Patterns in the minimal polynomial of sum of square roots

Question

Let $x_1,\dots,x_n$ be the roots of $x^n+\sum_{k=0}^{n-1}(-1)^{n-k}a_{n-k}x^k=0$, so $a_n=(-1)^nx_1\dots x_n$.

The minimal polynomial of $\sqrt{x_1}+\dots+\sqrt{x_n}$ is $$P_nQ_n=f_n^2-a_n\,g_n^2$$ where $$P_n=\prod_{\substack{\mu_1,\dots,\mu_n\in\{\pm1\}\\\mu_1\dots\mu_n=1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_n\sqrt{x_n}))\\ Q_n=\prod_{\substack{\mu_1,\dots,\mu_n\in\{\pm1\}\\\mu_1\dots\mu_n=-1}}(x-(\mu_1\sqrt{x_1}+\dots+\mu_n\sqrt{x_n}))$$ and $$f_n=\frac{P_n+Q_n}2\\ g_n=\frac{Q_n-P_n}{2\sqrt{x_1\dots x_n}}$$

Note that the leading term of $P_n,Q_n$ are $x^{2^{n-1}}$,
so the leading term of $f_n$ is $x^{2^{n-1}}$.
This answer proved the leading term of $g_n$ is $2^{n-1}(n-1)!x^{2^{n-1}-n}$.

Note that $P_n,Q_n$ are swapped under each automorphism $\phi_i:\sqrt{x_i}\mapsto-\sqrt{x_i}$,
so $f_n,g_n$ are unchanged under each automorphism $\phi_i:\sqrt{x_i}\mapsto-\sqrt{x_i}$,
so $f_n,g_n$ contain no square roots, so $f_n,g_n$ are symmetric polynomials in $x_1,\dots,x_n$, so they can be expressed as polynomials in elementary symmetric polynomials $a_1,\dots,a_n$.

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Consider the difference of $f_n$ with the minimal polynomial for $n-1$: $$\tag1\label1 f_n-(f_n|_{a_n=0}) $$ For $n=2,3$, $\eqref{1}=0$.

My question is: How to prove the following?

For $n\ge4$, $\eqref{1}=O(x^{2^{n-1}-2n})$ and the leading coefficient is divisible by $64$.

For $n=4$, $\eqref{1}=64$

For $n=5$, $\eqref{1}=10240a_5x^6+\ldots$

For $n=6$, $\eqref{1}=1712128a_6x^{20}+\ldots$

For $n=7$, $\eqref{1}=345899008a_7x^{50}+\ldots$

For $n=8$, $\eqref{1}=86249439232a_8x^{112}+\ldots$

For $n=9$, $\eqref{1}=26408252342272a_9x^{238}+\ldots$

This answer proved a special case $a_1=\dots=a_{n-1}=0,a_n=-(-1)^n$ of this problem, so the $n$ roots of $x^n-1=0$ are $x_k = \exp(k \frac{2 \pi i}{n})$, so we take $\sqrt{x_k}= \exp(k \frac{\pi i}{n})$ and $f_n|_{a_n=0}=x^{2^{n-1}}$.

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Some thoughts:

\eqref{1} is divisible by $a_n$.

Since $P_n$ and $Q_n$ are homogeneous of total degree $2^{n-1}$ in ($n+1$) variables $x,\sqrt{x_1},\dots,\sqrt{x_n}$, then $f_n$ and $g_n$ are homogeneous of total degree $2^{n-1}$ in ($n+1$) variables $x,\sqrt{x_1},\dots,\sqrt{x_n}$, then \eqref{1} is homogeneous of total degree $2^{n-1}$ in ($n+1$) variables $x,\sqrt{x_1},\dots,\sqrt{x_n}$.

The total degree of $a_n=(\sqrt{x_1})^2\dots(\sqrt{x_n})^2$ in ($n+1$) variables $x,\sqrt{x_1},\dots,\sqrt{x_n}$ is $2n$.

