local rings of order p 5
Andrzej Nowicki 21.07.2018, acl-11
Abstract
We present explicit descriptions of a certain class of finite commutative local rings of order p5, where p is a small prime. If R is a ring from this class and M is its unique maximal ideal, then char(R) = p2, pM = 0, M3 = 0, p ∈ M2 and |M2| = p2. Known results of Corbas and Williams play the main role in our paper.
1 Introduction
Throughout this paper all rings are assumed to be commutative with unity. We are interested in descriptions, up to isomorphism, of finite rings of small orders. Every finite ring is Artinian, and it is (see for example [2], [16]) a finite product of local rings. The orders of finite local rings are powers of primes. Hence, for a classification of finite rings of a given order, we need to know a classification of finite local rings of order pn, where p is a prime and n is a positive integer. There are several important papers ( [20], [13], [17], [12], [18], [15], [23], [1], [14], [11] [19]) concerning finite rings (not necessary commutative) of orders pn with n 6 4. In 2000, B. Corbas and G. D.
Williams published two interesting papers [8] and [9]. In the first their paper we find a classification up to isomorphism of all local rings (commutative and non-commutative) for n 6 4. In the second their paper [9] there is such classification for n = 5. Thanks to this paper we are ready to write a full list of local rings (not necessary commutative) of order p5 for almost all p.
There is only one especially difficult case arising in characteristic p2. We say that rings of this case are specific. More exactly, let R be a local ring of order p5 and let M be its unique maximal ideal. We say that R is specific if char (R) = p2, pM = 0, M3 = 0, p ∈ M2 and |M2| = p2. In [5], some results on specific rings are given. A full description of specific rings is based on some, rather extensive, matrix descriptions given by Graham D. Williams in [21]. We are interested only in commutative cases.
The authors of the mentioned paper [9] do not repeat this description. They inform only that if p 6= 2 then there are 10 distinct specific rings, and there are 6 such rings for p = 2. Similar information, without presentations, we find in the papers of M.
Behboodi and R. Beyranvand ([3], [4]).
In this paper we present explicit descriptions of specific rings for small p. We apply the mentioned results of Williams. The papers [21], [8] and [9] play an important role in our paper.
2 Initial restrictions and observations
For all a, b, c, d, u, v ∈ Zp, we denote by hh
at + bp, ut + vp, ct + dpii
the ring Zp2[x, y, t]/I, where I is the ideal of Zp2[x, y, t] equals
(x2− at − bp, xy − ut − vp, y2− ct − dp, t2, xt, yt, px, py, pt).
It is not difficult to show (see [9] 696, [21]) that every specific ring R is of the above form hh
at + bp, ut + vp, ct + dpii
where a, b, c, d, u, v are elements of Zp such that the symmetric matrices a u
u c
,
b v v d
are nonzero and linearly independent over Zp, that is, the triples A = (a, u, c) and B = (b, v, d) are linearly independent over Zp. Lemma 2.1. Let R =hh
at + bp, ut + vp, ct + dpii
be a specific ring.
(1) If a 6= 0, then R is isomorphic to a ring R0 of the formhh
t, u0t + v0p, c0t + d0pii . (2) If a = 0 and u 6= 0, then R is isomorphic to a specific ring R0 of the form hh
bp, t, c0t + d0pii .
Proof. (1). Assume that a 6= 0 and denote by s the inverse of a in Zp. Put R0 =hh
t, u0t + v0p, c0t + d0pii
, where: u0 = su, v0 = v − sbu, c0 = sc and d0 = d − sbc.
It is clear that R0 is specific. Let ϕ be the ring automorphism of Zp2[x, y, t] defined by ϕ(x) = x, ϕ(y) = y and ϕ(t) = st − sbp. Then we see that ϕ(I) = I0, where I and I0 are the ideals of Zp2[x, y, t] associated with R and R0, respectively. Thus, by Proposition 3.2, the rings R and R0 are isomorphic.
(2). Assume that a = 0, u 6= 0 and denote by s the inverse of u in Zp. Put R0 =hh
bp, t, c0t + d0pii
, where: c0 = sc and d0 = d − scv. It is clear that R0 is specific.
Let ϕ be the ring automorphism of Zp2[x, y, t] defined by ϕ(x) = x, ϕ(y) = y and ϕ(t) = st − svp. Then we see that ϕ(I) = I0, where I and I0 are the ideals of Zp2[x, y, t]
associated with R and R0, respectively. Thus, by Proposition 3.2, the rings R and R0 are isomorphic.
Lemma 2.2. If p > 3, then every specific ring is isomorphic to a specific ring of the form hh
t, u0t + v0p, c0t + d0pii
with u0, v0, c0, d0 ∈ Zp. Proof. Let R = hh
at + bp, ut + vp, ct + dpii
. If a 6= 0, the this assertion follows from Lemma 2.1(1). If c 6= 0, then we use the automorphism x 7→ y, y 7→ x, t 7→ t, and we have the previous case.
Now assume that a = c = 0. In this case u 6= 0 (because the triple (a, u, 0) is nonzero). Repeating the proof of Lemma 2.1(2) we see that R is isomorphic to the specific ring S = [bp, t, dp] = Zp2[x, y, t]/I, where I = (x2 − dp, xy − t, y2 − dp, t2, tx, ty, px, py, pt). Consider the automorphism ϕ of Zp2[x, y, t] defined by ϕ(x) = x + y, ϕ(y) = y, ϕ(t) = t. Put h3 = y2− dp, h2 = xy − (t − dp), h1 = x2+ 2t + (d − b)p.
Then we have
ϕ(y2− dp) = y2 − dp = h3,
ϕ(xy − t) = (x + y)y − t = xy + y2− t = xy + (y2− dp) − (t − dp) = h2+ h3, ϕ(x2− bp) = (x + y)2− bp = x2+ 2xy + y2− bp
= x2+ 2(xy − t + dp) + 2t − 2dp + (y2 − dp) + dp − bp
= h1+ 2h2+ h3.
