Searching algebraic conditions using MAGMA

Emanuela Ughi

Abstract

An algebraic condition about points which are covered by a secant of certain arcs is obtained using MAGMA

1 Notations

Throughout this paper, p is an odd prime, p ≡ 3(4), so that -1 is a non square in GF (q); q = p^{2}, so that GF (q) = GF (p)(i) , where i^{2} = −1; the generic element of GF (q) will be written as a + ib, where a, b ∈ GF (p).

We will perform our construction in AG(2, q), and we will consider also AG(2, q) as a subset of P G(2, q) in the usual manner.

Now, let Q be a fixed point of AG(2, q), Q = (x_{1} + ix_{2}, y_{1} + iy_{2}).

We will suppose that Q ∉ axes X = 0, Y = 0, so that (x_{1}, x_{2}) ≠ (0, 0) and (y_{1}, y_{2}) ≠ (0, 0) .

Definition 1.1  K(Q) = \{(ρ (x_{1} + ix_{2}) , \frac{1}{ρ} (y_{1} + iy_{2})) |ρ ∈ GF (p)\}
Definition 1.2  K′(Q) = K(Q) ∪ X_{∞} ∪ Y_{∞}

2 The problem

Given points

Q = (x_{1} + ix_{2}, y_{1} + iy_{2})

P = (x + iy, z + it)

let us suppose that P is such that

zy_{2} − ty_{1} ≠ 0andxx_{2} − yx_{1} ≠ 0

(This hypothesis corresponds to the fact that P does not belong to any one of horizontal or vertical lines touching K(Q) ).

When P is covered by a secant of K(Q) ?

Here I used the program logq1.doc (appended) to prove that

P iscoveredbyasecantofK(Q) ⇔ deltaisasquareinGF(p)

where

delta:=(xˆ 2*zˆ 2*x1ˆ 2*y2ˆ 2 - 2*xˆ 2*zˆ 2*x1*y1*x2*y2 + xˆ 2*zˆ 2*y1ˆ 2*x2ˆ 2 -

2*xˆ 2*z*t*x1ˆ 2*y1*y2 + 2*xˆ 2*z*t*x1*y1ˆ 2*x2 - 2*xˆ 2*z*t*x1*x2*y2ˆ 2 +

2*xˆ 2*z*t*y1*x2ˆ 2*y2 + xˆ 2*tˆ 2*x1ˆ 2*y1ˆ 2 + 2*xˆ 2*tˆ 2*x1*y1*x2*y2 +

xˆ 2*tˆ 2*x2ˆ 2*y2ˆ 2 + 2*x*y*zˆ 2*x1ˆ 2*y1*y2 - 2*x*y*zˆ 2*x1*y1ˆ 2*x2 + 2*x*y*zˆ 2*x1*x2*y2ˆ 2 - 2*x*y*zˆ 2*y1*x2ˆ 2*y2 - 2*x*y*z*t*x1ˆ 2*y1ˆ 2

