yj=vj-tj. Then ui=xi+si, vj=yj+tj, xi∈miℤ and yj∈mjℤ. Hence
(8)uivj-sitj=siyj+xitj+xiyj=xitj+(si+xi)yj∈[miℤ+mjℤ]
and so
(9)(uivj-sitj)wijk∈[miℤ+mjℤ]wijk∀k(1≤k≤n).
Now by condition (*) we have
(10)(uivj-sitj)wijkak=0∀k(1≤k≤n).
It follows that
(11)∑k=1n(∑i,j=1nsitjwijk)ak=∑k=1n(∑i,j=1nuivjwijk)ak,
that is,
(12)∑i=1nsiai·∑j=1ntjaj=∑i=1nuiai·∑j=1nvjaj.
Thus the operation “·” is well-defined.
(2) Suppose the operation “·” is well-defined. Then
(13)(∑i=1n‍siai+∑i=1n‍uiai)·(∑j=1n‍tjaj)=(∑i=1n‍(si+ui)ai)·∑j=1ntjaj=∑k=1n(∑i,j=1n(si+ui)tjwijk)ak=∑k=1n(∑i,j=1nsitjwijk)ak+∑k=1n(∑i,j=1nuitjwijk)ak=(∑i=1nsiai)·(∑j=1ntjaj)+(∑i=1nuiai)·(∑j=1ntjaj).
On the other hand(14)(∑i=1nsiai)·(∑j=1ntjaj+∑i=1nvjaj)=(∑i=1nsiai)·(∑j=1n(tj+vj)aj)=∑k=1n(∑i,j=1nsi(tj+vj)wijk)ak=∑k=1n(∑i,j=1nsitjwijk)ak+∑k=1n(∑i,j=1nsivjwijk)ak=(∑i=1nsiai)·(∑j=1ntjaj)+(∑i=1nsiai)·(∑j=1nvjaj).
Thus “·” is a distributive operation.
(3) By definition of “·”, we have
(15)ai·aj=∑k=1nwijkak∀1≤i,j≤n
and so
(16)(ai·aj)·al=(∑α=1nwijαaα)·al=∑k=1n(∑α=1n‍wijαwαlk)ak
and also
(17)ai·(aj·al)=ai·(∑α=1nwjlαaα)=∑k=1n(∑α=1n‍wjlαwiαk)ak.
Now, it is clear that “·” is associative if and only if (ai·aj)·al=ai·(aj·al) for all 1≤i,j,l≤n, if and only if
(18)∑α=1nwijαwαlk≡mk∑α=1nwjlαwiαk∀1≤i,j,k,l≤n·Theorem 4.
Forn∈ℕ, let
(19)R1=〈a1,…,an:miai=0,aiaj=∑k=1nwijkak,i,j=1,…,n〉,R2=〈a1,…,an:miai=0,aiaj=∑k=1nzijkak,i,j=1,…n〉
be two presentations with suitable wijk,zijk∈ℤ, (1≤i,j,k≤n). Assume pij∈ℤ where 1≤i,j≤n and f is the following map:
(20)f:R1→R2,f(∑i=1nsiai)=∑j=1n(∑i=1nsipij)bj.
Then (1)
f is well-defined if and only if for 1≤i,j≤n,
(21)mj∣mipij.(2)
Iff is well-defined, then
(a)
f is an ℤ-module homomorphism,(b)
f is one to one if and only if for each s1,s2,…,sn∈ℤ,mj∣∑i=1m‍sipij implies that mj∣sj for all 1≤j≤n, and(c)
f is a ring homomorphism if and only if for 1≤i,j,k≤n,
(22)∑α,β=1npiαpjβwαβk≡mk∑l=1nzijlplk.Proof.
(1) (⇒). Assume s∈miℤ, where 1≤i≤n. Then by definition f we have
(23)∑j=1nspijbj=f(sai)=f(0)=0.
It follows that spijbj=0 for all 1≤j≤n. Therefore, mipijℤ⊆mjℤ, for all 1≤j≤n and so for 1≤i,j≤n,
(24)mj∣mipij.
(1) (⇐). Suppose ∑i=1n‍siai=∑i=1n‍tiai where si,ti∈ℤ. Then si-ti∈miℤ for all i (1≤i≤n) and so (si-ti)pij∈mipijℤ. Now by our hypothesis mj∣(si-ti)pijthatis,sipijbj=tipijbj for all 1≤i,j≤n. Therefore,
(25)∑j=1n(∑i=1nsipij)bj=∑j=1n(∑i=1ntipij)bj.
This means that
(26)f(∑i=1nsiai)=f(∑i=1ntiai)
and hence f is well-defined.