Hence, ϕ(I) = J , where J is the ideal associated with the specific ring S0 =hh
− 2t + (b − d)p, t − dp, dpii .
This implies (see Proposition 3.2) that the rings S and S0 are isomorphic. Since p > 3, we have −2 6= 0 and so, using again Lemma 2.1(1), we finish this proof.
Lemma 2.3. Let w1 = at+bp, w2 = ut+vp, w3 = ct+dp, where a, b, c, d, u, v ∈ Zp, and let k ∈ Zp. Then the specific rings
hh
w1, w2, w3
ii and
hh
w1, w2−kw1, w3−2kw2+k2w1
ii are isomorphic.
Proof. Consider the automorphism x 7→ x, y 7→ y + kx, t 7→ t, and use the same method as in the proofs of the previous lemmas.
Now we introduce the following new notation. Let D
w, r, s E
:= Zp2[x, y]
.
x3, xy + wp, y2− rx2+ sp, px, py , where w, r, s ∈ Zp. Observe that if w 6= 0 or s 6= 0, then
D w, r, s
E
is a specific ring.
Proposition 2.4. If p > 3, then every specific ring is isomorphic to a ring of the form D
w, r, sE
with w 6= 0 or s 6= 0.
Proof. Let R be a specific ring. We already know (see Lemma 2.2) that R is isomorphic to a specific ring R0 of the formhh
t, u0t + v0p, c0t + d0pii
, where u0, v0, c0, d0 ∈ Zp. Using Lemma 2.3 for k = u0, we may assume that u0 = 0. Moreover, using an automorphism of Zp2[x, y, t] given by x 7→ x, y 7→ y and t 7→ −t, we may assume that R = Zp2[x, y, t]/I, where I =
x2+ t, xy + wp, y2+ rt + sp, t2, tx, ty, px, py, pt
, for some w, r, s ∈ Zp. Since R is specific, we have either w 6= 0 or s 6= 0. Now consider the ring homomorphism f : Zp2[x, y, t] → Zp2[x, y] defined by f (x) = x, f (y) = y, f (t) = −x2. This homomorphism is surjective, and f (I) = J , where J is the ideal of Zp2[x, y] equals
x3, xy +wp, y2−rx2+sp, px, py
. Thus, by Proposition 3.2), the rings R andD
w, r, sE are isomorphic.
Proposition 2.5. Let R =D
w, r, sE
and p > 3. If w 6= 0, then R is isomorphic to the specific ring
D
1, w2r, s E
.
Proof. Put I := (x3, xy + wp, y2− rx2+ sp, px, py). Let f be an automorphism of Zp2[x, y] given by f (x) = wx, f (y) = y. Then f (I) = J , where J =
x3, xy + p, y2− rw2x2+ sp, px, py
. Hence, by Proposition 3.2, the rings Zp2[x, y]/I = D
w, r, sE and Zp2[x, y]/J =D
1, w2r, sE
are isomorphic.
Colorary 2.6. If p > 3, then every specific ring is isomorphic to a specific ring D
w, r, sE
, where either w = 0 or w = 1.
Now we know that if p > 3, then for a description of all specific rings of order p5, we need to find a classification, up to isomorphism, of all rings
D w, r, s
E
where either w = 1 or w = 0 with s 6= 0.
Repeating the above proofs and propositions we obtain the following isomorphisms.
Proposition 2.7. hh
t + bp, ut + vp, ct + dpii
≈D
w, r, sE
, where w = ub − v,
r = c − u2
s = bc − d + 2uv − 2u2b.
If ub − v 6= 0, then hh
t + bp, ut + vp, ct + dpii
≈D
1, r(ub − v)2, sE .
3 Isomorphisms of specific rings
First we present some propositions concerning isomorphisms of rings.
Proposition 3.1. Let σ be an automorphism of a ring R, and let I be an ideal of R.
Then σ(I) is also an ideal of R, and the rings R/I and R/σ(I) are isomorphic.
Proof. Since σ is surjective, σ(I) is an ideal of R. Let ϕ : R/I → R/σ(I) be a mapping defined by ϕ(x + I) = σ(x) + σ(I), for all x ∈ R. It is clear that ϕ is a well defined isomorphism of rings.
Proposition 3.2. Let R and S be rings, I an ideal of R and J an ideal of S. Assume that the rings R/I and S/J are finite, and they have the same order. If there exists a surjective homomorphism of rings f : R → S such that f (I) = J , then the rings R/I and S/J are isomorphic.
Proof. Let f : R → S be a surjective homomorphism such that f (I) = J . Let ϕ : R/I → S/J be a mapping defined by ϕ(r + I) = f (r) + J for all r ∈ R.
Observe that ϕ well defined and it is a surjective homomorphism of rings. Since the rings R/I and S/J are of the same finite cardinality, ϕ is also injective. Therefore, ϕ is an isomorphism.
Let us recall that D
w, r, sE
= Zp2[x, y]/J with J =
x3, xy + wp, y2 − rx2 + sp, px, py
, where w, r, s ∈ Zp, and either w 6= 0 or s 6= 0. Every element of D
w, r, sE has a unique presentation of the form au2+ bu + cv + e, with u = x + J and v = y + J , where a, b, c ∈ Zp and e ∈ Zp2.
Proposition 3.3. Let R1 = D
w1, r1, s1E
and R2 = D
w2, r2, s2E
be specific rings of order p5. The rings R1 and R2 are isomorphic if and only if there exist elements b1, b2, c1, c2 ∈ Zp satisfying, in the field Zp, the following five conditions:
(∆1) b1c2− b2c1 6= 0, (∆2) b1b2 + r1c1c2 = 0,
(∆3) w2 = w1(b1c2+ b2c1) + s1c1c2, (∆4) r2(b21+ r1c21) = b22+ r1c22,
(∆5) s2 = s1(c22− r2c12) + 2w1(b2c2− r2b1c1) .