+

2*x*y*z*t*x1ˆ 2*y2ˆ 2 - 8*x*y*z*t*x1*y1*x2*y2 + 2*x*y*z*t*y1ˆ 2*x2ˆ 2 -

2*x*y*z*t*x2ˆ 2*y2ˆ 2 - 2*x*y*tˆ 2*x1ˆ 2*y1*y2 + 2*x*y*tˆ 2*x1*y1ˆ 2*x2 -

2*x*y*tˆ 2*x1*x2*y2ˆ 2 + 2*x*y*tˆ 2*y1*x2ˆ 2*y2 + 4*x*z*x1ˆ 2*y1ˆ 2*x2*y2 +

4*x*z*x1ˆ 2*x2*y2ˆ 3 + 4*x*z*y1ˆ 2*x2ˆ 3*y2 + 4*x*z*x2ˆ 3*y2ˆ 3 - 4*x*t*x1ˆ 2*y1ˆ 3*x2 - 4*x*t*x1ˆ 2*y1*x2*y2ˆ 2 - 4*x*t*y1ˆ 3*x2ˆ 3 - 4*x*t*y1*x2ˆ 3*y2ˆ 2 + yˆ 2*zˆ 2*x1ˆ 2*y1ˆ 2 + 2*yˆ 2*zˆ 2*x1*y1*x2*y2 + yˆ 2*zˆ 2*x2ˆ 2*y2ˆ 2 + 2*yˆ 2*z*t*x1ˆ 2*y1*y2 - 2*yˆ 2*z*t*x1*y1ˆ 2*x2 + 2*yˆ 2*z*t*x1*x2*y2ˆ 2 - 2*yˆ 2*z*t*y1*x2ˆ 2*y2 + yˆ 2*tˆ 2*x1ˆ 2*y2ˆ 2 - 2*yˆ 2*tˆ 2*x1*y1*x2*y2 + yˆ 2*tˆ 2*y1ˆ 2*x2ˆ 2 - 4*y*z*x1ˆ 3*y1ˆ 2*y2 - 4*y*z*x1ˆ 3*y2ˆ 3 - 4*y*z*x1*y1ˆ 2*x2ˆ 2*y2 - 4*y*z*x1*x2ˆ 2*y2ˆ 3 + 4*y*t*x1ˆ 3*y1ˆ 3 + 4*y*t*x1ˆ 3*y1*y2ˆ 2 + 4*y*t*x1*y1ˆ 3*x2ˆ 2 + 4*y*t*x1*y1*x2ˆ 2*y2ˆ 2) Moreover,

delta= 0 ⇔ PbelongstoatangenttoCinapointofK(Q)

The idea is that 2 distinct points of K(Q) covering by their secant P correspond to 2 distinct solutions (in b) (in GF (p)) of the equation

rr = Resultant(ff, ee, a) = 0

so that

rr = x*z*b*x1*y2 - x*z*b*y1*x2 - x*t*b*x1*y1 - x*t*b*x2*y2 + x*y1ˆ 2*x2 + x*x2*y2ˆ 2 +

y*z*b*x1*y1 + y*z*b*x2*y2 + y*t*b*x1*y2 - y*t*b*y1*x2 - y*x1*y1ˆ 2 -

y*x1*y2ˆ 2 - z*bˆ 2*x1ˆ 2*y2 - z*bˆ 2*x2ˆ 2*y2 + t*bˆ 2*x1ˆ 2*y1 + t*bˆ 2*y1*x2ˆ 2

and that 2 distinct such solutions exist ⇔ ⇔ discriminant of rr = 0 respect to b = delta = square in GF (p) The idea works when

1) qw1 = Coefficient(ff, a, 1) and qw2 = Coefficient(ee, a, 1) are not both 0

2) rr has actually degree 2. In this case indeed the condition ’rr = 0’ is equivalent to the condition :

thesistem \{ \begin{matrix}ee = 0 \\ ff = 0 \end{matrix} hassomesolutionsinGF(p) × GF(p)

We must remark that the equation rr = 0 can not have the solution 0, because

F actorization(Evaluate(rr, b, 0)) = (y_{1}^{2} + y2_{2})(xx_{2} − yx_{1})

which is ≠ 0 in our hypothesis, because (y_{1}^{2} + y_{2}^{2}) ≠ 0 because -1 is a non square in GF (p). Moreover, the secant we are looking for does exist if and and only if rr = 0 has 2 distinct solutions.

Proof of 1)

F actorization(Resultant(qw1, qw2, b)) = (x_{1}^{2} + x_{2}^{2}) ∗ (zy_{2} − ty_{1})

which is ≠ 0 in our hypothesis, because (x_{1}^{2} + x_{2}^{2}) ≠ 0 because -1 is a non square in GF (p).