But \eqref{1} is divisible by $a_n$, so the total degree of $\frac{\eqref{1}}{a_n}$ in ($n+1$) variables $x,\sqrt{x_1},\dots,\sqrt{x_n}$ is $2^{n-1}-2n$.

Answer 1

Here is how i understand the question and the framework.

Let $n\ge 2$ be a fixed natural number. We consider the polynomial ring $R$ over $\Bbb Q$ generated by transcendental variables $y_1,y_2,\dots,y_n$, its field of fractions $L$, and the subfield $K$ of $L$ generated by $y_1^2,y_2^2,\dots,y_n^2$: $$ \begin{aligned} R &= \Bbb Q[y_1,y_2,\dots,y_n]\ ,\\ K &= \Bbb Q(y_1^2,y_2^2,\dots,y_n^2)\ ,\\ L &= \Bbb Q(y_1,y_2,\dots,y_n)\ . \end{aligned} $$ Let $L[X]$ be the polynomial ring over $L$ in the transcendental variable $X$.

The extension $L:K$ is algebraic, of degree $2^n$, in a tower of quadratic extensions of degree two, and the Galois group $G$ is generated by the morphisms $(\psi_j)_{1\le j\le n}$. Here, $\psi_j$ sends $y_j$ to $-y_j$ and invariates all other generators of $L$.

(Think of $y_j$ as $\sqrt{x_j}$ in OP terminology.)

We consider the index set $J=\{+1,-1\}^n$ of all $n$-tuples $\mu$ with entries in the set of signs $\{+1,-1\}$. Each $\mu\in J$ has a product of the components, $\Pi\mu$, and we define $$ \begin{aligned} J(P) &= \{\ \mu\in J\ :\ \Pi \mu=+1\ \}\ ,\\ J(Q) &= \{\ \mu\in J\ :\ \Pi \mu=-1\ \}\ ,\\[2mm] P &= \prod_{\mu\in J(P)}(X-\mu\cdot y)\ ,\qquad\text{ where } \mu\cdot y:=\mu_1y_1+\mu_2y_2+\dots+\mu_ny_n\ ,\\ Q &= \prod_{\mu\in J(Q)}(X-\mu\cdot y)\ ,\\ f &=\frac 12(P+Q)\ . \end{aligned} $$ The sets $J(P), J(Q)$ have each the cardinality $$d:=2^n/2\ ,$$ so $P,Q,f$ have degree $d$.

The given definition of $P,Q$ make them invariant under the Galois group $G$. The symmetric group $S=S(n)$ of substitutions / permutations of the generators also acts.

So as mentioned, the coefficients of the $X$-powers in $P,Q,f$ are invariated by $S$, so they are polynomials in the elementary symmetric polynomials in the $y$-variables. The standard notation for these elementary symmetric polynomials is $e_1,e_2,\dots, e_n$, so i will also use this notation. We write $E_1,E_2,\dots,E_n$ for $e_1,e_2,\dots,e_n$ computed in the squares $y_1^2,y_2^2,\dots,y_n^2$. By abuse of notation, we also allow $E_1,E_2,\dots,E_n$ to be independent variables, potentially with no relation to the symmetric functions.

So there are "universal polynomials" $U(P),U(Q), U(f)$ in the variables $X,E_1,E_2,\dots,E_n$ with: $$ \begin{aligned} P &= \prod_{\mu\in J(P)}(X-\mu\cdot y)=U(P)(X;E_1,E_2,\dots,E_n)\ ,\\ Q &= \prod_{\mu\in J(P)}(X-\mu\cdot y)=U(Q)(X;E_1,E_2,\dots,E_n)\ ,\\ f &= \frac 12(P+Q)=U(f)(X;E_1,E_2,\dots,E_n)\ . \end{aligned} $$

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In the question the following polynomial $F$ is considered, and recall that by abuse we use now $E_1,E_2,\dots,E_n$ as independent variables with no connection to the initial $y$-variables: $$ F:= U(f)(X;E_1,E_2,\dots,E_{n-1},E_n) -U(f)(X;E_1,E_2,\dots,E_{n-1},0) \ . $$ We have to prove:

Claim: The $X$-degree of $F$ is $d-2n$, its $X$-coefficient in this degree is $$cE_n\ ,\qquad c \in \Bbb Z\ $$ and $c$ is a number divisible by $64$.