Proof. Denote by u and v the cosets of x and y in R1, respectively. Denote by µ and ν the cosets of x and y in R2, respectively. Then we have: u3 = 0, uv = −w1p, v2 = r1u2 − s1p, pu = 0, pv = 0, µ3 = 0, µν = −w2p, ν2 = r2µ2− s2p, pµ = 0 and pν = 0. Observe that u2v = 0 and µ2ν = 0.
Assume that f : R2 → R1 is an isomorphism of rings. Let:
f (µ) = a1u2+ b1u + c1v + d1p, u = f (α1µ2+ β1µ + γ1ν + δ1p) , f (ν) = a2u2+ b2u + c2v + d2p, v = f (α2µ2+ β2µ + γ2ν + δ2p) , where all the coefficients belong to Zp. Then we have
f (µ2) = (b21+ r1c21) u2− (s1c21+ 2b1c1w1) p, f (ν2) = (b22+ r1c22) u2− (s1c22+ 2b2c2w1) p,
f (µν) = (b1b2+ r1c1c2) u2− (w1b1c2+ w1b2c1+ s1c1c2) p,
Since −w2p = µν, we have −w2p = f (−w2p) = f (µν) and so, we obtain the equality
−w2p = (b1b2+ r1c1c2) u2− (w1b1c2+ w1b2c1+ s1c1c2) p.
which implies (∆2) and (∆3). By the same way, starting from the equality s2p = r1µ2− ν2, we obtain (∆4) and (∆5). Moreover,
u = f (α1µ2+ β1µ + γ1ν + δ1p)
= α1
(b21+ r1c21) u2 − (s1c21+ 2b1c1w1) p + β1
a1u2+ b1u + c1v + d1p +γ1
a2u2+ b2u + c2v + d2p + δ1p.
Comparing the coefficients of u and v, we get β1b1+ γ1b2 = 1, β1c1+ γ1c2 = 0. By a similar way, starting from v = f (α2µ2+ β2µ + γ2ν + δ2p), we obtain β2b1 + γ2b2 = 0 and β2c1+ γ2c2 = 1. Therefore,
β1 γ1 β2 γ2
· b1 c1 b2 c2
= 1 0 0 1
and so, we have (∆1).
Now assume that b1, b2, c1, c2 are elements of Zp satisfying (∆1) − (∆5). Let ϕ : Zp2[x, y] → Zp2[x, y] be a linear automorphism given by ϕ(x) = b1x + c1y, ϕ(y) = b2x + c2y. It is easy to check that ϕ(J2) = J1, where J1, J2 are the ideals of Zp2[x, y]
defined by
J1 = (x3, xy + w1p, y2− r1x2+ s1p, px, py) , J2 = (x3, xy + w2p, y2− r2x2+ s2p, px, py) .
This implies, by Proposition 3.1, that the rings R1 and R2 are isomorphic, because R1 = Zp2[x, y]/J1 and R2 = Zp2[x, y]/J2. This completes the proof.
Very often we will apply the above proposition. Let R1 = hw1, r1, s1i, R2 = hw2, r2, s2i, and assume that f : R2 → R1 is an isomorphism. Let b1, b2, c1, c2 be elements of Zp satisfying (∆1) − (∆5), then we will say that they are basic elements of f .
Proposition 3.4. Assume that f : R2 → R1 is an isomorphism of rings, where R1 = D
w1, r1, s1
E
and R2 = D
w2, r2, s2
E
. If b1, b2, c1, c2 are basic elements of f , then b21 + r1c21 6= 0, and then r2 = bb222+r1c22
1+r1c21. Moreover, c22+ r2c21 6= 0,
c2b2+ r2c1b1 = 0,
r1(c22+ r2c12) = b22+ r2b21,
d2w1 = w2(b1c2+ b2c1) + s2c1b1,
d2s1 = s2(b21− r1c12) + 2w2(r1c1c2− b1b2) , where d = b1c2− b2c1 6= 0.
Proof. Suppose b21+ r1c21 = 0. Let d := b1c2− b2c1 6= 0. Then, by (∆2), we have r1c1d = r1c1b1c2 − (r1c21)b2 = r1c1b1c2+ b21b2 = b1(b1b2+ r1c1c2) = b1· 0 = 0.
But d 6= 0, so r1c1 = 0. If c1 = 0, then b1 = 0 (because b21+ r1c21 = 0) and then d = 0;
a contradiction. If r1 = 0 then b1 = 0 and, by (4), b2 = 0 so, d = 0; a contradiction.
Hence, b21+ r1c21 6= 0 and, by (∆4), we have r2 = (b22+ r1c22)/(b21+ r1c21).
We know, by (∆1), that d 6= 0 so, we have the invertible 2 × 2 matrices M = b1 b2
c1 c2
and M−1 = 1 d
c2 −b2
−c1 b1
.
Hence, c1/d, −b2d/, −c1d/, b1/d are basic elements of the isomorphism f−1 : R1 → R2, and hence, the remaining relations follow from Proposition 3.3.
Using the above propositions and a computer we are ready to present some full descriptions of all specific rings for small primes p > 3. In every case we obtain 10 pairwise nonisomorphic rings.
Very often we will write (w, r, s) instead ofD
w, r, sE
. Using a computer we see that we have always 10 triples (w, r, s). Exactly 5 of them are with w = 0, and exactly 5 have w = 1. In every case we see triples (0, 0, 1), (0, 1, 1), (1, 0, 0) and (0, 0, a) where a is a non-square modulo p. We will say that a description satisfying these conditions is standard.
4 Descriptions of some cosets
We assume that p > 3, and we denote by Kp the set of all nonzero squares of Zp. We denote by K0p the set of non-squares of Zp. Moreover, we denote by ε the smallest (with respect to ordinary inequality) element of Kp0. Note that |Kp| =
K0p
= p−12 . Several times we will use the following well known lemma (see for example [10]).
