HIRAMINE, YUTAKA

Osaka City University Repository

Hiramine, Y.
Osaka J. Math.
22 (1985), 61-69
A GENERALIZATION OF HALL QUASIFIELDS
YUTAKA HIRAMINE
(Received March 26, 1984)
1. Introduction
Let Q=Q(+, °) be a right quasifield which satisfies the following condi-
tions :
(1.1) Q is a two dimensional left vector space over its kernel K with a
basis {1, λ}.
(1.2) There exist two mappings r and s from K*=K— {0} into K such
that every element ξ=a-\-b\ of Q not in K satisfies the equation ξ2—r(b)ξ—
s(b)=0.
(1.3) Each element of K commutes with all the elements of Q.
Several examples of such Q are known. For example, the Hall quasi-
fields satisfy the conditions above, where r and s are constant functions and
the quadratic polynomial x2—rx— s is irreducible over K. Moreover, the
quasifields which correspond to the spread sets constructed by Narayana Rao
and Satyanarayana [3] also satisfy the conditions above, where r(x)=3x~1
ί
 s(x)=
2χ-2andK=GF(52"-1).
The purpose of this paper is to study the quasifields satisfying the condi-
tions (1. !)-(!. 3). In §2 we prove the following theorem which gives a con-
dition for £?(+, °) to be a quasifield.
Theorem 1. Let K be a field and let r and s be mappings from K* into
K such that (i) of— r(u)x— s(u) is irreducible over K for each u^K^ and (ii) v2
—r(x)v—s(x)=wx has a unique solution in K* for each v^K, w^K*. Let
Q={x+y\\x, y^K} be a left vector space over K. If a multiplication o on Q
is defined by
\jsx-\-tx\ ify=Q ,
where F(x, y)=χ?— r(y)x— s(y), then Q(+, o) is a quasifield which satisfies (1.1)-
(1.3).
Let K=GF(q) and let Φ^ be the set of the ordered pairs (r, s) such that
r and s satisfy (i) and (ii) of Theorem 1. The spread set Σ
r
,
s
 which corre-
62 Y. HlRAMINE
sponds to (r, s)^Φ
κ
 is defined as follows: S
r
,
s
={M(#, y)\x, y^K}, where
ϊ) for X^K and M(x> ^=(/(*, y) g(x, y)) for X^K and y^K*'
Here/(#, y)= —y"1(x2—r(y)x—s(y)) and #(#, y)= —x+r(y).
Let τr
r>s be the translation plane constructed by 2ΓfS and set L(#, ^)={(^,
vM(x, y))\v<=KxK} for A?, j eJ^, L(oo)={(0, 0, v)\v^KxK}. Let G be
the linear translation complement of τr
r>s and set A={L(x, 0)\x^K} U {L(oo)}
and Ω= {L(x, y)\x^Ky y^K*}. In §3 we prove the following theorem.
Theorem 2. If G^OO^CO.O) ^ transitive on Ω, then r(x)=axn and s(x)=
bx2" for some a, b&K and n with Q<n<
:
q—2.
The Hall planes and the planes of Narayana Rao and Satyanarayana satisfy
the condition of this theorem. But an element of Φ# is not always represented
in this form (Remark 3.6.).
Throughout the paper notations are standard and taken from [1] and [2].
All sets and groups are finite except in §2.
2. Proof of Theorem 1
Let Q be a set with two binary operations +> ° satisfying the assumption
of Theorem 1. Since Q is a left vector space, the following holds.
Lemma 2.1. (Qy +) is an abelian group.
Lemma 2.2. Let a, b, c, d^K and assume a+b\ φ 0 and c+d\ Φ 0. Then
the equation (a+b\) (x+y\)=c+d\ (2.1)
has a unique solution for x+y\ in Q*—Q— {0}.
Proof. (2.1) is equivalent to
ax— by-\of— r(y)x— s(y)) = c , (2.2)
b r(y)+ay—bx — d if 3>ΦO
or ax = c, bx = d if y = 0 . (2.3)
By the second equation of (2.2),
b r(y) = bx—ay+d (2.4)
Substituting this into the first equation of (2.2), we have
y~1(dx+bs(y)) = c. (2.5)
Hence b s(y)-{-dx=cy. By this and the second equation of (2.2),
d2-bd r(y)-b2 s(y) = (ad-bc) y . (2.6)
GENERALIZATION OF HALL QUASIFIELDS 63
Therefore (2.2) is equivalent to (2.4) and (2.6) when δΦO.