(2) (a) Assumer,si,ti∈ℤ, for i=1,…,n. Clearly,
(27)f(r∑i=1nsiai+∑i=1ntiai)=f(∑i=1n(rsi+ti)ai)=∑j=1n(∑i=1n(rsi+ti)pij)bj=∑j=1n(∑i=1nrsipij)bj+∑j=1n(∑i=1ntipij)bj=rf(∑i=1nsiai)+f(∑i=1ntiai).
Thus f is an ℤ-module homomorphism.
(2) (b) By definitionf, (b) is clear.
(2) (c) By (2) (a),f is always a group homomorphism. Thus it is sufficient to show that f(aiaj) = f(ai)f(aj) for each i,j(1≤i,j≤n). However by multiplications of R1, R2 and definition of f we have
(28)f(aiaj)=f(∑l=1nwijlal)=∑k=1n(∑l=1n‍wijlplk)bk
and also
(29)f(ai)f(aj)=(∑s=1npisbs)(∑t=1npjtbt)=∑k=1n(∑s,t=1npispjtzstk)bk.
Thus f is a ring homomorphism if and only if for all 1≤i,j,k≤n,
(30)∑α,β=1npiαpjβwαβk≡mk∑l=1nzijlplk.Corollary 5.
Letn be a positive integer and let Ai=〈ai〉 (1≤i≤n) be cyclic additive groups of orders mi, respectively. If
(31)R1=〈a1,…,an:miai=0,aiaj=∑k=1nwijkak,i,j=1,…,n〉,R2=〈a1,…,an:miai=0,aiaj=∑k=1nzijkak,i,j=1,…,n〉
are two presentations with suitable wijk,zijk∈ℤ(1≤i,j,k≤n), then R1≅R2 if and only if there exist pij∈ℤ (1≤i,j≤n) such that(1)
mj∣mipij, for 1≤i,j≤n;(2)
for allsi∈ℤ and 1≤j≤n, mj∣∑i=1m‍sipij implies that mj∣sj for all 1≤j≤n, and(3)
∑α,β=1n‍piαpjβwαβk≡mk∑l=1n‍zijlplk, for 1≤i,j,k≤n.Proof.
It is clear by Theorem4.


## 3. Finite Rings of Orderp6 with Characteristic p6 and p5

In this section, we first classify all finite rings of orderp6 with characteristic p6. Then this classification is extended to the rings order pn with characteristic pn.Proposition 6.
(i) LetR=〈a〉 be an additive cyclic group of order p6 and w∈ℤ. Then R = 〈a:p6a=0,a2=wa〉 is always a ring.
(ii) LetR1=〈a:p6a=0,a2=wa〉 and R2=〈a:p6a=0,a2=za〉 be two presentations with w,z∈ℤ. Assume q∈ℤ and f is the following map:
(32)f:R1→R2,f(ra)=rqa.
Then (1)
f is well-defined;(2)
f is a ring homomorphism if and only if for wq≡p6zq2;(3)
f is one to one if and only if for each r∈ℤ,p6∣rq implies that p6∣r.Proof.
(i) It is straightforward by relations(*) and (**) in Theorem 3.
(ii) The proof is obtained by simplification of the relations in Theorem4.Since the function introduced in preceding proposition is multiplication by someq∈ℤ, and is always well-defined, we have the following corollary.Corollary 7.
LetR1=〈a:p6a=0,a2=wa〉 and R2=〈a:p6a=0,a2=za〉 be two presentations with w,z∈ℤ. Then R1≅R2 if and only if there exists q∈ℤ such that(1)
wq≡p6zq2;(2)
for eachr∈ℤ,p6∣rq implies that p6∣r.Proof.
It is clear by Proposition6.Theorem 8.
For any prime numberp, there are exactly seven rings of order p6 whose additive group is cyclic, only one of these rings have identity. However all of these are commutative.Proof.
LetR1=〈a:p6a=0,a2=wa〉 and R2=〈a:p6a=0,a2=za〉 be two presentations with w,z∈ℤ. The relation (2) in Corollary 7 implies that p is relatively prime to q and so by the relation (1) in Corollary 7, we deduce that w≡p6zq. Since the relation w≡p6zx has a solution, gcd(p6,z)∣w. Now we proceed by cases.
Case1. gcd(p,w)=1, that is, w is relatively prime to p. Since gcd(p6,z)∣w, z is also relatively prime to p. Thus R1≅R2 where both w and z are relatively prime to p.