Lemma 4.1.
(1) If p ≡ 1 (mod 4), then −1 ∈ Kp and −ε ∈ K0p. (2) If p ≡ 3 (mod 4), then −1 ∈ K0p and −ε ∈ Kp. (3) If p 6= 5 then 5 ∈ Kp ⇐⇒ p ≡ ±1 (mod 10).
(4) Let 0 6= m ∈ Zp. If −m ∈ Kp0, then there exist a, b ∈ Zp such that ε = a2+ mb2.
Let us recall (see Corollary 2.6) that every specific ring is isomorphic to a specific ring
D w, r, s
E
, where either w = 0 or w = 1. We denote by Tp the set of all triples (w, r, s) such that w ∈ {0, 1}, r, s ∈ Zp and either w 6= 0 or s 6= 0. This set has exactly 2p2− p elements. If a = (w1, r1, s1) and b = (w2, r2, s2) are elements of Tp, then we will say that they are equivalent, and we will write a ∼ b, if the rings D
w1, r1, s1E and D
w2, r2, s2E
are isomorphic. For a given t ∈ Tp we denote by [t]p the coset of t, that is, [t]p = {t0 ∈ Tp; t0 ∼ t} .
We fix the following notations:
t1 = (0, 0, 1), t2 = (0, 0, ε), t3 = (0, 1, 1), t6 = (1, 0, 0).
Proposition 4.2. For every prime p > 3 we have:
(a) [t1]p = n
(w, 0, s); w ∈ {0, 1}, s ∈ Kp
o . (b) [t2]p =n
(w, 0, s); w ∈ {0, 1}, s ∈ K0po . (c) [t6]p =n
t6o .
Proof. Let a = (w, r, s) ∈ Tp, and a ∼ t1. Let b1, b2, c1, c2 be basic elements. We have: w1 = 0, r1 = 0, s1 = 1, w2 = w, r2 = r, s2 = s. Moreover, we have: b21 6= 0, b1b2 = 0, rb21 = b22, and s = c22 − rc21. Hence, b1 6= 0, b2 = 0, c2 6= 0, r = 0, and consequently s ∈ Kp. Thus, we proved (a). The proofs of (b) and (c) are similar.
Now we will describe the coset [t3]p. Look at several examples.
[t3]3 = = n
(0, 1, 1), (0, 1, 2), (1, 1, 0)o , [t3]5 = n
(0, 1, 1), (0, 1, 4), (1, 4, 1), (0, 4, 4), (1, 1, 0)o ,
[t3]7 =
(0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 1, 5), (0, 1, 6), (0, 2, 1), (0, 2, 2), (0, 2, 3), (0, 2, 4), (0, 2, 5), (0, 2, 6), (0, 4, 1), (0, 4, 2), (0, 4, 3), (0, 4, 4), (0, 4, 5), (0, 4, 6), (1, 1, 0), (1, 1, 2), (1, 1, 5), (1, 2, 0), (1, 2, 1), (1, 2, 6), (1, 4, 0), (1, 4, 3), (1, 4, 4)
.
We know (see Proposition 3.3) that a triple (w, r, s) belongs to [t3]p if and only if there exist basic elements, that is, there exist elements b1, b2, c1, c2 ∈ Zp satisfying, in the field Zp, the following five conditions: b1c2 − b2c1 6= 0, b1b2 + c1c2 = 0, w = c1c2, r (b21+ c21) = b22+ c22, s = c22− c21. Moreover (see Proposition 3.4), b21+ c21 6= 0. Observe that always (1, 1, 0) ∈ [t3]p. In fact, take b1 = c1 = c2 = 1, b2 = −1, and apply Proposition 3.3.
Proposition 4.3.
[t3]p =
M ∪n
(0, r, s); r, s ∈ Kpo
, if p ≡ 1 (mod 4),
M ∪n
(0, r, s); r ∈ Kp, 0 6= s ∈ Zp
o
, if p ≡ 3 (mod 4), where M is the set of all triples of the form
1, r, j −rj
, where r, j are elements of Kp such that j2+ r 6= 0 in Zp.
Proof. Put X = {(0, r, s); r ∈ Kp, 0 6= s ∈ Zp} , Y = {(0, r, s); r, s ∈ Kp} . Let t = (0, r, s) be a triple belonging to [t3]p, and let b1, b2, c1, c2 ∈ Zp, be basic elements as in Proposition 3.3. Then c1c2 = 0 and b1b2 = 0. Hence, c1 = 0 or c2 = 0.
Assume that c1 = 0. Then b1 6= 0, c2 6= 0 and b2 = 0, and we have: s = c22, r = (c2/b1)2 and so, r, s ∈ Kp.
Now assume that c2 = 0 Then c1 6= 0, b2 6= 0 and b1 = 0. Hence, r = (b2/c1)2, so r ∈ Kp, and s = −rc21 = −(b2/c1)2c21 = −b22 = (−1) · b22. If p ≡ 1 (mod 4), then −1 is a square modulo p, and this implies that s ∈ Kp. If p ≡ 3 (mod 4), then −1 ∈ Kp0, and this implies that s ∈ K0p.
In each case, r ∈ Kp, and we see that if p ≡ 1 (mod 4) then [t3]p ⊂ Y , and if p ≡ 3 (mod 4) then [t3]p ⊂ X. Moreover, now it is clear that we may construct the elements b1, b2, c1, c2, needed for the opposite inclusions.
Now assume that (1, r, s) ∈ [t3]p, and let b1, b2, c1, c2 ∈ Zp be basic elements. Then c1c2 = 1 and b1b2 = −1 so, these elements are nonzero, and c1 = c−12 , b1 = −b−12 in Zp. Moreover, r = bb222+c22
1+c21 = (b2c2)2, s = c22− rc21 = c2− (b2cc22)2 2
= c22 − b22. Hence, r ∈ Kp, and s = j −rj where j = c22 ∈ Kp. Since b21 + c21 6= 0, we have j2 + r 6= 0 in Zp. Thus, we proved that (1, r, s) ∈ M.
