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Let $f(x,y)=0$ and $g(x,y)=0$ be bivariate polynomial equations where the polynomials have the same degree, say, $N\geq 3$. Furthermore, both of them have the same terms but different coefficients. For example, $f(x,y) = a_1x^2y^3 + a_2xy^2 + a_3$ and $g(x,y) = b_1x^2y^3 + b_2xy^2 + b_3$. How may common roots of $f(x,y)=0$ and $g(x,y)=0$ are there? I am not interested in finding the common roots but the number. Can Bézout's Theorem help?

Thanks a lot.

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Bezout's theorem would imply that if $f,g$ had no common factors, then the number of common roots is at most $N^2$. –  Mohan May 22 '13 at 17:21
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And the resultant $R(f,g)$ would decide if $f$ and $g$ had any common roots at all (in an extension field). Are the polynomials over a field ? –  Dietrich Burde May 22 '13 at 18:37
    
Hm, what if you compute the Gröbner basis for such varieties? Do you get anything interesting? (This would be quite nice to experiment with a bit, to get some feeling for the problem). –  Per Alexandersson May 22 '13 at 20:03
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Since you fix the terms, the Bernstein-Kushnirenko theorem (or Bernstein-Khovanskii-Kushnirenko) would be more precise than Bezout's theorem. –  quim May 22 '13 at 20:08
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I don't think much mathematical background is needed. The $\Delta_i$ is just the convex hull of the set of exponents of the terms involved in the $p_i$, inside $\mathbb{R}^2$, so in your case they are both equal to the triangle with vertices (2,3), (1,2), (0,0). The coefficients don't matter. Vol denotes "mixed volume" but I think that when all $\Delta_i$ are equal, this is just ordinary volume. Have a look at arXiv:0812.4688. BTW, just as Bezout, this only gives an upper bound on the number of solutions, which is the exact number if the coefficients are "general enough". –  quim May 23 '13 at 8:58

1 Answer 1

I reread the previous posts. The answers are very interesting but the question is really hilarious. When we have the same terms in each equation (with different coefficients in a field $K$) then the problem is essentially a linear one. For instance, here, we obtain (generically) a sole solution in $K$ for $x^2y^3,xy^2$. Then a value for $x^2y^4$, then a value for $y$, then a value for $x$.

The moral of the story is that some trivial matters can lead to exciting discussions and take time before downgrading certain issues to stackexchange !

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The problem is linear only in the sense that finding the roots of the characteristic polynomial of a matrix is linear (that is, not at all). –  S. Carnahan Sep 21 '13 at 13:11
    
Dear Carnahan, I don't understand one word you write. We have a linear system of $2$ equations in the $2$ unknowns $x^uy^v,x^ry^s$. Then we obtain (generically) one solution in $K$ ($x^uy^v=?,x^ry^s=?$). We deduce easily (according to the values of $u,v,r,s$) the solutions in $x,y$, using $p^{th}$ roots. Can you solve a characteristic polynomial with radicals ? –  loup blanc Sep 21 '13 at 14:02
    
Moreover, and it is funny, the second part of the calculation is essentially: solving a linear system in $log(x),log(y)$.:) –  loup blanc Sep 21 '13 at 14:50
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Given the statement of the problem, you always get two linear equations. If there are more terms in the polynomials $f$ and $g$, you will get more unknowns. Your method only seems to work when you have 2 unknowns. –  S. Carnahan Sep 21 '13 at 14:52

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