Singular

Hi! I am working on arithmetic geometry but I never used any computer program. I've tried with Singular, but i've got a lot a troubles, so I post my question here, if some one can help me.

I ask for solutions x(t),y(t),z(t),u(t) \in \mathbb{C}(t) ( complex function field of one variable) of the following system:

(1) x^2 - t^3 * u^2 + t^3 = (t^6 * u^2 - t)*y^2
(2) x^2 - 2*(t^3)*(u^2) + 1/(t^3) = (t^3)*(t^6*u^2 - t)*z^2

Can some one, who is familiar with Singular, compute it? I am very grateful. Thank in advance

Best,
Thi

Ps: If the system has no solutions, then what is happen if i replace t^3 by t^5 resptively t^6 by t^10 in the equations.

Hello Thi,

your problems seems to be quite hard but not because of the Groebner basis computations. In fact, these do not help much in your case:
You can do:
ring r = (0,t), (x,y,z,u), lp;
poly f = x2-t3*u2+t3-t6*u2*y2-t*y2;
poly g = x2-2*t3*u2+1/t3-t3*(t6*u2-t)*z2;
ideal i = f,g;
std(i);
to compute a Groebner basis over Q(t)[x,y,z,u]. But this will basically give you your initial system:
_[1]=(-t9)*y2u2+(-t4)*y2+(t12)*z2u2+(-t7)*z2+(t6)*u2+(t6-1)
_[2]=(t3)*x2+(-t12)*z2u2+(t7)*z2+(-2t6)*u2+1
I have tried to write down solutions for each SINGLE equation, and it's not so hard. But here's the problem:
You may solve the second equation, e.g. by x=-1/t, z=i/t3, u=1/t3. But plugging this into the first equation gives you
y2=(t6+t-1)/(t4+t3)
which has no solution in C(t). So the problem is that you must ensure to be able to compute the square root inside C(t). And I do not see how SINGULAR can help you much with that kind of problem. Here's an idea: Replace x,y,z,u by quotients of polynomials, then multiply your equations by the divisor polynomials and obtain a new system which only involves polynomials in C[t]. You may then try to produce necessary conditions in terms of degrees etc. I have not tried this but maybe it's worth a try?!
Frank

Hi Frank! Thank you very much for your reply. It seems to me that this system has no solution.

But can you try one more time with Singular, if i have

(1) x^2 - t^3 * u^2 + t^3 = (t^6 * u^2 - t^3)*y^2
(2) x^2 - 2*(t^3)*(u^2) + 1/(t^3) = (t^3)*(t^6*u^2 - t^3)*z^2

i.e. i replace t^3 in the right hand side instead of t.

By the way, i just want to know if the system

(1) x^2 - t^3 * u^2 + t^3 = (t^6 * u^2 - t^3)*y^2
(2) x^2 - 2*(t^3)*(u^2) + 1/(t^3) = (t^3)*(t^6*u^2 - t^3)*z^2

has a solution? If yes, then what is an example for a solution there?

I'm sory that i've forgotten to say that both sides of these 2 equations should not be equal to zero.

Hi,
Singular can decide your question by using primdecGTZ or absPrimdecGTZ from primdec.lib (provided the computations are not too hard).
In case your system has a solution of the form

x=a(t), y=b(t), z=c(t), u=d(t) with a,b,c,d in Q(t) (Q the rationals)

primdecGTZ will find it. If a,b,c,d are in C(t) (C the complex numbers) absPrimdecGTZ will find it in an extension field of Q. Here are two typical examples:

ring r = (0,t),(x,y), dp;
ideal i = x^2 - (1-t)^2, (y- t)^3;
LIB"primdec.lib";
option(prot); //just to see the progress of the computation
primdecGTZ(i);
//->
[1]:
[1]:
_[1]=y3+(-3t)*y2+(3t2)*y+(-t3)
_[2]=x+(t-1)
[2]:
_[1]=y+(-t)
_[2]=x+(t-1)
[2]:
[1]:
_[1]=y3+(-3t)*y2+(3t2)*y+(-t3)
_[2]=x+(-t+1)
[2]:
_[1]=y+(-t)
_[2]=x+(-t+1)
The solutions are defined by setting each polynomial to 0. Hence you have two solutions:
[1][2]: y=t,x=1-t and [2][2]: y=t,x=t-1 ( [1][1] and [2][1] are the primary ideals ).

ring r = (0,t),(x,y), dp;
ideal i = x^2 - (1-t)^2, y^2 + 2*t^2;
list Sol = absPrimdecGTZ(i);
def S = Sol[1];
setring S; absolute_primes;
//->
[1]:
[1]:
_[1]=2*a2+1
_[2]=y+(-2t)*a
_[3]=x+(-t+1)
[2]:
2
[2]:
[1]:
_[1]=2*a2+1
_[2]=y+(2t)*a
_[3]=x+(t-1)
[2]:
2
You have again two solutions but in the field Q(a)(t), where a=sqrt(-2) defined by 2*a2+1=0:
y= +- (-2t)*a, x=+-(t-1).

In your example you have no solution in C(t):
ring r = (0,t),(x,y,z,u), dp;
ideal i = x^2 - t^3 * u^2 + t^3, (t^6 * u^2 - t^3)*y^2, x^2 - 2*(t^3)*(u^2) + 1/(t^3), (t^3)*(t^6*u^2 - t^3)*z^2;
list Sol = absPrimdecGTZ(i);
def S = Sol[1];
setring S; absolute_primes;
//->
[1]:
[1]:
_[1]=a
_[2]=z
_[3]=y
_[4]=(2025t6)*x2+(-289t6)*y2+(646t6)*yz+(-361t6)*z2+(-34t6)*yu+(38t6)*zu+(-t6)*u2+(4050t9-t6-2025t3+1)
_[5]=(t3)*x2+(2t6-1)
_[6]=(-91125t6)*x3+(-103275t6)*x2y+(-39015t6)*xy2+(-4913t6)*y3+(115425t6)*x2z+(87210t6)*xyz+(16473t6)*y2z+(-48735t6)*xz2+(-18411t6)*yz2+(6859t6)*z3+(-6075t6)*x2u+(-4590t6)*xyu+(-867t6)*y2u+(5130t6)*xzu+(1938t6)*yzu+(-1083t6)*z2u+(-135t6)*xu2+(-51t6)*yu2+(57t6)*zu2+(-t6)*u3+(-182250t9-135t6+91125t3+135)*x+(-206550t9-17t6+103275t3+17)*y+(230850t9+19t6-115425t3-19)*z+(-12150t9-t6+6075t3+1)*u
[2]:
1

You can still simplify the output by
interred(absolute_primes[1][1]);
//->
_[1]=a
_[2]=z
_[3]=y
_[4]=(t6)*u2+(t6-1)
_[5]=(t3)*x2+(2t6-1)
Since there is no plynomial of the form x+g(t) and u+h(t), there is no solution in C(t).

In addition to my previous comment, where I advertise primdecGTZ and absPrimdecGTZ, I like to remark that "solving" over a function field is inherent complicated and that primdecGTZ and absPrimdecGTZ are very expensive commands (using the option(prot); command shows this). Therefore it is often recommended just to compute a Groebner basis with std or groebner, inspect the equations, make some substitutions, recompute a Groebner basis etc..

What I want to say is: A combination of using its own brain together with Singular is usually the best strategy, and even more fun.

Enjoy!
Gert-Martin Greuel