Now let t =
1, r, j − rj
, where r, j ∈ Kp and j2 + r 6= 0 in Zp. Let r = u2 and j = v2, where u, v are nonzero elements of Zp. Put
b1 = −v
u, b2 = u
v, c1 = 1
v, c2 = v.
Then t ∈ [t3]p, because the above elements b1, b2, c1, c2 are basic.
Denote by ta the triple (1, 1, 1) ∈ Tp. We would like to find the smallest (with respect to lexicographic order), element of [ta]p. It is easy to check that (1, 1, 0) 6∈ [ta]p, and (1, 0, s) 6∈ [ta]p for all s ∈ Zp.
Proposition 4.4. The following two conditions are equivalent.
(1) (1, 1, 1) ∼ (0, r, s) for some r, s ∈ Zp, s 6= 0.
(2) p ≡ ±1 (mod 10).
Proof. (1) ⇒ (2). Assume that (0, r, s) ∼ (1, 1, 1) and b1, b2, c1, c2 are basic elements. Then, by (∆3) and (∆5), we have 1 = sc1c2 = s(c22 − c21). Hence, c1 6= 0, c2 6= 0, and c1c2 = c22− c21. Put x = c1, b = c2. Then x2− bx − b2 = 0 and this implies that 5 is a square modulo p. Hence, p ≡ ±1 (mod 10).
(2) ⇒ (1). Assume that p ≡ ±1 (mod 10). Then 5 is a square modulo p. Let 5 = a2, 0 6= a ∈ Zp. It is clear that a + 1 6= 0 in Zp. Take b1 = a+12 , b2 = −1, c1 = 1, c2 = a+12 , r = 1, s = a+12 . Then (0, r, s) ∼ (1, 1, 1) and b1, b2, c1, c2 are basic elements.
Thus, the triple ta = (1, 1, 1) is minimal in [ta]p if and only if p 6≡ ±1 (mod 10).
It follows from the above proof, that in the opposite case we have always a triple, smaller than (1, 1, 1), of the form (0, 1, s). Examples: (0, 1, 1) belongs to [ta]p for p = 11, 19, 29, 31, 59, 71, 79, 89, and (0, 1, 3) ∈ [ta]41, (0, 1, 2) ∈ [ta]61.
We already know (see Corollary 2.6) that every specific ring for p > 3 is isomorphic to a specific ring hw, r, si, where either w = 0 or w = 1.
Lemma 4.5. If p > 3 and s 6= 0, then (a) (0, 0, s) ∼ (1, 0, s−1),
(b) (0, 1, s) ∼ (1, s−2, 0),
(c) (0, r, s) ∼ (1, rs−2, s−1(1 − r)), if r 6= 0 and r 6= −1, (d) (0, −1, s) ∼ (1, −(2s)−2, 7(4s)−1), if p > 5.
Proof. We have two specific rings R1 = hw1, r1, s1i and R2 = hw2, r2, s2i. In every case we take the elements b1, b2, c1, c2 as in the table:
case w1 r1 s1 w2 r2 s2 b1 b2 c1 c2
a 0 0 s 1 0 s−1 1 0 1 s−1
b 0 1 s 1 s−2 0 1 −s−1 1 s−1
c 0 r s 1 rs−2 (1 − r)s−1 1 −rs−1 1 s−1 d 0 −1 s 1 −(2s)−2 7(4s)−1 2 (2s)−1 1 s−1
In every case the elements b1, b2, c1, c2 are basic. Thus, by Proposition 3.3, the rings R1 and R2 are isomorphic.
Note particular cases: (0, 0, 1) ∼ (1, 0, 1) ,
(0, 1, 1) ∼ (1, 1, 0) ∼ (0, s, s) ∼ 1, s−1, s−1− 1
for s 6= 0.
As a consequence of the above lemma and the previous propositions we obtain the following theorem
Theorem 4.6. If p > 5, then every specific ring is isomorphic to a specific ring of the form h1, r, si, where r, s ∈ Zp.
We will say that a specific ring R has a normal form if R is isomorphic to a specific ring of the form h1, r, si, where r, s ∈ Zp.
5 Standard descriptions
We will construct a set {t1, t2, . . . , t10} of representatives of T with respect to ∼.
In this set will appear triples t1 = (0, 0, 1), t2 = (0, 0, ε), t3 = (0, 1, 1), t6 = (1, 0, 0) introduced in Section 4. The important results of Graham D. Williams [21] will play the main role in our construction.
Let us recall that every specific ring R has a formhh
at + bp, ut + vp, ct + dpii
, where a, b, c, d, u, v are elements of Zp such that the triples A := (a, u, c) and B := (b, v, d) are linearly independent over Zp. It is known ([7], [22]) that we may assume that either A = (1, 0, 0) or A = (1, 0, 1) or A = (1, 0, ε). In every case we have a = 1.
5.1 The case A = (1,0,0)
If A = (1, 0, 0) then, by [21] Proposition 2, the triple B = (b, v, d) has one of the following three form: B1 = (0, 0, 1), B2 = (0, 0, ε), B3 = (0, 1, 0). Thus, in this case we have three specific rings:
hh
t, 0, 1p ii
, hh
t, 0, εp ii
, hh
t, 1p, 0 ii
, which are, by Proposition 2.7, isomorphic to: D
0, 0, −1E ,D
0, 0, −εE , D
1, 0, 0E
, respectively.
If p ≡ 1 (mod 4), then −1 is a square modulo p. In this case we have −1 ∈ Kp and
−ε ∈ K0p and so, by Proposition 4.2, (0, 0, −1) ∼ t1 and (0, 0, −ε) ∼ t2.