Here again we must control that the process works, that is to say that (er1, er2) ≠ (0, 0), where

er1 = Coefficient(qw1, b, 1)ander2 = Coefficient(qw2, b, 1).

Let us suppose by absurd that they are both 0.Then

\{ \begin{matrix}er1= − zx_{1} + tx_{2} = 0 \\ er2= − zx_{2} − tx_{1} = 0 \end{matrix}

but (z, t) ≠ (0, 0), so by Cramer’s rule (x_{1}^{2} + x_{2}^{2}) = 0 , which is absurd.

Proof of 2)

F actorization(h) = F actorization(Coefficient(rr, b, 2)) = (x_{1}^{2} + x_{2}^{2}) ∗ (zy_{2} − ty_{1}) ≠ 0

Let us recall our hypothesis on P: P does not belong to horizontal or vertical lines touching K(Q).

Let Π be the Baer subplane of P G(2, q) containing K′(Q). Π contains also all horizontal and vertical lines above.

Our point P ∉ Π. So there is one and just one line r through P which intersects Π in a line r′ = (r ∩ Π) of Π.

Moreover, r is not horizontal nor vertical, so that the equation rr = 0 has 2 solutions b_{1}, b_{2} in GF (p) or in a quadratic extension.Observe that K′(Q) = C_{Q} ∩ Π , then

deltaisasquareinGF(p) ⇔ b_{1}, b_{2}belongtoGF(p) ⇔ r′issecantK′(Q)

deltaisanon − squareinGF(p) ⇔ b_{1}, b_{2}belongtoGF(q) \ GF(p) ⇔ r′isexternaltoK′(Q)

Then,

delta= 0 ⇔ r′istangenttoK(Q) ⇔ ristangenttoCinapointofK(Q)

3 Logq1.doc

B¡x,y,z,t,a,b,x1,y1,x2,y2,i¿ :=PolynomialRing(Integers(),11);

/*Put Q:=(x1+ix2,y1+iy2);when P(x+iy,z+it) is covered by secants of K(Q)?

We must look for 2 points Pa=(a(x1+ix2,(1/a)*(y1+iy2)), and Pb=(b(x1+ix2,(1/b)*(y1+iy2)), such that P, Pa, Pb are on a line; we are looking for 2 distinct solutions, with a, b different from 0*/ aa:=x+i*y-(a*x1+i*a*x2); bb:=b*(a*(z+i*t)-(y1+i*y2)); cc:=x+i*y-(b*x1+i*b*x2); dd:=a*(b*(z+i*t)-(y1+i*y2)); w:=aa*dd-bb*cc; c1:=Coefficient(w,i,0); c2:= Coefficient(w,i,1); c3:= Coefficient(w,i,2); //OK because Degree(w,i)=2; ww:=c1-c3+i*c2; //ww=0 gives the condition ”The points are on a line”; print ww; print ””; print c1-c3; //real part of w; print ””; //imaginary part of w; print c2; print””; ff:=(c1-c3) div (a-b); ee:=c2 div (a-b); /*ff:=-x*y1 + y*y2 - z*a*b*x1 + t*a*b*x2 + a*x1*y1 - a*x2*y2 +

b*x1*y1 - b*x2*y2;

ee:=-x*y2 - y*y1 - z*a*b*x2 - t*a*b*x1 + a*x1*y2 + a*y1*x2 + b*x1*y2 + b*y1*x2;*/

rr:= Resultant(ff,ee,a); /*/ polynomial of degree 2 in b; are there 2 distinct solutions in b?;*/ print rr; h:=Coefficient(rr,b,2); k:=Coefficient(rr,b,1); j:=Coefficient(rr,b,0); delta:=k*k-4*h*j; print ””; print delta; print ””; qw1:=Coefficient(ff,a,1); qw2:=Coefficient(ee,a,1); er1:=Coefficient(qw1,b,1); er2:=Coefficient(qw2,b,1);