# Existence of non-trivial solution to non linear polynomial system

I need to find conditions for the existence of non-trivial solutions to a multivariable polynomial system in two cases:

The first case:

$f1: a_1x^2+a_2xy+a_3y^2+a_4z^2=0$

$f2: b_1x^2+b_2xy+b_3y^2+b_4z^2=0$

The second case:

$f3: a_1x^2+a_2xy+a_3y^2+a_3z^2+a_5xz=0$

$f4: b_1x^2+b_2xy+b_3y^2+b_3z^2+b_5xz=0$

(note that the coefficient of y2 and z2 are equal)

All the coefficient are reals number and the variable x,y,z should be real too.

I tried to use resultant and eliminating one variable, for even a more simple case:

$f5: a_1x^2+a_2xy+a_3y^2+a_3z^2=0$

$f6: b_1x^2+b_2xy+b_3y^2+b_3z^2=0$

The problem that i get different solution:

1: if I try to eliminate $z$, I get $RES(f5,f6,z)=x^2[(a_2c_1−a_1c_2)x+2(b_2c_1−b_1c_2)y]^2$ So there is always non trivial (x≠0 or y≠0) real x,y such that RES(f5,f6,z)=0

2: On the other hand if I eliminte $x$ I get: $\small RES(f5,f6,x)=[(a_2c_1−a_1c_2)^2+4(a_2b_1−a_1b_2)(b_2c_1−b_1c_2)]x^4+4(b_2c_1−b_1c_2)^2x^2z^2$

thus a non-trivial solution exist only if $(a_2c_1−a_1c_2)^2+4(a_2b_1−a_1b_2)(b_2c_1−b_1c_2)≤0$.

I tried to check this numerically and it seems like the second condition is the right one, so why the first attempt eliminating z is wrong?

Any help will be appreciated

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Your equations define conics in $\mathbb{P}^2$, so you always have solutions, at least over an algebraically closed field. In general you will find $4$ points, but it could be less in particular cases if the curves intersect with tangency. If you want to work over the field of reals (which seems to be because of inequalities), then your first conic need to have at least one point, in which case it is rational, so you can parametrise and obtain a polynomial in degree $4$. –  Jérémy Blanc Jun 14 '13 at 20:11
I didn't unerstand- why the first conic must have one rational point and how can I parametrise it. Can you please give an example? –  user34985 Jun 16 '13 at 18:47

Using Groebner bases (or even direct computing) we see, that the solutions of your system $f_5=f_6=0$ are as follows:

Case 1: $a_1b_2-a_2b_1=0$. Then it follows $y^2+z^2=0$. Over the real numbers this means $y=z=0$, and $a_1x=b_1x=0$. For $a_1=b_1=0$ we have always a solution with $x\neq 0$.

Case 2: $a_1b_2-a_2b_1\neq 0$. Then we obtain either $y=0$ and $x=z=0$, or if $y\neq 0$, then $$x=\frac{(a_3b_1-a_1b_3)(y^2+z^2)}{(a_1b_2-a_2b_1)y},$$

and $ry^2+sz^2=0$ with $$r:= a_1^2b_3^2 - a_1a_2b_2b_3 - 2a_1a_3b_1b_3 + a_1a_3b_2^2 + a_2^2b_1b_3 - a_2a_3b_1b_2 + a_3^2b_1^2$$ and $$s:= (a_1b_3 - a_3b_1)^2.$$ Here you have the case 2, i.e., either $y=z=0$ again (we have required $y\neq 0$, so this was solved earlier), or some non-trivial solution. The other systems can be solved too, using Groebner bases.

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thank you very much!! I am not familiar with the Groebner bases so I have a few more questions: 1. As far as I understood, Groebner bases give me a new set of equations, which theirs solution is exactly as the original equations. This is always true? is there any restrictions on the coefficient? 2. I see that (using computer software such as 'mathematica') for each ordering of the variable i get different bases. Can I just choose any order which will give me easiest set of equation? –  user34985 Jun 16 '13 at 18:09
3. If I choose an ordering (let say $x>y>z$) and get a condition, let's say about $x$ (as in the example). Is it possible to substitute it in the original equations, and get a new Groebner bases using different order of the variable (for exmple $z>x>y$)? Can I solve directly the original equations without using Groebner bases (after substitute $x$)? 4. Is there any recommended book about Groebner bases? I am not a mathematician so I don't familiar with all the terminology such as Ideal? A book with full examples solution will be the best one for me. –  user34985 Jun 16 '13 at 18:09
5. Is there a way to utilize the fact that my equations are quadratic forms? I know that for two equations with two variables, the Sylvester matrix give an immediate solution. I also found that there are known result for three quadratic equations with three variable (using resultant). Isn't any general result for any two homogeneous equations with three variable?? Thanks a lot, –  user34985 Jun 16 '13 at 18:09
Thank, just a few last clarification: 1.Regarding the first question, what about over $\mathbb{R}$, which is my case? what are the restrictions? what's are the points that it should be take care? 2.Regarding question 5- In fact I need for other problem, the general case of two quadratics forms. I tried to look on that most general case, but it seems that the equations derived from Groebner bases are hard, so if there some source which give a closed solution for that case, it would be great. Nonetheless i'm always happy to learn new stuff, and surely I will read yours lecture notes. –  user34985 Jun 16 '13 at 19:58