If p ≡ 3 (mod 4), then −1 ∈ Kp0 and −ε ∈ Kp and so, by Proposition 4.2, (0, 0, −1) ∼ t2 and (0, 0, −ε) ∼ t1. Therefore, in the case A = (1, 0, 0) we obtain three representatives: t1, t2 and t6.
5.2 The case A = (1,0,1), p = 4k +1
Now we assume that p ≡ 1 (mod 4), and A = (1, 0, 1). In this case −1 is a nonzero square modulo p (see Lemma 4.1). Let −1 = m2 with 0 6= m ∈ Zp. Consider the following 2 × 2 matrices
Q = m m 1 −1
, D = Q−1 =
" 1
2m 1 2 1 2m −12
#
, Dt =
" 1
2m 1 2m 1 2 −12
# .
Let Bi = DtCiD for i = 1, 2, . . . 5, where C1 = 1 0
0 1
, C2 = ε 0 0 ε
, C3 = 0 0 0 1
, C4 = 0 0 0 ε
, C5 = 1 0 0 ε
. We have 5 symmetric matrices:
B1 =
"
−12 0 0 12
#
, B2 =
"
−ε2 0 0 2ε
#
, B3 =
"
−14 −4m1
−4m1 14
#
, B4 =
"
−4ε −4mε
−4mε ε4
# ,
B5 =
"
−1+ε4 1−ε4m
1−ε 4m
1+ε 4
# .
We know from [21] Proposition 2 that by this way we obtain 5 new representatives:
hh
t − 12p, 0, t + 12pii
, hh
t − ε2p, 0, t + 2εpii , hh
t − 14p, −4m1 p, t +14pii , hh
t − ε4p, −4mε p, t +4εp ii
, hh
t − 1+ε4 p, −1−ε4mp, t + 1+ε4 p ii
. They give, by Proposition 2.7, the following representative triples
h1(p) = (0, 1, −1), h2(p) = (0, 1, −ε), h3(p) = 1, −161, −12 , h4(p) =
1, −16ε2, −2ε
, h5(p) =
1, −(1−ε)16 2, −1+ε2 .
Observe that (see Proposition 4.3) h1(p) ∼ t3 = (0, 1, 1) for all p > 3. It is easy to check that h2(p) ∼ (0, 1, ε) for all p > 3. Examples:
h3(5) = (1, 4, 2), h3(13) = (1, 4, 6), h3(17) = (1, 1, 8), h3(29) ∼ (1, 4, 10), h3(37) ∼ (1, 3, 5), h3(41) ∼ (1, 1, 18), h3(53) ∼ (1, 4, 14), h3(61) ∼ (1, 3, 7), h4(5) ∼ (1, 1, 1), h4(13) ∼ (1, 1, 3), h4(17) ∼ (1, 2, 3), h4(29) ∼ (1, 1, 5), h4(37) ∼ (1, 1, 12), h4(41) ∼ (1, 2, 19), h4(53) ∼ (1, 1, 7), h4(61) ∼ (1, 1, 22), h5(5) ∼ (1, 1, 2), h5(13) ∼ (1, 1, 1), h5(17) ∼ (1, 1, 1), h5(29) ∼ (1, 1, 2), h5(37) ∼ (1, 1, 1), h5(41) ∼ (1, 1, 3), h5(53) ∼ (1, 1, 1), h5(61) ∼ (1, 1, 2).
5.3 The case A = (1,0,1), p = 4k +3
Now we assume that p ≡ 3 (mod 4), and A = (1, 0, 1). In this case −1 is a non- square modulo p (see Lemma 4.1). Consider the field L = Zp(m), where m = √
−1, and take the following 2 × 2 matrices over L (the same as in the previous case):
Q = m m 1 −1
, D = Q−1 =
" 1
2m 1 2 1 2m −12
#
, Dt =
" 1
2m 1 2m 1 2 −12
# .
Let a, b be elements of Zp such that a2 + b2 = ε (see Lemma 4.1), and let Bi = DtCiD for i = 1, 2, where C1 = 1 0
0 1
, C2 = a + bm 0
0 a − bm
. that is, B1 =
"
−12 0 0 12
#
, B2 =
"
−a2 b2
b 2
a 2
#
. We know from [21] Proposition 2 that by this way we obtain 2 new representatives:
hh t − 1
2p, 0, t + 1 2p
ii ,
hh t − a
2p, b
2p, t +a 2p
ii .
They give, by Proposition 2.7, the triples g1(p) = (0, 1, −1), g2(p) =
1,b42, −a . Observe (see Proposition 4.3) that g1(p) ∼ t3 = (0, 1, 1) for all p > 3. Examples:
g2(3) ∼ (1, 1, 1), g2(7) ∼ (1, 1, 1), g2(11) ∼ (1, 1, 2), g2(19) ∼ (1, 1, 2), g2(23) ∼ (1, 1, 1), g2(31) ∼ (1, 1, 3), g2(43) ∼ (1, 1, 1), g2(47) ∼ (1, 1, 1).
5.4 The case A = (1,0, ε), p = 4k +1
Let p ≡ 1 (mod 4), and A = (1, 0, ε). In this case −ε is a non-square modulo p (see Lemma 4.1). Consider the field L = Zp(m), where m = √
−ε, and take the matrices D and Dt over L, the same as in the previous cases. We need two elements a, b ∈ Zp
(see Lemma 4.1) such that a2+ εb2 = ε. In our case the elements a = 0 and b = 1 are good. Let Bi = DtCiD for i = 1, 2, where C1 = 1 0
0 1
, C2 = m 0
0 −m
, that is, B1 =
"
−2ε1 0 0 12
#
, B2 =
"
0 12
1
2 0
#
. Now, by [21] Proposition 2, we obtain 2 new representatives: hh
t −2ε1p, 0, εt + x12pii , hh
t, 12p, εtii
. They give, by Proposition 2.7, the following representative triples f1(p) = (0, ε, −1), f2(p) = 1,ε4, 0 , Examples:
f1(5) = (0, 2, 1), f1(13) = (0, 2, 1), f1(17) = (0, 3, 1), f1(29) ∼ (0, 2, 1), f1(37) ∼ (0, 2, 1), f1(41) ∼ (0, 3, 1), f1(53) ∼ (0, 2, 1), f1(61) ∼ (0, 2, 1), f2(5) ∼ (1, 2, 0), f2(13) ∼ (1, 2, 0), f2(17) ∼ (1, 3, 0), f2(29) ∼ (1, 2, 0), f2(37) ∼ (1, 2, 0), f2(41) ∼ (1, 3, 0), f2(53) ∼ (1, 2, 0), f2(61) ∼ (1, 2, 0).
