# Preliminaries

Let $$n$$ be an integer such that $$n \geq3$$. Denote $$\left[ n \right] \equiv \{1,2, \ldots ,n \}$$. Let $$P$$ be a non-empty subset of $$\left[ n \right]$$ such that $$\left|P \right| \geq 3$$. Denote $$X_{P} \equiv \left( x_{i} \colon i \in P \right)$$ as an ordered alphabet. Let $$\mathbb{F}$$ be a field such that $$\operatorname{char} \left( \mathbb{F} \right) \neq 2$$. Denote $$\mathbb{F} \left[ X_{P} \right]$$ as the polynomial ring over $$\mathbb{F}$$ in the (commuting) letters of $$X_{P}$$.

## Polynomials as functions

Let $$P_{1},P_{2}$$ be non-empty subsets of $$P$$ such that $$\left|P_{1} \right| = \left|P_{2} \right|$$, then the polynomial $$f$$ in the letters of $$X_{P_{1}}$$, denoted by $$f \left( X_{P_{1}} \right) \in \mathbb{F} \left[ X_{P_{1}} \right]$$, and the polynomial $$f$$ in the letters of $$X_{P_{2}}$$, denoted by $$f \left( X_{P_{2}} \right) \in \mathbb{F} \left[ X_{P_{2}} \right]$$, will be considered equivalent as functions.

## Multivariate Polynomial to univariate polynomial and back

A (multivariate) polynomial in $$\mathbb{F} \left[ X_{P} \right]$$ can be uniquely written as a (univariate) polynomial in the letter/variable $$x_j$$ for some $$j \in P$$, or in other words as a polynomial in $$\left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right]$$. Conversely, a polynomial in $$\left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right]$$ can be uniquely written as a polynomial in $$\mathbb{F} \left[ X_{P} \right]$$. Denote $$f \left( X_{P \setminus \{ j \}} ;x_{j} \right)$$ as the polynomial in $$\left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ x_{j} \right]$$ which is equivalent to the polynomial $$f \left( X_{P} \right)$$, for any $$f \left( X_{P} \right) \in \mathbb{F} \left[ X_{P} \right]$$.

## Univariate resultants and multivariate polynomials

For any $$j \in P$$ and $$k \in P \setminus \{ j \}$$, consider the polynomials $$f \left( X_{P} \right) \in \mathbb{F} \left[ X_{P} \right]$$ and $$g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \}} \right) \in \mathbb{F} \left[ X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} \cup \{ k \} }\right]$$ and denote $$y \equiv x_{j} \equiv x_{k}$$. Then the (univariate) resultant of the polynomials $$f \left( X_{P \setminus \{ j \}} ;y \right) \in \left( \mathbb{F} \left[ X_{P \setminus \{ j \}} \right] \right) \left[ y \right]$$ and $$g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \in \left( \mathbb{F} \left[ X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \} }\right] \right) \left[ y \right]$$, with respect to the variable $$y$$, is well-defined, and shall be denoted by $$\operatorname{res}_{y} \left( f \left( X_{P \setminus \{ j \}} ;y \right) ,g \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)$$

# Recursive definition

The polynomial function $$q \left( X_{ \left[ n \right]} \right) \in \mathbb{F} \left[ X_{\left[ n \right]} \right]$$ is defined recursively by the condition $$q \left( X_{ \left[ 3 \right]} \right) \equiv -x_{1}^2-x_{2}^2-x_{3}^2+2x_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$$ and the rule $$q \left( X_{ \left[ n \right]} \right) = \operatorname{res}_{y} \left( q \left( X_{P \setminus \{ j \}} ;y \right) ,q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right)$$ for $$n \geq 4$$, any non-empty $$P \subseteq \left[ n \right]$$ such that $$\left|P \right| \geq 3$$ and any $$j \in P$$.

## Example

Suppose that $$n = 4$$, $$P = \{1,2,3\}$$ and $$j = 3$$. Then $$\left[ n \right] = \{1,2,3,4\}$$, $$\left| P \right| = 3 \geq 3$$ and $$j \in P$$; furthermore \begin{align} q \left( X_{P \setminus \{ j \}} ;y \right) & = q \left( X_{\{1,2,3\} \setminus \{ 3 \}} ;y \right) \\ & = q \left( X_{\{1,2\}} ;y \right) \\ & = -y^2+2 \left(x_{1}+x_{2} \right)y- \left( x_{1}-x_{2} \right)^2 \end{align} and \begin{align} q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) & = q \left( X_{\left( \{1,2,3,4\} \setminus \{1,2,3\} \right) \cup \{ 3 \}} ;y \right) \\ & = q \left( X_{\{3,4\}} ;y \right) \\ & = -y^2+2 \left(x_{3}+x_{4} \right)y- \left( x_{3}-x_{4} \right)^2 \end{align} Therefore $$q \left( X_{ \left[ 4 \right] } \right) = \operatorname{res}_{y} \left( q \left( X_{ \{1,2 \}} ;y \right) ,q \left( X_{ \{3,4 \}} ; y \right) \right)$$ is unambiguous.

