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DEMONSTRATIO MATHEMATICA
Vol. XLVII
No 4
2014
Jan Jakóbowski, Danuta Kacperek
HAVLIČEK–TIETZE CONFIGURATIONS IN VARIOUS
PROJECTIVE PLANES
Abstract. A. Lewandowski and H. Makowiecka proved in 1979 that existence of
the Havliček–Tietze configuration (shortly H ´ T) in the desarguesian projective plane is
equivalent to existence in the associated field, a root of polynomial x2 ` x ` 1, different
from 1. We show that such a configuration exists in every projective plane over Galois
field GF pp2 q for p ‰ 3. As it has been demonstrated, in a projective plane over arbitrary
field F , each hexagon contained in H ´ T, satisfies the Pappus–Pascal axiom, even if F is
noncommutative. Moreover, such a hexagon either is pascalian or has exactly one pair of
opposite sides intersecting at a point collinear with two points not belonging to these sides.
In particular, all such hexagons are pascalian iff charF “ 2. For the (noncommutative)
field of quaternions, we have determined the set of all roots of the mentioned polynomial.
Every H ´ T is the special Pappus configuration, in which three main diagonals of the
hexagon are concurrent.
1. Preliminary results
Definition 1.1. [6, p. 180] In a projective plane, a set of four triangles
forms the configuration of Havliček–Tietze (shortly pH ´ Tq) if:
a) No two of them are with a common vertex;
b) Each two of them are in six-fold homology;
c) The centers of homologies of any two triangles are the vertices of the other
two, the axes being the corresponding opposite sides.
Definition 1.2. [6, p. 180] A pair of triangles T1 , T2 is special six-fold
homologic if:
1. T1 , T2 have no common vertex;
2. T1 , T2 are in six-fold homology;
3. The centers of homologies are vertices of a new pair of triangles T3 , T4 ;
2010 Mathematics Subject Classification: 51A20, 51A30, 51E15, 12F05, 12Y05.
Key words and phrases: Galois field, Havliček–Tietze configuration, homologic triangles,
pascalian hexagon, projective plane, skew field of quaternions.
DOI: 10.2478/dema-2014-0078
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J. Jakóbowski, D. Kacperek
4. If a center of any arbitrary homology of triangles T1 , T2 is one of the
vertices of a triangle Ti , i “ 3, 4, then the axis of this homology is incident
with the remaining vertices of Ti .
Theorem 1.3. [6, p. 181–182] For every pair of special six-fold homologic
triangles T1 , T2 there exists another pair of triangles T3 , T4 such that these
four triangles form the pH ´ Tq.
The first Havliček–Tietze configuration was discovered in the projective
plane of order 4 and described in [3].
Theorem 1.4. [6, p. 182–184] In arbitrary desarguesian projective plane
π there exists an pH ´ Tq if and only if π is isomorphic with a plane over
a field F containing a root a ‰ 1 of the polynomial x2 ` x ` 1. Then T1 “
tA1 , A2 , A3 u, T2 “ tB1 , B2 , B3 u, T3 “ tC1 , C2 , C3 u, T4 “ tD1 , D2 , D3 u, with
A1 “ p1, 0, 0q, A2 “ p0, 1, 0q, A3 “ p0, 0, 1q, B1 “ pa, 1, 1q, B2 “ p1, a, 1q,
B3 “ p1, 1, aq, C1 “ pa, 1, aq, C2 “ pa, a, 1q, C3 “ p1, a, aq, D1 “ p1, a2 , aq,
D2 “ pa2 , 1, aq, D3 “ p1, 1, 1q, form the required pH ´ Tq.
Corollary 1.5. In a projective plane over a field F with characteristic 3
there exists no pH ´ Tq.
Definition 1.6. [7, p. 185] A hexagon pP1 , P2 , . . . , P6 q is pascalian if each
two opposite sides Pi Pi`1 and Pi`3 Pi`4 (subscripts modulo 6) intersect at a
point collinear with vertices not belonging to these sides.
