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I posted this in math.stackexchange, but this might be a better place. Assume that $f(X,Y,Z,V,W)\in \mathbb{Z}[X,Y,Z,V,W]$ is some polynomial and assume that $f(x,y,z,v,w)=0$. I would like to know if there is some way to figure out if there are non-trivial constants in $\mathbb{Q}$, s.t.

$$f(a_1+a_2x+a_3y+a_4z,b_1+b_2x+b_3y+b_4z,c_1+c_2x+c_3y+c_4z,v,w)=0$$

Of course a trivial solution to this is $a_2=1$, $a_i=0$, $i\neq 2$ etc. The variables $x,y,z,v,w$ are considered as variables, i.e. I'm looking for a ''generic'' different solution that only depends on the polynomial not on the particular solution. In other words, $\mathbb{Q}[x,y,z,v,w]\cong Q[X,Y,Z,V,W]/(f)$.

Say, I want to assume that the new solution has a different value for $X$. This means that one would need an inequality

$$a_1+a_2x+a_3y+a_4z \neq x$$

I define the polynomial

$$g(a_1,a_2,a_3,a_4,a)=a(a_1+a_2x+a_3y+a_4z - x) - 1,$$

over the coefficient ring $\mathbb{Z}[x,y,z]$, where $a$ is a new variable. Clearly, the trivial choice of $a_1,\ldots,a_4$'s will not satisfy this. Next define another polynomial

$$h(a_1,\ldots,a_4,b_1,\ldots,b_4,c_1,\ldots,c_4) = f(a_1+a_2x+a_3y+a_4z,b_1+b_2x+b_3y+b_4z,c_1+c_2x+c_3y+c_4z,v,w)$$

Then a different solution to the original polynomial will be a solution to the system $$\left\{\begin{array}{l}g = 0\\ h = 0\end{array} \right. $$

over $\mathbb{Q}$. Is there a way to compute this with Mathematica? The documentation on polynomial rings and Grobner basis looked a bit unclear...

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  • $\begingroup$ (1) A specific example might make this more amenable to analysis. Withouty an example it's a bit difficult to follow. (2) You might try this. Remove the constants a1,b1,c1 (I suspect they are not really needed). Set a2=b3=c4=1. Expand the polynomial. One term in this should be f(x,y,z,v,w), which you can set to zero. Now look for nonzero parameter values that force what remains to vanish. $\endgroup$ – Daniel Lichtblau Sep 25 '14 at 23:35

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