# Is this well-defined?

Is the polynomial function defined above well-defined, in the sense that it is invariant (not even up to sign) of the choice of a subset $$P$$ and its element $$j$$?

Small examples are evidence for the positive answer.

## Some ideas

It is not hard to show by induction that the degree of $$q \left( X_{ \left[ n \right] };y \right) \in \left( \mathbb{F} \left[ X_{\left[ n \right] \setminus \{ i \}} \right] \right) \left[ y \right]$$ for any $$i \in \left[ n \right]$$ is $$2^{n-2}$$, so to obtain that, in general, \begin{align} \operatorname{res}_{y} \left( q \left( X_{P \setminus \{ j \}} ;y \right) ,q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) \right) & = \operatorname{res}_{y} \left(q \left( X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}} ; y \right) , q \left( X_{P \setminus \{ j \}} ;y \right) \right) \end{align} But this is obviously not enough. Small examples are evidence to $$q \left( X_{ \left[ n \right] } \right)$$ being a symmetric function in $$X_{ \left[ n \right] }$$, so $$q \left( X_{ \left[ n \right] } \right)$$ must be a multi-symmetric function in $$X_{P \setminus \{ j \}},X_{\left( \left[ n \right] \setminus P \right) \cup \{ j \}}$$, but I don't know how to prove the former claim.

## Sidenotes

I have in the past posted another question regarding this object, but that discussion is irrelevant to the current question, so I will not link it here.

Notice that $$\let\eps\varepsilon q(x_1,x_2,x_3) =-\bigl(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(\sqrt{x_1}-\sqrt{x_2}+\sqrt{x_3}\bigr) \bigl(-\sqrt{x_1}-\sqrt{x_2}+ \sqrt{x_3}\bigr)\\ =-\prod_{\eps_1,\eps_2\in\{\pm1\}}\left( \sqrt x_3+\sum_{I<3}\eps_i\sqrt{x_i}\right).$$

So the roots of this polynomial (as a one in $$x_3$$) are exactly $$\left(\sum_{i <3}\eps_i\sqrt{x_i}\right)^2$$ (each root appears in the above list twice, but it is a simple root in $$q$$).

Now a straightforward induction shows that $$q(x_1,\dots,x_n) =\prod_{\eps_1,\dots,\eps_{n-1}\in\{\pm1\}}\left( \sqrt x_n+\sum_{I and the (simple) roots of this polynomial (as a one in $$x_n$$) are exactly $$\left(\sum_{i

The inductive step here is just an application of the standard formula $$\mathop{\mathrm{res}}(A,B) =a_0^kb_0^n\prod_{i,j}(\lambda_i-\mu_j),$$ where the $$\lambda_i$$ and the $$\mu_j$$ list all the roots of $$A(x)=a_0x^n+\dots \quad\text{and}\quad B(x)=b_0x^k+\dots,$$ respectively. (Here we use that the degrees are powers of two!)

Remark. Surely, the polynomial defined by the above formulas, is symmetric in all variables, since its degree (in $$\sqrt{x_i}$$) is divisible by 4.

• Thanks, Ilya. It seems that your answer is dependent on the condition of the recursion, which is the definition of $q(x_1,x_2,x_3)$, which in turn enables one to "find" the general formula, thus prove that it is a symmetric function. I was hoping for an answer which doesn't rely on "solving" the recursion, or in other words an answer which does not depend on the condition of the recursion, but perhaps that is "too hard". Commented Dec 10, 2021 at 17:25
• Hm, well. Do you have a reasonable hope that tge same holds wothin a more general set up, where an exact formula is absent? Commented Dec 10, 2021 at 20:43
• I do. Suppose that instead $q(x_1,x_2,x_3) \equiv -x_{1}^2-x_{2}^2-x_{3}^2+2x_1x_2+2x_1x_3+2x_2x_3-4x_1x_2x_3$. Now a sort of formula for the n-th case is much harder to come up with, but it is possible, and the definition works in this case also. So I was hoping for a way to show that the function is well defined, without resorting to "solving" the recursion, in order to "catch" both cases, the one in the original question and the one in the current comment, together. But perhaps that is unrealistic. Commented Dec 11, 2021 at 11:47