Case2. gcd(p2,w)=p. Then the relation (p6,z)∣w shows that either gcd(z,p)=1 or z∈{p,2p,3p,…,(p-1)p}. But the relation w≡p6zq follows that w≡pzq and hence p∣zq because gcd(p2,w)=p. Since gcd(p,q)=1, p∣z. Therefore we obtain that z∈{p,2p,3p,…,(p-1)p} and so gcd(p2,z)=p. Thus R1≅R2 where both gcd(p2,w)=p and gcd(p2,z)=p.
Case3. gcd(p3,w)=p2. Then the relation gcd(p6,z)∣w shows that either gcd(z,p)=1 or z∈{p,2p,3p,…,(p-1)p,p2,2p2,3p2,…,(p-1)p2}. But the relation w≡p6zq follows that w≡p2zq and hence p2|zq because gcd(p2,w)=p. Since gcd(p,q)=1, p2∣z. Therefore we obtain that z∈{p2,2p2,3p2,…,(p-1)p2} and so gcd(p3,z)=p2. Thus R1≅R2 where both gcd(p3,w)=p2 and gcd(p3,z)=p2.
Case4. gcd(p4,w)=p3. Then the relation gcd(p6,z)∣w shows that either gcd(z,p)=1 or
(33)z∈{p,2p,3p,…,(p-1)p,p2,2p2,3p2,…,(p-1)p2,p3,2p3,3p3,…,(p-1)p3}.
But the relation w≡p6zq follows that w≡p3zq and hence p3∣zq because gcd(p4,w)=p3. Since gcd(p,q)=1, p3∣z. Therefore we obtain that z∈{p3,2p3,3p3,…,(p-1)p3} and so gcd(p4,z)=p3. Thus R1≅R2 where both gcd(p4,w)=p3 and gcd(p4,z)=p3.
Case5. gcd(p5,w)=p4. By a similar argument, we deduce R1≅R2 where both gcd(p5,w)=p4 and gcd(p5,z)=p4.
Case6. gcd(p6,w)=p5. By a similar argument, we deduce R1≅R2 where both gcd(p6,w)=p5 and gcd(p6,z)=p5.
Case7. gcd(p6,w)=p6. By a similar argument, we deduce R1≅R2 where both gcd(p6,w)=p6 and gcd(p6,z)=p6.An analysis similar to that in the proof of Theorem8 shows that there are exactly n+1 rings of order pn with characteristic pn. These rings are
(34)〈a:pna=0,a2=wa〉,w
∈
{
1
,
p
,
p
2
,
…
,
p
n
}We can obtain the following result by applying relations(*) and (**) in Theorem 3, for the rings of order p6 with characteristic p5. Also, this manner may be used for obtaining all finite rings of order p6 with characteristic pk, where 1≤k≤4.Corollary 9.
LetR=A1⊕A2 where A1 = <a1> and A2 = <a2> be cyclic groups of order p and p5, respectively. Assume wijk∈ℤ where 1≤i,j,k≤2. Then R=〈a1,a2:pa1=p5a2=0,aiaj=∑k=12‍wijkak,i,j=1,2〉 is a ring if and only if the following hold.(1)
p4∣w112,p4∣w122 and p4∣w212;(2)
w111w122≡p5w121w112;(3)
w211w212≡p5w121w122≡p5w221w112;(4)
w211w112≡p5w111w212+w112w222;(5)
w211w122+w212w222≡p5w121w212+w122w222;(6)
w221w122≡p5w212w221.Application and simplification of the relations (1), (2), and (3) in Corollary5, for the rings of order p6 with characteristic p5, give us the following result.Corollary 10.
Let(35)R=〈a1,a2:pa1=p5a2=0,aiaj=∑k=12wijkak,i,j=1,2〉,S=〈a1,a2:pa1=p5a2=0,aiaj=∑k=12zijkak,i,j=1,2〉
be two rings with suitable integers wijk and zijk satisfied in Theorem 3, for 1≤i,j,k≤2. Then R≅S if and only if there exist pij∈ℤ such that the following hold.(1)
p4∣p12;(2)
Ifp∣s1p11+s2p21 and p5∣s1p12+s2p22, then p∣s1 and p5∣s2;(3)
p112≡pz111(p11+p21);(4)
p11p21w111+p11p22w121≡pz121(p11+p21);(5)
p212w111+p22p21w211+p222w221≡pz221(p11+p21);(6)
p112w112≡p5z112p22;(7)
p21p11w112+p22p11w212+p22p12w222≡p5z212p22;(8)
p11p21w112+p11p22w122+p12p22w222≡p5z122p22;(9)
p212w112+p22p21(w212+w122)+p222w222≡p5z222(p12+p22).In the following, one gives an algorithm for computing the finite rings whose abelian group is isomorphic toℤm1⊕ℤm2⊕⋯⊕ℤmn. We represent the algorithm in the form of two sub-algorithms. In Algorithm 11, one lets (m1,m2,…,mn) be the input and A be the set of all sequences (wijk) that satisfy the condition 0≤wijk≤mk-1, for any 1≤i,j,k≤n. Then one has the sequences (wijk) that satisfy the conditions (*) and (**) in Theorem 3 as the outputs of the algorithm. In fact, any sequence (wijk) that is an output of this algorithm shows a ring as the following:
(36)〈a1,a2…,an:miai=0,aiaj=∑k=1nwijkak,i,j=1,…n〉.
