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Fix $\ell \geq 3$, $r \geq 2$ and $1 \leq k \leq \ell - 1$ and $z_1, \ldots, z_\ell \in \mathbb{C}$ with $z_i \neq 0$ for all $i$ and $z_i \neq z_j$ for all $i \neq j$. Now consider the (irreducible, non-homogeneous) polynomials

$q_i = z_{\ell}x_i^r - z_ix_{\ell}^r - (z_{\ell} - z_i)$ for $1 \leq i < \ell$ and

$p = z_\ell x_0^{r + k} - x_1\cdots x_k(1 - x_{\ell}^r)$

in $R = \mathbb{C}[x_0, \ldots, x_{\ell}]$. I am interested in the ideal $I = (q_1, \ldots, q_{\ell - 1}, p) \subset R$. After having computed examples using Sage/Singular, I think that $I$ is prime, but I don't see how to prove this. Does anyone know some method which could be applied to this ideal to check if it is prime?

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    $\begingroup$ how about the familiar theorem that an ideal is prime iff the quotient ring is an integral domain? $\endgroup$
    – vidyarthi
    Commented Oct 25, 2016 at 9:14
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    $\begingroup$ You could try examining the Groebner bases (with respect to various term orders) to see if you can spot some pattern. If you can describe an explicit Groebner basis then you should have a good shot at proving that the ideal is prime. $\endgroup$
    – Timothy Chow
    Commented Oct 26, 2016 at 1:10
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    $\begingroup$ @TimothyChow Thank you for the hint. I just checked and it turns out that the basis given above already is a Groebner basis. I know about the algorithm of Gianni-Trager-Zacharias, but I don't see how knowing a Groebner basis makes this algorithm easier. Maybe I am missing something obvious here (probably due to my lack of understanding of commutative algebra). $\endgroup$
    – Nils Amend
    Commented Oct 26, 2016 at 9:52
  • $\begingroup$ Out of curiosity, why do you want to know this ideal is prime? $\endgroup$
    – Floresza
    Commented Oct 26, 2016 at 20:41
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    $\begingroup$ @NilsAmend : It's not necessarily a slam dunk, but what you can try to do is to step through a primality testing algorithm in slow motion and see if you can make sense of what's happening, in a way that lets you prove "in words" that the result will always be prime. (Finding a Groebner basis in the first place is often a big stumbling block in this plan, but you've already gotten over that hurdle.) $\endgroup$
    – Timothy Chow
    Commented Oct 27, 2016 at 2:51

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