One of the possible formulations of the Pappus–Pascal proposition has
the following form [7, p. 185], [2, p. 151]:
(P) An arbitrary hexagon with two pairs of opposite sides intersecting at
a point collinear with the vertices not belonging to these sides is pascalian.
2. New results
Corollary 2.1. If a is a root of x2 ` x ` 1 then a´1 is a root of this
polynomial, too. Moreover a´1 “ a2 .
Remark 2.2. It follows from [6, p.
centers of homologies
˜
¸
˜
A1 A2 A3
A1
p1q
, p2q
B3 B2 B1
B2
˜
¸
˜
A1 A2 A3
A1
p4q
, p5q
B2 B3 B1
B3
181] that C1 , C2 , C3 , D1 , D2 , D3 are
A2
A3
B1
B3
A2
A3
B1
B2
¸
˜
,
p3q
¸
˜
,
p6q
A1
A2
A3
B1
B3
B2
A1
A2
A3
B1
B2
B3
¸
,
¸
respectively.
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Havliček–Tietze configurations in various projective planes
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Corollary 2.3.
a) The following quadruples of points are collinear: tA1 , B3 , C1 , D2 u,
tA2 , B2 , C1 , D3 u, tA3 , B1 , C1 , D1 u, tA1 , B2 , C2 , D1 u, tA2 , B1 , C2 , D2 u,
tA3 , B3 , C2 , D3 u, tA1 , B1 , C3 , D3 u, tA2 , B3 , C3 , D1 u, tA3 , B2 , C3 , D2 u.
b) Every hexagon contained in pH ´ Tq consists of two triangles forming the
pH ´ Tq.
From Corollary 2.3, the following is immediate
Proposition 2.4. Every pH ´ Tq with exactly vertices of four triangles
taken, and with all sides of the triangles omitted, is the Pappus configuration,
where the three main diagonals of the hexagon are concurrent (see Figure 1).
Thus it is a configuration of type p123 , 94 q (cf. [1, p. 15]).
A3
C3
A1
D3
B3
C2
C1
D1
C3
B1
D2
B2
C2
A2
Fig. 1. H ´ T modified as special Pappus configuration
The following result is some generalization of Theorem 5 given in [7,
p. 186] characterizing the projective plane of order 4.
Proposition 2.5. Let An , Bn , Cn , Dn , n “ 1, 2, 3, be points given in
Theorem 1.4, X, Y P tA, B, C, Du, X ‰ Y , q, r, s, t, u, v P t1, 2, 3u, and
q ‰ s ‰ u ‰ q, r ‰ t ‰ v ‰ r. Then:
a) Every hexagon pXq , Yr , Xs , Yt , Xu , Yv q is pascalian.
b) Every hexagon containing the same points as in item a), but in other
order, is pascalian iff char F “ 2. Otherwise such a hexagon contains
exactly one pair of opposite sides intersecting at a point collinear with the
two points not belonging to these sides.
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J. Jakóbowski, D. Kacperek
Proof. a): By Definition 1.1, triangles tXq , Xs , Xu u and tYr , Yt , Yv q are in
six-fold homology. Therefore, among six permutations of pYr , Yt , Yv q there
exist those inducing the homologies
˜
¸ ˜
¸ ˜
¸
Xq Xs Xu
Xq Xs Xu
Xq Xs Xu
Yr
Yv
Yt
Yt
Yr
Yv
Yv
Yt
Yr
each with its center. From the first homology, it follows that the opposite
sides Xq Yr and Yt Xu of pXq , Yr , Xs , Yt , Xu , Yv q intersect at a point collinear
with Xs and Yv . Analogously, the opposite sides Yr Xs and Xu Yv intersect
at a point collinear with Yt and Xq (the second homology). The opposite
sides Xs Yt and Yv Xq intersect at a point collinear with Xu and Yr (the third
homology).
