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Given a set of polynomials equations in variables $x_1\dots x_n$ and a set of linear inequalities $L_k(x_1\dots x_n)\ne0$, is the set of solutions an algebraic set? If it is, what is the corresponding set of polynomials?

PS: if the number of solutions with the inequalities included is finite, the answer to the first question is yes. What are the corresponding polynomials?

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  • $\begingroup$ The solution is in general not an algebraic set, for example $k\backslash\{0\}$, where $k$ is your field. Such sets (defined by polynomial equations and inequalities) are called semialgebraic. $\endgroup$
    – Alexandre Eremenko
    Commented Oct 4, 2023 at 8:32
  • $\begingroup$ Ah right, let me restate the question. $\endgroup$
    – Alm
    Commented Oct 4, 2023 at 9:16
  • $\begingroup$ Such sets will at least be constructible sets, which is a broader class than algebraic sets. en.wikipedia.org/wiki/Constructible_set_(topology) $\endgroup$
    – Terry Tao
    Commented Oct 4, 2023 at 16:31

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This is to answer the question in PS.

When the number of solutions is finite, each $L_k$ takes on them a finite number of nonzero values. Let $S_k$ denote the set of these values. Then we can replace each $L_k(x_1,\dots,x_n)\ne 0$ with a single polynomial equation: $$\prod_{s\in S_k} (L_k(x_1,\dots,x_n) - s)=0.$$ These equations together with the original polynomial equations define the same set of solutions.

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  • $\begingroup$ Sorry, I meant that the number of solutions is finite with the inequalities included. $\endgroup$
    – Alm
    Commented Oct 4, 2023 at 15:39
  • $\begingroup$ Yes, this is how I interpreted the question. What is your concern? $\endgroup$
    – Max Alekseyev
    Commented Oct 4, 2023 at 15:42
  • $\begingroup$ Ah ok, it is right. I am wondering if the set of polynomials generates a radical ideal if the original polynomials without inequalities do. $\endgroup$
    – Alm
    Commented Oct 4, 2023 at 16:15

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