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Let us say I have a system of 3 polynomials, f1(x,y,z), f2(x,y,z), f3(x,y,z). How to find the resultant of these 3 polynomials? What I mean is: is there any special method to do this? Does the Macaulay resultant apply to this kind of problem?

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Are your polynomials homogeneous? If so, then yes, the Macauly resultant is what you want. The Macauly resultant of $f_1,f_2,f_3$ is a single polynomial in the coefficients of $f_1,f_2,f_3$ that will vanish if and only $f_1,f_2,f_3$ have a common zero in $\mathbb{P}^2(K)$ (for an algebraically closed field $K$). This even works if $K$ has characteristic $p$.

It follows in general from elimination theory that there is some polynomial ideal of the coefficients whose vanishing implies the existence of a common zero, but I don't know an intrinsic reason why the case of $n$ homogeneous polynomials in $n$ variables gives a principal ideal.

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  • $\begingroup$ yet it is a principal ideal $\endgroup$ Feb 4, 2014 at 13:33
  • $\begingroup$ Do you have a reference for the first statement in your second paragraph? I'm particularly interested in the case where we have $n+1$ inhomogeneous polynomials in $n$ variables, over $\mathbb{C}$. (Or perhaps I'm misunderstanding the scope of that sentence.) $\endgroup$ Sep 29, 2014 at 15:52
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    $\begingroup$ @JasonDeVito I think you have to homogenize, then check if the common solutions are points "at infinity". For example, do $ax+b$ and $cx+d$ have a common root. The only sensible polynomial condition to check is $ad-bc=0$. But this says that they have a common root if $a=c=0$, which is true only in the sense that if $a=c=0$, then they both vanish at the point at infinity in $\mathbb P^1$. $\endgroup$ Sep 29, 2014 at 23:22
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    $\begingroup$ @Freeman. Yes, that's why I said a non-trivial zero in projective 2-space, which rules out the trivial zero at $(0,0,0)$. It also identifies zeros that are scalar multiples of one another, which is the natural way to count zeros in this setting. $\endgroup$ Sep 20, 2018 at 0:54
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    $\begingroup$ @Freeman. There is no algebraic expression of the coefficients whose vanishng implies a real solution. I guess you're asking if there are inequalities, for example for the polynomial $ax^2+bx+c$, there is an inequality involving the discriminant which says if there is a real root. I do not know the answer in general. But if you look at a textbook on real algebraic geometry, this question is likely to be discussed. $\endgroup$ Sep 20, 2018 at 11:19
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See HAL : hal-00912907, version 1 , An Introduction to Trägheitsformen. An explicit method is developed there for particular cases.

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  • $\begingroup$ Typing "hal-00912907" into Google leads to hal.archives-ouvertes.fr/docs/00/91/29/07/PDF/Abdallahpaper.pdf which is loading very, very slowly on my computer. $\endgroup$ Feb 3, 2014 at 23:36
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(Please could you put this answer together with the preceding one. I do not know, myself, how it works.) There is An algorithm to compute resultants, by Marc Chardin , published in Effective Methods in Algebraic Geometry, Edit.Teo Mora, Carlo Traverso (Birkhäuser). The case of three polynomials is studied explicitely : "Abstract: We here give a method to calculate the resultant of three polynomials in terms of a square- free decomposition and resultants of two polynomials. After that, we show how the subresultant algorithm enables us to avoid many calculations. In the last part, we study the possible extension to the general case of n homogeneous polynomials in n variables."

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