b): In every considered hexagon there are three points denoted by letter
X and three denoted by Y . If the order of points is different from that given
in a) then two X-s (or equivalently two Y -s) must be neighboring. One
takes sides of a hexagon in cyclic order, so we may assume that X is in the
first and in the second place. Therefore, there are only the following possibilities of such (ordered) hexagons: pX, X, X, Y, Y, Y q, pX, X, Y, Y, X, Y q,
pX, X, Y, X, Y, Y q. We have to consider three triples of lines in any hexagon:
two opposite sides and the diagonal passing through two points not belonging
to these sides. In each mentioned possibility, there is the following property:
one triple contains three lines of the shape XY , this is a triple of concurrent
lines; each of the remaining two triples contains one line of the shape XY ,
one line of the shape XX and one Y Y – such three lines are not concurrent,
unless charF “ 2. The same arguments as in a) work in the case of former
triples . In the case of latter triples, it remains to show that the lines Xq Xs ,
Yr Yt , and Xu Yv are concurrent if and only if charF “ 2. Easy calculation
gives the required result.
Example 2.6. We shall consider two hexagons, say pB1 , B2 , B3 , C2 , C3 , C1 q
and pA1 , A2 , D2 , D3 , A3 , D1 q. Thus, we obtain the following lines and points
for the former hexagon:
B1 B2 “ b3 : x1 ` x2 ` a2 x3 “ 0;
b3 X c1 “ tp0, ´a2 , 1qu;
C2 C3 “ c1 : x1 ` ax2 ` x3 “ 0;
B3 C1 : ´ax2 ` x3 “ 0;
p0, ´a2 , 1q P B3 C1
ô ´a ¨ p´a2 q ` 1 “ 0 ô a3 ` 1 “ 0 ô 1 ` 1 “ 0 ô charF “ 2;
B2 B3 “ b1 : a2 x1 ` x2 ` x3 “ 0;
b1 X c2 “ tp1, 0, ´a2 qu;
C3 C1 “ c2 : x1 ` x2 ` ax3 “ 0;
C2 B1 : x1 ´ ax3 “ 0;
p1, 0, ´a2 q P C2 B1
ô 1 ´ a ¨ p´a2 q “ 0 ô 1 ` a3 “ 0 ô 1 ` 1 “ 0 ô char F “ 2;
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Havliček–Tietze configurations in various projective planes
B3 C2 : x1 ´ x2 “ 0; C1 B1 : x1 ´ ax2 “ 0;
C3 B2 : ax1 ´ x2 “ 0; p0, 0, 1q P C3 B2 .
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B3 C2 X C1 B1 “ tp0, 0, 1qu;
For the latter hexagon we have:
A1 A2 “ a3 : x3 “ 0; D3 A3 : x1 ´ x2 “ 0; A1 A2 X D3 A3 “ tp1, 1, 0qu;
D2 D1 “ d3 : x1 ` x2 ` x3 “ 0; p1, 1, 0q P d3 ô 1 ` 1 ` 0 “ 0 ô char F “ 2;
A2 D2 : x1 ´ ax3 “ 0; A3 D1 : x1 ´ ax2 “ 0; A2 D2 X A3 D1 “ tpa, 1, 1qu;
D3 A1 : x2 ´ x3 “ 0; pa, 1, 1q P D3 A1 ;
D2 D3 “ d1 : a2 x1 ` x2 ` ax3 “ 0;
D1 A1 : x2 ´ ax3 “ 0;
D2 D3 X D1 A1 “ tp´p1 ` 1q, a2 , aqu;
A3 A2 “ a1 : x1 “ 0;
p´p1 ` 1q, a2 , aq P a2 ô 1 ` 1 “ 0 ô char F “ 2.