Assume δ=0 and d=0. Then Λ Φ O and £ΦO. Hence (2.1) has no solu-
tion in Q — K and has a unique solution dΓ^+Oλ in K*
Assume 6=0 and JΦO. Then 0ΦO. By (2.3), (2.1) has no solution in
K and by (2.2) it has a unique solution a~lc-}-a~ld\ in Q—K.
Assume 6ΦO and ad—bc=0. Then (2.6) is equivalent to (b~ld)2— r(y)
(b~ld)—s(y)=Q. By the assumption (i) of Theorem 1, (2.1) has no solution in
Q—K. Therefore it has a unique solution δ'^ +Oλ in K*.
Assume δΦO and ad— fcΦO. Then (2.1) has no solution in K by (2.3).
Since δΦO, (2.6) is equivalent to (b~ld)2—r(y) (b-ld)—s(y)=b-\ad—bc)y and
hence (2.6) has a unique solution y' in K9 by the assumption (ii) of Theorem 1.
Let x' be the unique solution of b r(y')—bx — ay' -\-d. Then x'-\-y' X is a unique
solution of (2.1).
Lemma 2.3. Let a, δ, c, d^K and assume α+fc\,Φθ, c+JλφO. Then
the equation
(x+y\) (a+b\) = c+d\ (2.7)
has a unique solution for x-\-y\ in Q9.
Proof. If b=Q, (2.7) has a unique solution a~lc-\-a~ld\. Assume iΦO.
Then (2.7) is equivalent to linear equations
xa-yb-\cf-r(b)a-s(b)) = c ,
xb+y(r(b)-a) = d . (2.8)
Since a(r(b)— a)— b(— b~\a2— r(b)a— s(b)))=— ί(δ)Φθ by the assumption (i)
of Theorem 1, (2.8) has a unique solution (x, .y)Φ(0, 0). Thus (2.7) has a
unique solution in Q*.
Proof of Theorem 1.
It follows immediately from the definition that Q(+, °) satisfies the fol-
lowing.
(2.9)
for all?, 97, /.eg. (2.10)
£0 = 0 f o r a U f e ρ . (2.11)
By Lemmas 2.1-2.3 and (2.9)-(2.11), Q is a weak quasifield. Since Q is a
finite dimensional vector space over K9 it is a quasifield by Theorem 7.3 of
[1]. Thus we have the theorem.
Suppose I K I < oo . The spread set Σ
r
,
s
= {M(xy y)\x, y^K} which
64 Y. HlRAMINE
corresponds to the above quasifield is defined as follows: Let K=GF(q) and
let M(x, J0=(y^
 y) g(Xy J)) be a 2x2 matrix over K. Define M(x, ;y)eΣr,s
if and only if \°(x+y\)=f(x, y)+g(x, 3>)λ. Then we have M(x, Q)=(Q ^
for xGK and M(*> y)=(X9 y)g(x, > where
andg(x, y) = —x+r(y) for x^K an
3. The proof of Theorem 2
Throughout this section let p be a prime and K=GF(q), q=pm. We use
the following notations.
K2= {k2\k<EΞK}
Φ
κ
 the set of ordered pairs (r, s) of r and $ which satisfy the condi-
tions (i) and (ii) of Theorem 1
M2(K) the set of 2 X 2 matrices over K
tr(M) the trace of a matrix M of M2(K)
det(M) the determinant of a matrix M of M2(K)
Let (r, ίjeφjr and Σ
r>5 the corresponding spread set defined in the last
paragraph of §2. Let π
rtS be the translation plane of order (f constructed from
Lemma 3.1. (i) Let M(x, y)eΣ
r
,
s
. Ify*Q,thenr(y)=tr(M(x,y))and
s(y)=-det(M(x,y)).
(ii) Let P, M<=ΞM2(K) with det(P)Φθ and set P~1MP=(* y\ Assume y*
0. Then P~1MP (=Σ
r§, ι/ ^ ^ onfy f/ r(y}=tr(p-lMP] and s(y)= -det(P~lMP).
Proof. By an easy computation we have (i).