5.5 The case A = (1,0,ε), p = 4k +3
In this last subsection we assume that p ≡ 3 (mod 4), and A = (1, 0, ε). In this case
−ε is a nonzero square modulo p. Let −ε = m2 with 0 6= m ∈ Zp, and let Q, D, Dt be
the matrices over Zp, as in the previous subsections. Let Bi = DtCiD for i = 1, 2, . . . 5, where
C1 = 1 0 0 1
, C2 = ε 0 0 ε
, C3 = 0 0 0 1
, C4 = 0 0 0 ε
, C5 = 1 0 0 ε
. We have 5 symmetric matrices:
B1 =
"
−2ε1 0 0 12
#
, B2 =
"
−12 0 0 2ε
#
, B3 =
"
−4ε1 −4m1
−4m1 14
#
, B4 =
"
−14 −4mε
−4mε 4ε
# ,
B5 =
"
−1+ε4ε 1−ε4m
1−ε 4m
1+ε 4
# .
We know from [21] Proposition 2 that by this way we obtain 5 new representatives:
hh
t − 2ε1p, 0, εt +12pii
, hh
t − 12p, 0, εt + ε2pii , hh
t − 4ε1p, −4m1 p, εt +14pii , hh
t − 14p, −4mε p, εt +4εpii , hh
t − 1+ε4ε p, −1−ε4mp, εt + 1+ε4 pii . They give, by Proposition 2.7, the following representative triples
e1(p) = (0, ε, −1), e2(p) = (0, ε, −ε), e3(p) = 1, −161 , −12 , e4(p) =
1, −ε162, −ε2
, h5(p) =
1, −(1−ε)16 2, −1+ε2 .
e1(3) ∼ (0, 2, 2), e1(7) ∼ (0, 3, 3), e1(11) ∼ (0, 2, 2), e1(19) ∼ (0, 2, 2), e1(23) ∼ (0, 5, 5), e1(31) ∼ (0, 3, 3), e1(43) ∼ (0, 2, 2), e1(47) ∼ (0, 5, 5), e2(3) ∼ (0, 2, 1), e2(7) ∼ (0, 3, 1), e2(11) ∼ (0, 2, 1), e2(19) ∼ (0, 2, 1), e2(23) ∼ (0, 5, 1), e2(31) ∼ (0, 3, 1), e2(43) ∼ (0, 2, 1), e2(47) ∼ (0, 5, 1), e3(3) ∼ (1, 2, 1), e3(7) ∼ (1, 3, 3), e3(11) ∼ (1, 2, 5), e3(19) ∼ (1, 2, 7), e3(23) ∼ (1, 5, 7), e3(31) ∼ (1, 3, 22), e3(43) ∼ (1, 2, 11), e3(47) ∼ (1, 5, 11), e4(3) ∼ (1, 2, 2), e4(7) ∼ (1, 3, 4), e4(11) ∼ (1, 2, 6), e4(19) ∼ (1, 2, 12), e4(23) ∼ (1, 5, 16), e4(31) ∼ (1, 3, 9), e4(43) ∼ (1, 2, 32), e4(47) ∼ (1, 5, 36), e5(3) ∼ (1, 2, 0), e5(7) ∼ (1, 3, 0), e5(11) ∼ (1, 2, 0), e5(19) ∼ (1, 2, 0), e5(23) ∼ (1, 5, 0), e5(31) ∼ (1, 3, 0), e5(43) ∼ (1, 2, 0), e5(47) ∼ (1, 5, 0).
6 Specific rings for odd p smaller than 100
We know from Theorem 4.6 that if p > 5 then every specific ring has a normal form. If p = 3 then there are specific rings without normal forms.
Example 6.1. If p = 3 then the specific rings D
0, 2, 1E
and D
0, 2, 2E
are nonisomor- phic, and they are without normal forms.
Proof. Let R1 = D
0, 2, 1 E
and R2 = D
0, 2, 2 E
. Suppose that R1 ≈ D 1, r, s
E for some r, s ∈ Z3. Let b1, b2, c1, c2 ∈ Z3 be basic elements of this isomorphism. Then,
by (∆4) and (∆2), we have the equalities b1b2 = c1c2 = 1. This implies that b1 = b2, c1 = c2 and so, we have a contradiction with (∆1).
Suppose that R2 ≈ D
1, r, sE
for some r, s ∈ Z3. Let b1, b2, c1, c2 ∈ Z3 be basic elements of this isomorphism. Then, by (∆4) and (∆2), we have the equalities b1b2 = c1c2 = 2. This implies that b1c2− b2c1 = 0 so, we have a contradiction with (∆1).
Now suppose that R1 ≈ R2 and let b1, b2, c1, c2 be basic elements. Then, by (∆3) and (∆5), we have c1c2 = 0 and c21+ c22 = 2 in Z3; a contradiction.