2.1. Havliček–Tietze configurations in finite projective planes
It follows from Theorem 1.4 that there exists infinite number of finite
projective planes containing pH ´ Tq. For every natural number u, if u2 `
u ` 1 “ upu ` 1q ` 1 “ p for some prime p, then u is a root of the considered
polynomial x2 ` x ` 1 in the field Zp . Thus, we conclude that configurations
pH ´ Tq exist in projective planes over the fields Z7 , Z13 , Z31 , Z73 , Z157 ,
Z211 , etc. One can use any algorithm to verify wether the number upu ` 1q `
1 “ p is prime or not.
If the polynomial x2 `x`1 does not have a root in Zp then, by definition,
the root exists in algebraic extension GF pp2 q of Zp [5, Theorem 10]. The
fields of characteristic 3 must be excluded since, in this case x2 ` x ` 1 “
px ´ 1q2 , i.e. only 1 is the root. Now, the following result is immediate.
Proposition 2.7. For every prime p ‰ 3 there exists pH ´ Tq in the
projective plane over the Galois field GF pp2 q.
Example 2.8. The polynomial x2 ` x ` 1 does not have a root in Z5 .
Consider the extension GF p52 q and let a be a root of x2 ` x ` 1. Then
GF p52 q “ tq ` ra; q, r P Z5 u and
p4 ` 4aq2 ` p4 ` 4aq ` 1 “ p1 ` 2a ` a2 q ` p4 ` 4aq ` 1 “ a2 ` a ` 1 “ 0,
so 4 ` 4a is the second root. The same arguments work for the both
roots, so let us consider only 4 ` 4a. We take the first pair of triangles
T1 “ tA1 , A2 , A3 u, T2 “ tB1 , B2 , B3 u, with A1 “ p1, 0, 0q, A2 “ p0, 1, 0q,
A3 “ p0, 0, 1q, B1 “ p4 ` 4a, 1, 1q, B2 “ p1, 4 ` 4a, 1q, B3 “ p1, 1, 4 ` 4aq
and independently of the conclusion of Theorem 1.4, verify whether these
triangles are six-fold homologic or not. Then compare our result with the
conclusion of Theorem 1.4. We shall use notation p1q,. . . ,p6q of the permutaUnauthenticated
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J. Jakóbowski, D. Kacperek
tions given in Remark 2.2. Simple calculation gives the following equations
of lines and their intersections:
Permutation p1q
A1 B3 : p1 ` aqx2 ` x3 “ 0;
A3 B1 : 4x1 ` p4 ` 4aqx2 “ 0.
A2 B2 : x1 ´ x3 “ 0;
The lines A1 B3 , A2 B2 and A3 B1 intersect in the common point C1 “
p4 ` 4a, 1, 4 ` 4aq.
A1 A2 : x3 “ 0;
A1 A3 : x2 “ 0;
A2 A3 : x1 “ 0;
B1 B2 : x1 ` x2 ` ax3 “ 0;
B1 B3 : x1 ` ax2 ` x3 “ 0;
B2 B3 : ax1 ` x2 ` x3 “ 0;
A1 A2 X B3 B2 “ tp1, 4a, 0qu;
A1 A3 X B3 B1 “ tp1, 0, 4qu; A2 A3 X B2 B1 “ tp0, 4a, 1qu.
The three points of intersection of corresponding sides lie on the line c1 :
ax1 ` x2 ` ax3 “ 0.
Permutation p2q
A1 B2 : 4x2 ` p4 ` 4aqx3 “ 0;
A3 B3 : x1 ´ x3 “ 0.
A2 B1 : 4x1 ` p4 ` 4aqx3 “ 0;
The lines A1 B2 , A2 B1 and A3 B3 intersect in the common point C2 “
p4 ` 4a, 4 ` 4a, 1q.
A1 A2 X B2 B1 “ tp1, 4, 0qu; A1 A3 X B2 B3 “ tp1, 0, 4aqu;
A2 A3 X B1 B3 “ tp0, 1, 4aqu.
