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I was wondering if it is well understood under what circumstances say three univariate polynomials $f(x),g(x),h(x)$ have a common root. In this situation, I can see that the resultant of each pair must vanish but that only ensures that each pair has a common root. Is there a way to generate a finite set of polynomials in the coefficients of $f,g,h$ which tells you when all 3 share at least one common root?

Would be interested in an answer for the more general (more than 3 polynomials) case too.

EDIT: After thinking about it a tad more here is a possibly interesting observation. If you have $n$ polynomials of degree at most $n$ then you can write it as a linear system in $x,x^2,...,x^n$. Using determinants, minors, etc you will be able to get $n-1$ necessary and sufficient relations between the coefficients which tells you when the polynomials share a common root. I would suspect that means in general if you have $k$ polynomials it might be possible to give $k-1$ polynomials in the coefficients which will be necessary and sufficient conditions for having a common root.

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    $\begingroup$ Their GCD is not a constant, if they have common root. $\endgroup$
    – Somnium
    Commented Apr 21, 2022 at 7:23
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    $\begingroup$ I think this may be complicated. For instance, if we take f(x) = $1+x+x^2+...x^(n_1-1)$, g(x) = $1+x+x^2+...x^(n_2-1)$, and h(x) = $1+x+x^2+...x^(n_3-1)$, it seems that there is a common factor if and only if the n's have a common divisor larger than 1. What would be the equivalent condition on the coefficients in this case? $\endgroup$
    – Derek
    Commented Apr 21, 2022 at 10:34
  • $\begingroup$ There is no formula for a GCD though no? I was looking for a situation where you can write down an ideal in the coefficients of the 3 polynomials and the ideal vanishes precisely on those points where those 3 polynomials share a common root. @Derek your example seems mysterious to me even for two polynomials. But we do know the resultant works there. $\endgroup$
    – Niareh
    Commented Apr 21, 2022 at 18:50
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    $\begingroup$ If all three polynomials are monic and their coefficients arbitrary indeterminates $a_1,\ldots,a_m, b_1,\ldots,b_n,c_1,\ldots,c_l$ you can just compute a gröbner base of $f,g,h$ in an elimination-termorder where $x > a_1,\ldots,c_l$. The elements of the gröbner base which are free of $x$ describe the values of $a_1,\ldots,c_l$ above which a common zero $x$ lies. $\endgroup$ Commented Apr 21, 2022 at 23:09
  • $\begingroup$ @Jürgen Böhm, nice! i think this generalises my answer posted below, no ? $\endgroup$ Commented Apr 21, 2022 at 23:25

2 Answers 2

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Assume that $f_n$ is monic. Then for indeterminates $u_1,\ldots,u_{n-1}$, all coefficients of the polynomial $Res_x(f_n,u_1f_1+\ldots+u_{n-1}f_{n-1})\in k[u_1,\ldots,u_{n-1}]$ vanish if and only if $f_1,\ldots,f_n$ have a common root [expanding out this resultant then gives the list of polynomials].

This is how elimination theory works - if $f_1(x_1,\ldots,x_m)=0,\ldots, f_n(x_1,\ldots,x_m)=0$ set-theoretically define a variety $V$, $f_n$ is monic in $x_m$ (which one ensure by applying a Noether normalization coordinate change), and $\pi:\mathbb{A}^m\to \mathbb{A}^{m-1}$ is the projection away from the last coordinate, then $\pi(V)$ is a variety in $\mathbb{A}^{m-1}$ defined by the coefficients of $Res_x(f_n,u_1f_1+\ldots+u_{n-1}f_{n-1})$. Repeatedly eliminating variables like this eventually gives you a finite morphism surjecting $V$ to $\mathbb{A}^{\dim(V)}$.

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  • $\begingroup$ Should there be plusses separating the $u_if_i$s instead of commas? $\endgroup$
    – Will Sawin
    Commented Apr 22, 2022 at 2:50
  • $\begingroup$ Yes, thanks, fixed! $\endgroup$ Commented Apr 22, 2022 at 3:12
  • $\begingroup$ This is indeed the first step of the proof of the fundamental theorem of elimination theory, a la Kronecker. The only reference I know is the first edition of Modern Algebra by van der Waerden. This was also used by Hilbert to describe unstable binary forms. $\endgroup$ Commented Apr 22, 2022 at 20:59
  • $\begingroup$ Excellent exactly what I was looking for. $\endgroup$
    – Niareh
    Commented Apr 22, 2022 at 23:29
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Well, the "naïve" answer (already implied in @Somnium's comment) is that a general method for finding the solutions of a system of simultaneous polynomial equations $$ f_1(x)=0, \ \ f_2(x)=0, \ \ \cdots \ \ f_n(x)=0 $$ is finding the $\gcd\bigl(f_1, f_2, \cdots, f_n\bigr)=g(x)$, of these polynomials:

Assuming that we are speaking about polynomials with complex coefficients and since $\mathbb{C}[x]$ is a PID, the ideal of $\mathbb{C}[x]$ generated by the polynomials $f_1, f_2, \dotsc, f_n$ is the ideal generated by their greatest common divisor: $$ \langle f_1, f_2, \dotsc, f_n\rangle=\langle\gcd\bigl(f_1, f_2, \dotsc, f_n\bigr)\rangle=\langle g\rangle. $$

Thus, the solution of the initial system, is equivalent to the solution of the single equation $$ g(x)=0. $$

Maybe there is no explicit formula for the gcd of a number of univariate polynomials (in terms of their coefficients) but there is certainly an algorithm: The $\gcd\bigl(f_1, f_2, \dotsc, f_n\bigr)$, can be found using the Euclidean division algorithm for finding the $\gcd$ of two univariate polynomials in $\mathbb{C}[x]$ together with the fact that for $n\geq 3$ $$ \gcd(f_1, f_2, \dotsc, f_n)=\gcd\bigl(f_1,\gcd(f_2, \dotsc, f_n)\bigr). $$

P.S.: I am not sure if this satisfies you as an answer but I was led to this description driven by the way the question has been posed and the subsequent discussion in the comments.

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  • $\begingroup$ I know you reference the comments generally, but, to be specific, @Somnium commented this as a solution. The questioner has indicated that it is unsatisfactory to them, though I agree with you that I'm not sure it should be unsatisfactory. \\ TeX note: please use $\langle g\rangle$ \langle g\rangle, not $<g>$ <g>. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Apr 21, 2022 at 23:02
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    $\begingroup$ @LSpice, you are right; actually the two comments you are mentioning were my motivation for posting this as an answer: OP's second comment led me to think that maybe Somnium's comment was not clearly perceived. Also: thanks for the edit and the TeX note. $\endgroup$ Commented Apr 21, 2022 at 23:41

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