Proposition 6.2. There are, up to isomorphism, exactly 10 specific rings of order 35 : R1 = Z9[x, y]/(x3, xy, y2+ 3, 3x, 3y) = D
0, 0, 1E , R2 = Z9[x, y]/(x3, xy, y2+ 6, 3x, 3y) = D
0, 0, 2E , R3 = Z9[x, y]/(x3, xy, y2− x2+ 3, 3x, 3y) = D
0, 1, 1E , R4 = Z9[x, y]/(x3, xy, y2− 2x2+ 3, 3x, 3y) = D
0, 2, 1E , R5 = Z9[x, y]/(x3, xy, y2− 2x2+ 6, 3x, 3y) = D
0, 2, 2E , R6 = Z9[x, y]/(x3, xy + 3, y2, 3x, 3y) = D
1, 0, 0E , R7 = Z9[x, y]/(x3, xy + 3, y2− x2+ 3, 3x, 3y) =
D 1, 1, 1
E , R8 = Z9[x, y]/(x3, xy + 3, y2− 2x2, 3x, 3y) = D
1, 2, 0E , R9 = Z9[x, y]/(x3, xy + 3, y2− 2x2+ 3, 3x, 3y) = D
1, 2, 1E , R10 = Z9[x, y]/(x3, xy + 3, y2− 2x2+ 6, 3x, 3y) = D
1, 2, 2E . Let us repeat that
D
w, r, sE
:= Zp2[x, y].
x3, xy + wp, y2− rx2+ sp, px, py ,
where w, r, s ∈ Zp. In the following examples we have 3 < p < 100. We write (w, r, s) instead of
D w, r, s
E
. The first set presents a standard description of specific rings of orders p5. Every triple t = (w, r, s), which appears in this set, is the minimal triple, with respect to a lexicographic order, in the coset [t]p. Observe that always we have 5 triples with w = 0 and 5 triples with w = 1.
The second set presents a normal description of specific rings of orders p5. Every element t = (w, r, s) of this set is the minimal triple (with respect to a lexicographic order) belonging to [t]p such that w = 1.
p = 5:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (0, 2, 1), (1, 0, 0), (1, 1, 1), (1, 1, 2), (1, 2, 0), (1, 4, 2).
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 2, 0), (1, 2, 1), (1, 4, 0), (1, 4, 2).
p = 7:
(0, 0, 1), (0, 0, 3), (0, 1, 1), (0, 3, 1), (0, 3, 3), (1, 0, 0), (1, 1, 1), (1, 3, 0), (1, 3, 3), (1, 3, 4)
(1, 0, 0), (1, 0, 1), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 3, 0), (1, 3, 2), (1, 3, 3), (1, 3, 4), (1, 3, 5)
p = 11:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 2, 1), (0, 2, 2), (1, 0, 0), (1, 1, 2), (1, 2, 0), (1, 2, 5), (1, 2, 6)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 2), (1, 2, 0), (1, 2, 1), (1, 2, 5), (1, 2, 6), (1, 2, 9)
p = 13:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (0, 2, 1), (1, 0, 0), (1, 1, 1), (1, 1, 3), (1, 2, 0), (1, 4, 6)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 1), (1, 1, 3), (1, 1, 5), (1, 2, 0), (1, 2, 1), (1, 4, 6)
p = 17:
(0, 0, 1), (0, 0, 3), (0, 1, 1), (0, 1, 3), (0, 3, 1), (1, 0, 0), (1, 1, 1), (1, 1, 8), (1, 2, 3), (1, 3, 0)
(1, 0, 0), (1, 0, 1), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 2), (1, 1, 8), (1, 2, 3), (1, 3, 0), (1, 3, 1)
p = 19:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 2, 1), (0, 2, 2), (1, 0, 0), (1, 1, 2), (1, 2, 0), (1, 2, 7), (1, 2, 12)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 2), (1, 2, 0), (1, 2, 1), (1, 2, 4), (1, 2, 7), (1, 2, 12)
p = 23:
(0, 0, 1), (0, 0, 5), (0, 1, 1), (0, 5, 1), (0, 5, 5), (1, 0, 0), (1, 1, 1), (1, 5, 0), (1, 5, 7), (1, 5, 16)
(1, 0, 0), (1, 0, 1), (1, 0, 5), (1, 1, 0), (1, 1, 1), (1, 5, 0), (1, 5, 2), (1, 5, 3), (1, 5, 7), (1, 5, 16)
p = 29:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (0, 2, 1), (1, 0, 0), (1, 1, 2), (1, 1, 5), (1, 2, 0), (1, 4, 10)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 2), (1, 1, 3), (1, 1, 5), (1, 2, 0), (1, 2, 1), (1, 4, 10)
p = 31:
(0, 0, 1), (0, 0, 3), (0, 1, 1), (0, 3, 1), (0, 3, 3), (1, 0, 0), (1, 1, 3), (1, 3, 0), (1, 3, 9), (1, 3, 22)
(1, 0, 0), (1, 0, 1), (1, 0, 3), (1, 1, 0), (1, 1, 3), (1, 3, 0), (1, 3, 2), (1, 3, 9), (1, 3, 10), (1, 3, 22)
p = 37:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 1, 2), (0, 2, 1), (1, 0, 0), (1, 1, 1), (1, 1, 12), (1, 2, 0), (1, 3, 5)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 1), (1, 1, 6), (1, 1, 12), (1, 2, 0), (1, 2, 1), (1, 3, 5)
p = 41:
(0, 0, 1), (0, 0, 3), (0, 1, 1), (0, 1, 3), (0, 3, 1), (1, 0, 0), (1, 1, 3), (1, 1, 18), (1, 2, 19), (1, 3, 0)
(1, 0, 0), (1, 0, 1), (1, 0, 3), (1, 1, 0), (1, 1, 1), (1, 1, 3), (1, 1, 18), (1, 2, 19), (1, 3, 0), (1, 3, 2)
p = 43:
(0, 0, 1), (0, 0, 2), (0, 1, 1), (0, 2, 1), (0, 2, 2), (1, 0, 0), (1, 1, 1), (1, 2, 0), (1, 2, 11), (1, 2, 32)
(1, 0, 0), (1, 0, 1), (1, 0, 2), (1, 1, 0), (1, 1, 1), (1, 2, 0), (1, 2, 1), (1, 2, 3), (1, 2, 11), (1, 2, 32)