The "only if" part of (ii) is an immediate consequence of (i). Assume
r(y}=tr(P~lMP) and s(y)=-d^t(P-1MP) and set P~1MP=(^ y\ Since tr
(p-lMP)=tr(M) and det(P'1MP)=έfeί(M), we have
r(y) = tr(M) = x+u (3.1)
and s(y) = —det(M)=—xu+yz . (3.2)
By (3.1), u— — x-\-r(y). Substituting this into (3.2) gives s(y)=of—r(y)x-\-yz.
As ^ΦO, *=-yr\*-r(y)*-t (y)}. Hence (j j)=( *^ ^  ^  y))eΣ,,s by
what we have mentioned in the last paragraph of §2.
Lemma 3.2. (i) The equation v2—r(x)v—s(x)=wx has a unique solution
GENERALIZATION OF HALL QUASIFIELDS 65
in K* for any v&K, w&K* if and only if (x—y)v2—(xr(y)—yr(x))v—(xs(y)—
ys(x))^pϋ for any vEiK and x, y&K*, x^y.
(ii) Assume p>2. Then (r, s)GΦ
κ
 if and only if the following two condi-
tions are satisfied.
(a) (r(y))2+4s(y)$K2foranyy(=K*.
(b) (xr(y)—yr(x))2+4(x—y) (xs(y)—ys(x})&K2 for any x, y^K*, x^py.
Proof. Assume (x — y)v2 — (xr(y) — yr(x))v — (xs(y) — ys(x))=0 for some
K and x, y^K*, x^y. Thdn x(v2—r(y)v—s(y))=y(v2—r(x)v~s(x)). Hence
(v2— r(y)v~ s(y))/y=(v2— r(x)v— s(x))/x. Put w=(v2— r(x)v— s(x))jx. Then w
Φθ as v
2
— r(x)v— s(x)^pO by the assumption (i) of Theorem 1 and the equation
v
2
— r(ξ)v— s(ξ)—wξ has at least two solutions for ξ.
Conversely, assume v2— r(x)v— s(x)=wx and v2— r(y)v— s(y)=wy for some
x, y&K*, xή=y. Then zoxy=y(v2— r(x)v— s(x))=x(v2— r(y)v— s(y)). This
gives (x — y)v2 — (^ 3;) — yr(x))v — (xs(y) — ys(x))=Q. Therefore (i) holds.
Assume p>2. Then it is well known that a quadratic equation aof-{-bx+
c=Q over K has no solution in K if and only if b2— 4ac<ξK2. Hence (ii)
follows immediately from (i).
Lemma 3.3. Assume \K \ >3 and let Py Q(ΞM2(K). If P+xQ(ΞΣrtSfor
any x^K, then either (i) Q—(f\ Q) and PeΣ
r
,
s
 or (ii) P and Q are scalar matrices.
Proof. Set S=Σ
Λ
 P=ίj\ i, j> *, /eX and Q=(a
c
 b\ a, b, c,
cx
Assume j=b=Q. Then P+xQ^Σ if and only if P+xQ is a scalar matrix.
Hence i-\-ax=l-\-dx and k+cx= 0. Since x is arbitrary, it follows that i=l,
a=d and k=c=0. Thus (ii) holds wheny=έ=0.
Assume yΦO and b=Q. By Lemma 3.1, P+xQ^Έ, if and only if r(j)=
tr(P+xQ) and s(j)= — det(P+Λ?Q). Hence
(3.3)
and s(j)=—adx2jr(jc—al—id)x+jk—il . (3.4)
Since (3.3) and (3.4) hold for all x^K, we have
Ό') = '+/, a+d=Q (3.5)
and s(j) = jk—il , jc-al-id =0, ad=0 (3.6)
Hence a=d=Q so jc=Q. As ΦO, c=0. Therefore (^ J)=(J Q) Thus W
holds when/ΦO and b=0.
Assume iΦO. Set χ=—b~1j. Then j-\-bx=Q and so P+xQ is a scalar
66 Y. HlRAMINE
matrix. Hence k—cb~lj=0 and i—ab~lj—l—db~lj. Setting zo=—b~lj, we
have j=—bw, k=—cw and l=i+aw—dw. Putting y=j-{-bx gives P-\-xQ=
(ΐ^^b-idy+aw+i)- ^ Lemma 3Λ> we have r(y)=b-\a+d)y+2
(aw+i) and s(y)=—b~\ad—bc)y2—b~\a-{-d) (aw+i)y—(azo+i)2. In particular
xr(y)-yr(x) = 2(i+aw) (x-y)
and xs(y)— ys(x)=(x— y) (b~\ad—bc)xy—(aw-\-i)2) . (3.7)
If p=2y then xr(y)— yr(x)=Q by (3.7). Hence we have a contradiction by
Lemma 3.2 (i).