In Algorithm 12, one represents the conditions required for the isomorphism of two rings based upon Theorem 4. In fact, one verifies the two sequences (wijk) and (zijk) obtained from Algorithm 11, under isomorphic conditions in Theorem 4. Therefore under Algorithms 11 and 12, one computes all the rings whose abelian group is isomorphic to ℤm1⊕ℤm2⊕⋯⊕ℤmn and are not isomorphic. We have implemented Algorithms 11 and 12 in C++ and presents a sample of its responses for the rings whose abelian group is isomorphic to ℤ2⊕ℤ32 in the following example.Algorithm 11.
(1) Input(m1,m2,…,mn)
(2) LetN=:{1,2,…,n}
(3) For all(i,j)∈N2, let λij=:gcd(mi,mj)
(4) LetA=:{w=(wijk)(i,j,k)∈N3; for all (i,j,k)∈N3,0≤wijk≤mk-1}
(5)B=:{}
(6) IfA=B, then go to end
(7) Choosew=(wijk)(i,j,k)∈N3∈A∖B
(8) LetB=:B∪{w}
(9) If there is(i,j,k)∈N3 such that λijwijk≇mk0, then go to step 6
(10) If there is(i,j,k,l)∈N4 such that ∑α=1n‍wijαwαlk≇mk∑α=1n‍wjlαwiαk, then go to step 6
(11) Printw=(wijk)(i,j,k)∈N3
(12) EndAlgorithm 12.
(1) Input(m1,m2,…,mn)
(2) LetN=:{1,2,…,n}
(3) For all(i,j)∈N2, let λij=:gcd(mi,mj)
(4) Inputw=(wijk)(i,j,k)∈N3 and z=(zijk)(i,j,k)∈N3 satisfied in Algorithm 11
(5) LetA=:{P=(pij)(i,j)∈N2; for all (i,j)∈N2,0≤pij≤mj-1}
(6)B=:{}
(7) IfA=B, then go to end
(8) ChooseP=(pij)(i,j)∈N2∈A∖B
(9) LetB=:B∪{P}
(10) If there is(i,j)∈N2 such that miPij≇mj0, then go to step 7
(11) LetC=:{}
(12) LetD=:{r=(rj)j∈N; for all j∈N,0≤rj≤mj-1,∑i=1n‍ripij≅mj0}
(13) Chooser∈D
(14) LetC=:C∪{r}
(15) If there isi∈N such that ri≇mi0, then go to step 7
(16) If there is(i,j,k)∈N3 such that ∑α,β=1n‍piαpjβwαβk≇mk∑l=1n‍zijlplk, then go to step 7
(17) Print two ringsR and S with representations w=(wijk)(i,j,k)∈N3 and z=(zijk)(i,j,k)∈N3, respectively, are isomorphic
(18) EndExample 13.
We compute the presentations of all finite rings of order26, whose abelian group is isomorphic to ℤ2⊕ℤ32. Let R be a such ring. Then we write the following presentation for R=〈a1,a2:2a1=25a2=0,aiaj=wij1a1+wij2a2,i,j=1,2〉:
(37)R=[w111w112w121w122w211w212w221w222].
The outputs of these algorithms for the rings of order26 whose abelian group is ℤ2⊕ℤ32 are listed below. 							(38)[00000000],[00000001],[00000002],[00000004],[00000008],[000000016],[000160000],[000160002],[000160004],[000160008],[000001600],[000001602],[000001604],[000001608],[0001601600],[0001601602],[0001601604],[0001601608],[00016016016],[016000000],[016000002],[016000004],[016000008],[0160000016],[0160160000],[0160160002],[0160160004],[0160160008],[01601600016],[00000010],[00000012],[00000014],[00000018],[0001601610],[0001601612],[0001601614],[0001601618],[00100001],[00001001],[00101001],[016101001],[10000000],[10000001],[10000002],[10000004],[10000008],[100000016].
Thus there exist47 rings of order 26 with characteristic 25.



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*Source: 467905-2013-01-15.xml*