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Consider an ideal $I = \langle f_1,\dotsc,f_n\rangle$ in the ring $k[x_1,\dotsc,x_m]$. Define the $i$-th elimination ideal to be $I_i = I \cap k[x_{i+1},\dotsc,x_m]$. For any two polynomials $f$ and $g$ in $I$, the resultant $\operatorname{Res}(f,g,x_1)$ belongs to the first elimination ideal $I_1$.

  1. Is there a way to represent $I_1$ as $\langle r_1,\dotsc,r_l\rangle$, where the basis $r_j$ are obtained as resultants of polynomials in $I$ with respect to $x_1$?
  2. Is there a way to represent each elimination ideal in this way? (I imagine by iterating the possible representation of point 1?)
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A bit long for a comment, so posted as an answer, although this is really a comment.

For 1) I would rather try the following. Assuming the $f$'s have the same degree in $x_1$, introduce formal variables $s_1,\dotsc,s_n$ and $t_1,\dotsc,t_n$, and take the binary resultant $\operatorname{Res}(s_1f_1+\dotsb+s_nf_n,t_1f_1+\dotsb+t_nf_n,x_1)$. This is a monstrous polynomial in the $s,t$ variables. I think the coefficients of this polynomial should be in $I_1$ and may give a basis. If the $f$'s don't have the same degree, you may have to make two copies of them, multiply one copy by powers of $x_1$ and the other copy by powers of $1-x_1$, so they all have the same degree, then do the above with $2n$ instead of $n$.

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    $\begingroup$ You had $s_1 f_1 + \cdots, s_n f_n$, which I'm pretty sure was supposed to be $s_1 f_1 + \dotsb + s_n f_n$. I edited accordingly. I hope that was all right. $\endgroup$
    – LSpice
    Apr 18, 2023 at 23:52
  • $\begingroup$ Thanks. But can I understand the coefficients of the $s,t$ variables in terms of Resultants of the f's? $\endgroup$ Apr 19, 2023 at 3:13
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    $\begingroup$ @LSpice: of course you are right. Thanks for fixing it. $\endgroup$ Apr 19, 2023 at 13:25
  • $\begingroup$ @giuliobullsaver: the way I see it, the construction I gave does give an understanding of the coefficients in terms of resultants of polyynomials in $I$. Please explain why you disagree, otherwise I don't know what you are looking for exactly. $\endgroup$ Apr 19, 2023 at 13:28
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    $\begingroup$ BTW the above is not original, it's just my recollection of my reading, a long time ago, of the eliminated chapter on elimination theory in the algebra book of van der Waerden. $\endgroup$ Apr 19, 2023 at 13:30

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