If p>2, then (xr(y)-yr(x))2+4(x-y) (xs(y)—ys(x))=^(x-y)2b~2 (ad-bc)
xy&K2 by Lemma 3.2 (ii). Let x be any element of K2— {0} and let j; be
any element of K—K2. Then clearly 4(#— y)2b~2xy&K2. Hence ad— be must
be an element of K2. From this x'y'^K2 for any #', y'&K*, x'Φy'. In
particular ίΓ2={0, 1}, which implies K=GF(3). This contradicts the assump-
tion.
Set L(x, y)=i(v, vM(x, y))\vEΞKxK} for x, y(ΞK, L(oo)={(0, 0, ϋ)|
v&KxK} and Δ={L(Λ?, 0)|^e^} U {L(°o)>, n={L(«, ^)|^eJ^, jeJί*}.
Then Δ U Ω is the set of lines of π
r>s through (0, 0, 0, 0). Let G be the linear
translation complement of π
r
,
s
 and set H=GL(eo)tL(0ιΰ, the stabilizer of the lines
L(oo) and L(0, 0) in G. Let <r=(rι /}) be a nonsingular 4x4 matrix, where
A, 5, C, D^M2(K). Then the following criterion is well known: σ is an
element of G if and only if the following conditions are satisfied.
(1) If C is nonsingular, then C-1Z)e2
rfi. (In this case L(oo)σ=L(uy v)y
where C~1D=M(u
ί
 ϋ)eΣ
r
.
β
.)
(2) If C is singular, then C— ( ^  ^  ) and D is nonsingular. (In this case
L(oo)
σ
=L(oo).)
(3) If A+M(x, y)C is nonsingular, then (A+M(x, y)C)~l (B+M(x, y)D)
eΣ
r
,
s
. (In this case L(x, y)σ=L(u, v\ where (A+M(x, y)C)~l (B+M(x, y)
(4) If A+M(x, y)C is singular, then A+M(x, y)C=(J JY (In this
case L(Λ?, ^)σ=L(oo).)
Lemma 3.4. Assume either r or s is not a constant function. Let A, B, C
and D be elements of M2(K) and set <r=(c r)) V <r^H> then the following hold.
(i) B=C=(® J) and A=("
c
 JY D=kA for some a, d, kϊΞK* and c^K.
(ii) r(a~1dky)=kr(y)9 s(a~1dky)=k2s(y). Moreover L(x, y)=L(k(xJΓa~1cy),
ka~ldy)for all x,
GENERALIZATION OF HALL QUASIFIELDS 67
Proof. Since σ fixes L(<χ>) and L(0, 0), B=D=(® Q\ Hence σ=(^ ^ )
and J^MDeΣ for any MeΣ. In particular A~lM(x, 0)D=xA~lD^^ for
each *eK If #-GF(3),*(l)=*(-l)Φθand (r(l)+r(-l))2-(r(l))2-(r(-l))2
= — (ί(dbl))— 1 by Lemma 3.2 (ii). Hence r and s are constant functions.
Applying Lemma 3.3 we have A~lD=k for some k^K*. Hence A~1MD=
kA-lMAt=S, for any Me2. Put -4=(*J) and let *eE.K, je^fiC*. Then
kA~lM(x, y)A=M(u, v) for some W<ΞJ£ and ΐ>eJ£*. Set M(x, y)=(xfΛ and
Q /"** ky\ (ab\ ίab$uv\ , V/ ^Since = we have
bkx+dky = av+bgf , (3.8)
bkf+dkg = cv+dg' (3.9)
and akx+cky = au+bf . (3.10)
Hence d(av)-b(cv)=d(bkx+dky)-b(bkf+dkg) by (3.8) and (3.9). From this
we have
(ad-k)v=k((lfy-1)tf+(Ά^
(3.11)
On the other hand, by Lemma 3.1 (i), we have
kr(y) = r(v)
and Vs(y) = s(v) (3.12)
We argue b—Q. Suppose έφO and set ψy—{v\r(v)=kr(y)} for
By (3.11), |Ψ,| >(?+l)/2 if ί>2 and |Ψ,| >qβ if p=2 for any yeΞXΛ Thus
|ψ,| >|.K:*|/2 for any ytΞK* so we have Ψ^ΠΨ.Φφ for all y, z^K*. This
implies that r is a constant function. Similarly s is also a constant function.