The three points of intersection lie on the line c2 : ax1 ` ax2 ` x3 “ 0.
Permutation p3q
A1 B1 : x2 ´ x3 “ 0;
A2 B3 : p4 ` 4aqx1 ` 4x3 “ 0;
A3 B2 : p4 ` 4aqx1 ` 4x2 “ 0.
The lines A1 B1 , A2 B3 and A3 B2 intersect in the common point C3 “
p1, 4 ` 4a, 4 ` 4aq.
A1 A2 X B1 B3 “ tp4a, 1, 0qu;
A2 A3 X B3 B2 “ tp0, 1, 4qu.
A1 A3 X B1 B2 “ tp4a, 0, 1qu;
The three points of intersection lie on the line c3 : x1 ` ax2 ` ax3 “ 0.
Permutation p4q
A1 A2 X B2 B3 “ tp1, 4a, 0qu;
A2 A3 X B3 B1 “ tp0, 1, 4aqu.
A1 A3 X B2 B1 “ tp4a, 0, 1qu;
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The lines A1 B2 , A2 B3 and A3 B1 intersect in the common point D1 “ pa, 4 `
4a, 1q and the three points of intersection of corresponding sides lie on the
line d1 : p4 ` 4aqx1 ` ax2 ` x3 “ 0.
Permutation p5q
A1 A2 X B3 B1 “ tp4a, 1, 0qu;
A2 A3 X B1 B2 “ tp0, 4a, 1qu.
A1 A3 X B3 B2 “ tp1, 0, 4aqu;
The lines A1 B3 , A2 B1 and A3 B2 intersect in the common point D2 “ p1, 4 `
4a, aq and the three points of intersection lie on the line d2 : x1 ` ax2 ` p4 `
4aqx3 “ 0.
Permutation p6q
A1 A2 X B1 B2 “ tp1, 4, 0qu;
A2 A3 X B2 B3 “ tp0, 4, 1qu.
A1 A3 X B1 B3 “ tp1, 0, 4q;
The lines A1 B1 , A2 B2 and A3 B3 intersect in the common point D3 “ p1, 1, 1q
and the three points of intersection lie on the line d3 : x1 ` x2 ` x3 “ 0.
Since the centers Ci , Di and the axes cj , dj satisfy the condition Ci P
cj ô i ‰ j and Di P dj ô i ‰ j, for i, j P t1, 2, 3u, it follows that triangles
T1 , T2 are special six-fold homologic. One can easily check that triangles
T3 “ tC1 , C2 , C3 u, T4 “ tD1 , D2 , D3 u complete T1 , T2 to some pH ´ Tq.
Note that p4 ` 4aq2 “ a in accordance with Corollary 2.1. Taking, e.g.
permutation p1q
˜
¸
C1 C2 C3
,
D3 D2 D1
we have obtained the following equations of lines and the following points of
intersection:
C1 D3 : 4x1 ` x3 “ 0;
C2 D2 : x1 ` p1 ` aqx3 “ 0;
C3 D1 : x1 ` 4ax3 “ 0;
C1 C2 : p4 ` 4aqx1 ` x2 ` x3 “ 0;
C1 C3 : x1 ` x2 ` p4 ` 4aqx3 “ 0; C2 C3 : x1 ` p4 ` 4aqx2 ` x3 “ 0;
D1 D2 : x1 ` x2 ` x3 “ 0; D1 D3 : p4 ` 4aqx1 ` x2 ` ax3 “ 0;
D2 D3 : ax1 ` x2 ` p4 ` 4aqx3 “ 0; C1 C2 X D3 D2 “ tp4a, 0, 1qu;
C1 C3 X D3 D1 “ tp1, 0, 4aqu; C2 C3 X D2 D1 “ tp4, 0, 1qu.
The lines C1 D3 , C2 D2 and C3 D1 intersect in the common point p0, 1, 0q “ A2 .
