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.



Polynomial function defined recursively by a resultant - is it well defined?
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
$$
 q(x\_1,\dots,x\_n) 
=\prod\_{\eps\_1,\dots,\eps\_{n-1}\in\{\pm1\}}\left(
 \sqrt x\_n+\sum\_{I<n}\eps\_i\sqrt{x\_i}\right),
$$
and the (simple) roots of this polynomial (as a one in $x\_n$) are exactly
$$
 \left(\sum\_{i <n}\eps\_i\sqrt{x\_i}\right)^2.
$$


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.


