Pencils of Conics in the Galois Fields of Order 2n

Pencils of Conics in the Galois Fields of Order 2nAuthor(s): Alan D. CampbellSource: American Journal of Mathematics, Vol. 49, No. 3 (Jul., 1927), pp. 401-406Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2370672 .

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Pencils of Conics in the Galois Fields of Order 2f. BY ALAN D. CAMPBELL.

We define a conic in a Galois field of order 2n [which field we write more briefly GF (2n) ] as the locus of all points whose coordinates x, y, z satisfy an equation of the form

(1) ax2 + by2 + cz2 + fyz + gzx + hXy 0,

whose coefficients and variables represent numbers in this GF(2n). The discriminant of (1) is

(2) A=fgh + af2 + bg2 + ch2.

A pencil of conics then has the equation

(3) XCI + ,uC2 O,

where CG = 0 and C2 =- 0 are two distinct conics. In this paper we derive the classes, or sets of classes, of projectively equivalent pencils and we give a typical pencil for each class, or set of classes. If there is an arbitrary coefficient in the typical pencil we say this pencil represents a set of classes, when- ever different values of this coefficient may give nonequivalent pencils and so represent distinct classes. The discriminant A of the general conic of a pencil we call the discriminant of the pencil. If we subject (3) to the transformation

(4) x a1x' + b1y' + c1z', y = a2x' + b2y' + c2z', z = a3,x + b3y' + c8z',

with determinant j a1, b2, C3 j $ 0, we get for the new pencil

K' = al, b2, c3 12 A.

(After such a transformation as this we drop the primes from the variables, without explicit mention of this fact).

We note that in a GF(2n) every number is a perfect square with just one square root, also (xzx + fly + yz)2 = cZ2x2 + /32y2 + y2z2. In any GF(2n) we have the following typical conics

(5) x2 + yz 0,

(6) xy = 0, 401



402 CAMPBELL: Pencils of Conics in the Galois Fields of Order 2n.

(7) $2 + y2 + asy 0

where Z2 + aZ + 1 0, Z =- /y, is irreducible in this GF(2n)*,

(8) x2 O.

We first divide the classes of pencils of conics into the following sets:

Set I. Each pencil of this set contains at least one double line.

Set II. Each pencil has no double line, but has at least one real line pair.

Set III. Each pencil has no double line and no real line pair, but has at least one conjugate imaginary line pair.

Set IV. Each pencil has no degenerate conics of any description.

Set I. We put (3) in a form (A) where al-x2 and C2 0 is of the form (1) with a O. If f g h 0, b;zO0 in (A), we put

x=x', y= (1/b?2)y" + (C2/bl/2) z', z z',

and we get the following class with a typical pencil and its discriminant

(9) Class 1, 2+ y2 _ 0, A_0.

If h =O , f =O in (A),t we put

x=x', y= (1/h) y' + (g/h)z', z-z',

and we get a pencil (A') with C1' x2 and a' g = ' 0, h' 1 in C2'.

If c' 0 0 in (A'), we obtain by an obvious transformation

(10) Class 2, _a2+,(z2 + xy) =0,

If c'=O, b'#O in (A'), we have

(11) Class 3, X2+,(y2+Xy) =0, A 0.

The pencil (11) has only one double line and so is not equivalent to (9).

If b' =c'O0 in (A'), we have

(12) Class 4, X2+p$y == AF0.

* See L. E. Dickson, " Linear Groups," p. 199.. t If h = f=O, g AzO, we put X = a', y =z, z=y' and we have again the case

h 0 0, f =0.



CAMPBELL: Pencils of Conics in the Galois Fields of Order 2n. 403

The pencil (12) has just one double line and so is not equivalent to (9). Any transformation (4) that is to send (11) into (12) must have bi c1 = c2 ? 0, b2c.3 O. But then (4) sends C2 of (11) into a22x'2 + b2 2y2 * which has a term in y#2P whereas (12) has no such term.

If f 7 0 in (A), we can reduce this pencil to a form (A"), with C1'-X2 and C2' having a' = h' = g' = O, f' = 1. If C2' 0 is reducible to (7) we get

(13) Class 5, AX2 + (y2 + Z2 + ayz) =0 p a.2AX12.