The three pairs of corresponding sides of the triangles T3 and T4 intersect at
points which lie on the line x2 “ 0 – the side A1 A3 of triangle T1 .
The same arguments work for the remaining permutations of vertices.
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2.2. Havliček–Tietze configurations in infinite projective planes
Certainly, H ´ T does not exist in the projective plane over the field of
real numbers, and it exists in the plane over the field of complex
? numbers.?In
´1` 3i
2
the latter case, the polynomial x ` x ` 1 has two roots
and ´1´2 3i .
2
Therefore, we conclude that H ´ T does not exist in the projective plane
over the field of complex rational numbers, and it exists in the plane over
the field of complex constructible numbers (cf. [8]).
Now, we shall determine all roots of x2 ` x ` 1 in the skew field of
quaternions [4, p. 5]. We are looking for elements a “ u ` vi ` wj ` zk such
that a2 ` a ` 1 “ 0, where u, v, w, z are real and the following equalities hold
for basic elements:
i2 “ j 2 “ k 2 “ ´1,
ij “ k “ ´ji,
jk “ i “ ´kj,
ki “ j “ ´ik.
Using these properties, we receive the following consideration:
pu ` vi ` wj ` zkq2 ` pu ` vi ` wj ` zkq ` 1 “ 0;
u2 ` uvi ` uwj ` uzk ` vui ` v 2 i2 ` vwij ` vzik ` wuj ` wvji
` w2 j 2 ` wzjk ` zuk ` zvki ` zwkj ` z 2 k 2 ` u ` vi ` wj ` zk ` 1 “ 0;
u2 ` uvi ` uwj ` uzk ` vui ´ v 2 ` vwk ´ vzj ` wuj ´ wvk
´ w2 ` wzi ` zuk ` zvj ´ zwi ´ z 2 ` u ` vi ` wj ` zk ` 1 “ 0;
pu2 ´ v 2 ´ w2 ´ z 2 ` u ` 1q ` puv ` vu ` wz ´ zw ` vqi
` puw ´ vz ` wu ` zv ` wqj ` puz ` vw ´ wv ` zu ` zqk “ 0;
$
$
2 ´ v 2 ´ w 2 ´ z 2 ` u ` 1 “ 0,
’
’
u
u2 ´ v 2 ´ w2 ´ z 2 ` u ` 1 “ 0,
’
’
’
’
’
’
& 2uv ` v “ 0,
& vp2u ` 1q “ 0,
’ 2uw ` w “ 0,
’ wp2u ` 1q “ 0,
’
’
’
’
’
’
% 2uz ` z “ 0,
% zp2u ` 1q “ 0.
a) 2u ` 1 “ 0. Then u “ ´ 12 and, by the first equation, we have v 2 `
w2 ` z 2 “ 34 . If now v 2 “ 43 then we obtain exactly the complex roots
already found.
general case, we have infinite number of solutions for
´ ? In
? ¯
3
3
v, w, z P ´ 2 , 2 (cf. [4, p. 263],
b) 2u ` 1 ‰ 0. Then v “ w “ z “ 0 and u2 ` u ` 1 “ 0 but the last equation
does not have a real solution.
Therefore, we have proved the following proposition
Proposition 2.9. A quaternion a “ u`vi`wj`zk is a root of polynomial
x2 ` x ` 1 if and only if u “ ´ 12 and v 2 ` w2 ` z 2 “ 43 . If it is the case then
|a| “ 1.
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Havliček–Tietze configurations in various projective planes
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It is well known that all projective planes over commutative fields satisfy
Pappus–Pascal axiom [2, p. 158]. Equivalently, we may use the condition
(P). Because of Corollary 2.4 and Proposition 2.5, every hexagon belonging
to H ´ T in the projective plane over the noncommutative field of quaternions also satisfies (P). Of course, also in this case, there exist the H ´ T-s
containing hexagons, which are pascalian and those which are not.
