Consider a polynomial in $d$ variables of degree $N>1$. When $d=1$, it is a well-established fact that such a polynomial can be expressed as a product of polynomials, each of degree 1. However, for $d>1$, this is no longer the case. A $d$-variate polynomial of degree $N$ can be decomposed into a product of irreducible polynomials, with the sum of their degrees equating to $N$. To elucidate the structure of these irreducible components, we introduce the following notation:

  • A partition of $N$ is denoted by $\lambda = (1^{m_1}, \ldots, \lambda_1^{m_{\lambda_1}})$, where $m_i$ represents the number of irreducible polynomials of degree $i$ in the decomposition of $P$.

  • A multipartition of $l(\lambda)$ into $N$ components is denoted by $\mu = (\mu^{(1)}, \ldots, \mu^{(N)})$, where each component $\mu^{(i)}$ is itself a partition of $m_i$, representing the multiplicities of the irreducible polynomials of degree $i$ in the decomposition of $P$.

  • The concatenation of the partition $\lambda$ and its associated multipartition $\mu$ is written as $\lambda_{\mu}$.


We now define the generalized coincident root locus, denoted by $X_{\lambda_\mu}$, associated with a partition $\lambda$ and a multipartition $\mu$ as follows. For $\lambda \vdash N$ and $\mu \vdash_N l(\lambda)$ with $\mu^{(i)} \vdash m_i(\lambda)$, $X_{\lambda_\mu}$ is the set of all polynomials that decompose into irreducible components in accordance with $\lambda_\mu$:

$X_{\lambda_\mu}\equiv $

$ \{ P \in \mathbb{C}[x_0,...,x_{d-1}]_N ~|~ P=\prod_{i=1}^{l(\lambda)} \prod_{j=1}^{l(\mu^{(i)})} P_{i,j}^{\mu_{i,j}} \text{, with } P_{\alpha,\beta} \in \mathbb{C}[x_0,...,x_{d-1}]_\alpha,~ \beta \in \mathbb{N}\}$

where the $P_{\alpha,\beta}$ are irreducible and distinct.


The primary objective is to construct the set $X_{\lambda_\mu}$ in Macaulay2, with the aim of determining the ideal $I = I(X_{\lambda_\mu})$ and computing a Gröbner basis for this ideal. Guidance and methodologies for implementing this in Macaulay2 are sought, as I am at the preliminary stage of familiarity with Macaulay2.

Nb : Should there be an alternative tool that facilitates a more streamlined approach than Macaulay2, and provided that you are adept at utilizing this alternative for a comprehensive response, I would deem such a response to be acceptable.


To fix ideas, let the irreducible decomposition of a $d$-variate polynomial $P$ of degree 12 reads $P=P_{1,1}^2\cdot P_{2,1} \cdot P_{2,2}^2 \cdot P_{4,1}$, where $P_{i,j}$ denotes the $j$-th degree $i$ polynomial constituting the irreducible decomposition of $P$. The associated $\lambda_{\mu}$ reads $(1^2,2^3,4^1)_{((2),(1,2),(1))}$. Therefore, $P\in X_{\lambda_\mu}$.

  • $\begingroup$ I'm not quite sure I understand your set $X_{\lambda_\mu}$. Could you give a bit of context, and maybe a simple example for $d = 2$? If I understand correctly, you consider all polynomials, where the factorisation in irreducibles has a specific number of factors with specific degree, which appear with specific multiplicity? Tbh, I suspect that the $X_{\lambda_mu}$ will generate the unit ideal. $\endgroup$ Jun 13, 2023 at 11:38
  • $\begingroup$ @red_trumpet Thanks for your comment. I added an example, could you please tell me if it gives you more clarity? $\endgroup$
    – Baloo
    Jun 13, 2023 at 12:55
  • $\begingroup$ Yep, that is what I thought you meant. So I think my answer applies, and $X_{\lambda_\mu}$ generates indeed the unit ideal. $\endgroup$ Jun 13, 2023 at 13:08

1 Answer 1


I suspect the $X_{\lambda_\mu}$ will generate the unit ideal (i.e. the full ring). If I understand correctly what you are after, then $$P \in X_{\lambda_\mu} \implies P_a := P(x_0 - a_0, \dotsc, x_{d-1} - a_{d-1}) \in X_{\lambda_\mu}$$ for all $a = (a_0, \dotsc, a_{d-1}) \in \mathbb C^d$. Geometrically, if you consider the zero set $$V(P) = \{\,x \in \mathbb C^d : P(x) = 0\,\}$$ this corresponds to a translation, i.e. $V(P_a) = V(P) + a.$ But then, since $V(P) \subsetneq \mathbb C^d$, you have $$\bigcap_{a \in \mathbb C^d} V(P_a) = \bigcap_{a \in \mathbb C^d} (V(P) + a) = \emptyset$$ By Hilbert's Nullstellensatz, the ideal generated by all the $P_a, a \in \mathbb C^d$ is the unit ideal. So in total we have $$\mathbb C[x_0, \dotsc, x_{d-1}] = I(P_a | a \in \mathbb C^d) \subset I(X_{\lambda_\mu}) \subset \mathbb C[x_0, \dotsc, x_{d-1}].$$

  • $\begingroup$ Since I am a physicist rather than a mathematician, let me clarify my interpretation of your answer for confirmation. It appears that, through the definition of $X_{\lambda_\mu}$, we generate a set of polynomials $\{P_a\}_a$ with no common element in their zero sets. Invoking the Weak Nullstellensatz, this set seemingly generates the unit ideal $I(X{\lambda_\mu})$. Is this accurate? If so, could you elucidate why $I(X_{\lambda_\mu})$ is generated by the $P_a$? Since this is the ideal I am looking for. By the way, sorry if this is quite trivial. $\endgroup$
    – Baloo
    Jun 13, 2023 at 13:39
  • $\begingroup$ As a supplementary request, my intention is to employ Macaulay2 for an empirical examination of this matter. While your response is insightful and much appreciated, is it feasible to corroborate your assertion through the explicit computation of this ideal using Macaulay2? $\endgroup$
    – Baloo
    Jun 13, 2023 at 13:42
  • 1
    $\begingroup$ @Baloo The $P_a$ might a priori not generate $I(X_{\lambda_\mu})$, but as they are contained in $X_{\lambda_\mu}$, you get an inclusion of ideals, and the smaller one is the full ring. As for applications, I'm not sure how feasible this method is to construct $1$ as a linear combination of the $P_a$. Maybe try an easy example, like $P = xy$? $\endgroup$ Jun 13, 2023 at 16:10
  • $\begingroup$ Thank you for your eloquent response. However, an aspect leaves me perplexed: if $I(X_{\lambda_\mu})$ constitutes the unit ideal for any $\lambda,\nu$, what implications does this hold for the set of equations that define a specific $X_{\lambda_\mu}$? Could this potentially suggest that $X_{\lambda_\mu}$ does not qualify as an algebraic variety? If so, what about the Zariski closure of a given $X_{\lambda_\mu}$ ? $\endgroup$
    – Baloo
    Jun 14, 2023 at 7:34
  • $\begingroup$ @Baloo Why would you think of $X_{\lambda_\mu}$ as a variety? That is a set of polynomials, and by $I(X_{\lambda_\mu})$ I simply mean the ideal generated by those polynomials. $\endgroup$ Jun 14, 2023 at 9:05

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