If 02' 0 in (A") is reducible to (6), we get

(14) Class 6, Ax2+uyz=0, ;A X12.

The pencil (14) has a real line pair and so is nonequivalent to (13).

Set II. We put (3) in a form (B) where C1-xy and C2 has h=0. If cg #& 0 in (B) * and ax2 + CZ2 + gzx is factorable, we can reduce (B) to a form (B') with C0'=- xy and a' - 0 c'g' # 0. If in (B') b'y2 + f'yz + c'z2

is factorable and f' # 0, we can reduce (B') to a form (B") with CG" = xy and c"f"g" 7 0, a" = b" = h" = 0, and we obtain

(15) Class 7, Axy+,(z2+yz+zx)-=, A &o(X+p).

If in (B') f' 0 0 and the terms in y and z alone are not factorable, we obtain

(16) Class 8, Axy + ((Z2 + y2 + ayz + ZX) = 0, ,- L (X2 + ,2 + (X)j

where oc makes the quadratic factor in A irreducible.

If in (B') f' = 0, we have

(17) Class 9, Axy+u(z2+zx)=O A=;iX21U.

If in (B) neither the three terms in x and z alone nor the three terms in y and z alone are factorable, we arrive at

(18) Set 10, AXy+,(X2+y2+Z2+ ayZ+P#ZX) = O=

af/ # 0, where X2 + Z2 + /3zx and y2 + Z2 + ayz are irreducible, with

* We must have g p 0 or f z 0 or fg - 0, otherwvise C2 0 is a double line. If g =0, cf ; 0, we can put a = y', y = w', z = z' and get the case cg z 0.




A 14x2 + pa/.?X1 + (W2 + 8t2)f2].

If the quadratic factor of A is irreducible, then (18) may possibly be equivalent to (16). EHowever in this case (4) must have b, = c = 0, a1 7 0, a2 = C2=-0, b2 #A 0, c3, $ O, a,2 + a32 +18a1a3 0 [because (16) has no term in x2]. But this last equation has only the solution a1 = a8 = 0. If the quadratic factor of A is reducible, then (18) may possibly be equivalent to (15). However (4) must then have

a1a2 = 0, b1b2 O= , a,2 + a22 + a32 + aa2a3 + /3a1a3 = 0,

b,2 + b2 + b3 + xb2b3 + f3blb3 = 0,

since (15) has no terms in x2 and y2. But then we have a1 0 O (say), a, 0, which gives us a2 + a32 + a1a3 0 (an irreducible equation).

If in (B) c 0, f $& 0, we have a pencil reducible to. (17). If in (B) c f =0, b 0, we have

(19) Class 11, Axy+ (y2+zx)=O,

If in (B) c-f bb=0,we obtain

(20) Class 12, Axy+ xz 0, A 0.

If in (B) f = 0, c P 0, and the three terms in x and z alone in C2 are not factorable, we get

(21) Class 13, Axy + JU (x2 + z2 + azx) =0 A == X2.

The line pair C2 0 in (21) is conjugate imaginary, so this pencil is non. equivalent to (17T).

Set III. We put (3) in a form (C) with Cl = 0 of the f9rm (7) and with h = O in C2. We must have f # O or g9 0 or fg =A O, otherwise C2=O

is a double line. We suppose g :7 0 (the case g = 0, f #F 0 is reducible to this). We must also have c-! 0, otherwise (C) contains a real line pair. We now transform (C) to a pencil (C') with Cl' = 0 of the form (7) and with f' = h'= , c' #0 O, g'=-1. Also in (C') either a' 0 or b' 0 or a'b' 7 O, otherwise C2' = 0 is a real line pair.

If b'5 O0 in (C'), we get

(23) Set 14, (x2 + y2 + fxy) + (y2 + Z2 + YXZ) ,

/?y # 0, A . (f2X2 + y2 t + y2y2). We assume for this set that y makes the quadratic factor of A irreducible.