This contradicts the assumption. Therefore b=Q.
From (3.8) and (3.10), v==a~1dky and u=kx+a'lcky. Hence r(a~1dky)=
kr(y\ s(a~ldky)=k2s(y) by (3.12) and L(x,y)σ=L(u, v)=L(kx+a~lckyy a~ldky).
Thus lemma holds.
Lemma 3.5. Set Ωy={L(x, y)\x(ΞK} for y£ΞK* and H^( 0$, α e j?C*, ί: e K } . ΓA<?w ίίi C H. Moreover ^  acts on Ωy and is transi-c a/
tive on Ωyfor eachy& K*.
Proof. Let σ=(^ ^^Hi Since A~lM(x, y}A=(*+a~lcy y\ A~1M
(x, y)A^Σ by Lemma 3.1 (ii) so σ^H. Moreover L(x, y) σ=L(x+a~lcy, y).
Since a^K* and c&K are arbitrary, we have the lemma.
Y. HlRAMINE
Proof of Theorem 2.
ff-2
Any mapping from K* into K can be uniquely written in the form Σ £<#'',
1-2 . .-2 . *-°
O^C/^^—2. Set r(3>)=Σ c^y1 and s(;y)=Σ ^vy' We may assume
ί=0 ι=0
that r or s is not a constant function. By Lemma 3.4 (ii), L(0, ΐ)σ=L(a~1ck,
a~
ldk), where σ=(
n
 7
 Δ ), A=(a A By Lemma 3.5, H is transitive on ίl if\L/ /C./1/ \C d /
and only if K^ faΓ1^! ^  kAl^H> Λ=^c 2)*' Set h=a~ldk Then> bY
Lemma 3.4,
r(Ay) - ad-lhr(y) (3.13)
and j(Ay) - (ad~l)2h2s(y) (3.14)
Suppose /ί is transitive on Ω. Then, for any h^K*, there exist α and
d in Jf£* which satisfy (3.13) and (3.14) simultaneously. Hence Σ £( λy=Σ
«-« »-* . _
ί = 0
 .
ί=0
Cgod^hy* and Σ dih
tyt=^Σ df(ad l f t f y % . Therefore c^f^c^ad lh and diht=
ί=0 f=0
d^ad'^h2 for all / with 0</<<7—2. If £WΦO and cM=t=0 for some m, n with
0</w, w^ί—2, then hm~1—hn~1=ad~1 and so Aw~w=l for any h€ΞK*. Thus
m=n,so that we have r(y)=c
n
yn. By a similar argument above, we have s(y)=
dty* for some t with ()<£<£—2.
Since (r(Ay))2/Wy))2=ί(Ay)My) by (3.13) and (3.14), ftΐ*f*\c\?*=dW\
dty*. From this h
2n
~
t
=\ for any h^K*. Thus *==2# (mod j— 1).
REMARK 3.6. An element of Φ
κ
 is not always represented in the form
(r(x)y s(x)), r(x)=axn, s(x)=bx2n. We list some of such examples below, which
were obtained by a computer search using Lemma 3.2.
(i) K—GFΠ}, r(x)
(ii) K=GF(ll), r(x)=ί
(iii) ^-GF(11), r(x)=2x9+6x*+4x7+3x*+&, )^-
References
[1] D.R. Hugese and F.C. Piper: Projedtive planes, Springer-Verlag, Berlin-Hei-
delberg-New York, 1973.
[2] H. Lϋneburg: Translation planes, Springer-Verlag, Berlin-Heidelberg-New York,
1980.
[3] M.L. Narayana Rao and K. Satyanarayana: A new class of square order planes,
J. Combin. Theory Ser. A 35 (1983), 33-Φ2.
GENERALIZATION OF HALL QUASIFIELDS 69
Department of Mathematics
College of General Education
Osaka University
Toyonaka, Osaka 560
Japan



A generalization of Hall quasifields

Hiramine, Yutaka

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A generalization of Hall quasifields

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