Example 2.10. In accordance with Theorem 1.4 and Proposition 2.9 take
the root a “ ´ 21 ` 12 i ` 21 j ` 12 k and hexagons with vertices Bi , Di , i “ 1, 2, 3.
Then a2 “ a´1 “ ´ 12 p1 ` i ` j ` kq and
ˆ
˙
ˆ
˙
1
1
B1 “
p´1 ` i ` j ` kq, 1, 1 , B2 “ 1, p´1 ` i ` j ` kq, 1 ,
2
2
ˆ
˙
1
B3 “ 1, 1, p´1 ` i ` j ` kq ,
2
˙
ˆ
1
1
D1 “ 1, ´ p1 ` i ` j ` kq, p´1 ` i ` j ` kq ,
2
2
ˆ
˙
1
1
D2 “ ´ p1 ` i ` j ` kq, 1, p´1 ` i ` j ` kq , D3 “ p1, 1, 1q.
2
2
In order to illustrate Proposition 2.5a) and Proposition 2.4, let us consider
the hexagon pB2 , D3 , B1 , D2 , B3 , D1 q. We have the following three triples of
lines – two opposite sides and the diagonal through remaining two points:
1
B2 D3 : x1 ´ x3 “ 0; D2 B3 : p1 ´ i ´ j ´ kqx2 ` x3 “ 0;
2
1
B1 D1 : x1 ` p1 ´ i ´ j ´ kqx2 “ 0;
2
˙
ˆ
1
1, ´ p1 ` i ` j ` kq, 1 P B2 D3 X D2 B3 X B1 D1 ;
2
1
D3 B1 : x2 ´ x3 “ 0; B3 D1 : p1 ´ i ´ j ´ kqx1 ` x3 “ 0;
2
1
D2 B2 : p1 ´ i ´ j ´ kqx1 ` x2 ` “ 0;
2
ˆ
˙
1
´ p1 ` i ` j ` kq, 1, 1 P D3 B1 X B3 D1 X D2 B2 ;
2
1
1
B1 D2 : p1 ` i ` j ` kqx1 ` x3 “ 0; D1 B2 : p1 ` i ` j ` kqx2 ` x3 “ 0;
2
2
B3 D3 : x1 ´ x2 “ 0;
ˆ
˙
1
1, 1, ´ p1 ` i ` j ` kq P B1 D2 X D1 B2 X B3 D3 .
2
Moreover, p0, 0, 1q P B1 D1 X D2 B2 X B3 D3 in accordance with Proposition 2.4.
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J. Jakóbowski, D. Kacperek
Corollary 2.11. H ´ T exists in the projective plane over the skew field
of rational quaternions.
References
[1] B. Grünbaum, Configurations of Points and Lines, AMS, Providence, Rhode Island,
2009.
[2] P. Dembowski, Finite Geometries, Springer–Verlag, Berlin–Heidelberg–New York,
1968.
[3] K. Havliček, J. Tietze, Zur Geometrie der endlichen Ebene der Ordnung n “ 4,
Czechoslovak Math. J. 21(96) (1971), 157–164.
[4] T. Y. Lam, A First Course in Noncommutative Rings, Springer–Verlag, New York,
Inc., 1991.
[5] S. Lang, Algebra, Addison–Wesley Publishing Company, 1970.
[6] A. Lewandowski, H. Makowiecka Some remarks on Havliček–Tietze configuration,
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[7] A. Lewandowski, H. Makowiecka A geometrical characterization of the projective plane
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J. Jakóbowski, D. Kacperek
FACULTY OF MATHEMATICS AND COMPUTER SCIENCE
UNIVERSITY OF WARMIA AND MAZURY IN OLSZTYN
10-719 OLSZTYN, POLAND
E-mail: [email protected]
[email protected]
Received March 11, 2013.
Communicated by J. Smith.
Unauthenticated
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