CAMPBELL: Pencils of Conics in the Gatlois Fields of Order 2n. 405

If the quadratic factor of A of (23) is reducible, then we take for C0-= 0 and C2' = 0 in (C') two conjugate imaginary line pairs with vertices at (0, O, 1) and (0, 1, 0) respectively.* We obtain

(24) Set 15, x(x2 + y2 + axy) + (X2 + Z2 + xz) = 0

oc,3 #0, CG = 0 and C2 = 0 reducible to (7), A-, L( a2l + ,2Pu). The thirid degenerate conic of (24), which must be a conjugate imaginary line pair, i's

(#2/a2 + 1)X2 + (32/a2)y2 + Z2 + (32ba)xy + lXZ = 0.t

This reduces to the form (17) with a/(a2 + f2)1/2 instead of '.'

If b'= 0 in (C') we get Set 15 again.

We note that there is no pencil with just one (or with just two) con-, jugate imaginary line pairs for its degenerate conics.

Set IV. We reduce the discriminant of (3) to the form

(25) +

which is to be an irreducible cubic. Then we put (3) in a form (D) with C1 = x2 + yz. We must have g9 # 0 or h # 0 or gh 0 in C2 of (D), otherwise (D) has a double line. We suppose g 7' 0 (the case g 9 O, h /& 0 is reducible to this). Now we ,reduce (D) to a form (D') with C10 x2 -j-+ b"y2 + yz and with f' = h' 0, g' =I . in C2', by transformations on x, y, z alone. The discriminant A- X3 + b'/L3 + a'X21Z + b"XIA2 must be of the form

(25), so we have b'= , a' I, b"-O. We get

(26) Set 16, X(x2 + yz) + (X2 + ay2 + iZ2 + XZ) =0,

with (25) as its discriminant.

Suppose now that in GF(2n) we have 2n = 3k + 1 and the pencil (3) has a discriminant reducible to A - X3 + a3, where a e cube. We put the discriminant in this form, we reduce C0 = 0 to the form (5); then our pencil has a discriminant A-- X3 + A213 + CX2 + (fg + h2)Xy2, where A2 belongs to C2 =O So we must have c 0, also f - 0 or g = 0 or fg =A 0 (otherwise the pencil has a degenerate conic). We suppose f :& 0 (the case f 0,

* A pencil containing two conjugate imaginary line pairs with the same vertex contains also a double line.

t This set evidently does not exist in GP (2). 7




9 7& 0 is reducible to this). We perform some evident transformations and get the pencil into the form

A(a1x2 + z2 + xy) + pu(a2X2 + b2y2 + yZ) =0,

with A - A3 + a2j3 + a1,ju2. So we have a, =0, and we get

(27) Set 17, A(z2+xy) +,L(x2+Iy2+Yz) =

A - X3 + u3, a 76 cube.

We now append a list of the classes of pencils of conics, together with their typical pencils and their discriminants.

Class 1, Ax2 +,uy2 An ^_0.

Class 2, Ax2 + (z2 + Xy) 0, -

Class 3, AX2 + jU(y2 + Xy) 0, _0.

Class 4, Xx2 +uXy =0, A 0. Class 6, Ax2 + A(y2 + z2 + ayz) o0 A 2XL2

Class 6, Ax + ,uyz=o, A X=u2.

Class 7, Axy + A (Z2 + yz + XZ) A ^ (X+u + )

Class 8, Axy +,du(z2 + y2 + ayz + XZ) =0, A (X2 +2 +X/) Class 9, Axy + A (Z2 + ZX) = A0 X2,

Set 10, Xxy +,u(x2 + y2 + z2 + ayz + xz) 0, A [X2 + aA#X/,+ (a2 + f2) 2j

'Class 11, Xxy+,4y2+zx)=0, =

Class 12, Xxy +xzz 0, A0. Class 13, Axy+ (x2+z2+ axz) 0, A=X2 . Set 14, A(x2 + y2 + fxy) + (y2 + z2 + yxz) O,

A = /3(2X2 + y2Xy + y2l2).

Set 16, X(x2+y2+~xy) +u(x2+z2+I3xz)=-0, A-AXu( o2X + /32,).

Set 16, AX(x2 + yz) + x2 + ay2 + fz2 + XZ) =0 A-X3 + A2 + (3

Set 17, A(z2+xy) +p(X2 +/y2 +yz) O,

A - XI + u3.

SYAousB, NEw YoRx.



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Pencils of Conics in the Galois Fields of Order 2n