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Proof of the claim:

• We first look which are the changes in $P,Q,f$ when we replace $X$ by $-X$. There are two cases, $n$ even, and $n$ odd. In each case we use the substitution $\nu=-\mu$, where the minus sign is propagated to each component. So we have $\Pi\nu=(-1)^n\Pi\mu$. In the even case $\mu,\nu$ belong to the same component $J(P)$ or $J(Q)$, in the odd case they are in different components. So we compute: $$ \small \begin{aligned} &n\text{ even:} & P(-X) & = \prod_{\mu\in J(P)}(-X-\mu\cdot y) = \prod_{\mu\in J(P)}(X+\mu\cdot y) = \prod_{\nu\in J(P)}(X-\nu\cdot y) =P(X)\ ,\\ && Q(-X) & = \prod_{\mu\in J(Q)}(-X-\mu\cdot y) = \prod_{\mu\in J(Q)}(X+\mu\cdot y) = \prod_{\nu\in J(Q)}(X-\nu\cdot y) =Q(X)\ ,\\[2mm] &n\text{ odd:} & P(-X) & = \prod_{\mu\in J(P)}(-X-\mu\cdot y) = \prod_{\mu\in J(P)}(X+\mu\cdot y) = \prod_{\nu\in J(Q)}(X-\nu\cdot y) =Q(X)\ ,\\ && Q(-X) & = \prod_{\mu\in J(Q)}(-X-\mu\cdot y) = \prod_{\mu\in J(Q)}(X+\mu\cdot y) = \prod_{\nu\in J(P)}(X-\nu\cdot y) =P(X)\ . \end{aligned} $$ So in both cases $f=\frac 12(P+Q)$ satisfies $f(X)=f(-X)$, so it lives only in even $X$-degrees.

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• $F$ collects from $U(f)$ exactly the monomials that contain $E_n$ to the power at least one. For instance monomials like $E_n$, $E_1E_2^2E_n^3$, $E_1^{100}E_2E_3^{27}E_n^4$ may survive. Seen as polynomial in the $y$-variables we use the weight $2j$ for $E_j$. For $X$ we use the weight one. With this weight system, $U(P),U(Q),U(f), F$ have total weight $d=2^n/2$. The maximal $X$-degree where $E_n$ is present is $d-2n$, and the corresponding monomial is $$ c\; E_nX^{d-2n}\ .$$ This shows the wanted relation $F\in O(X^{d-2n})$.

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• It remains to show the wanted divisibility property for $c$, which is in fact the main problem. The key observation is that if we consider the $E$-variables $E_1,E_2,\dots,E_n$ with no relation to the $y$-variables, then the term $cE_nX^{d-2n}$ is also present in the specialization with zero of the $E$-variables, all but the last one: $$ U(f)(X;0,0,\dots,0,E_n)=X^d+c\;E_n X^{d-2n}+\text{(lower terms in degrees $<d-2n$) .}$$ When we also specialize $E_n=0$ and subtract we obtain $$ \begin{aligned} F(X;0,0,\dots,0,E_n) &=U(f)(X;0,0,\dots,0,E_n)- U(f)(X;0,0,\dots,0,0) \\ &=c\;E_n X^{d-2n}+\text{(lower terms in degrees $<d-2n$) .} \end{aligned} $$So we need the term in $X$-degree $d-2n$ from $U(f)(X;0,0,\dots,0,E_n)$.

  And it is enough to further specialize $E_n$ to a value like $\pm 1$ to isolate $c$. So we compute the simpler expression, which is an $X$-polynomial: $$F(X;0,0,\dots,0,\pm 1)\ .$$

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• For this we may specialize the initial values of $y_1,y_2,\dots,y_n$ so that the squares $y_1^2,y_2^2,\dots,y_n^2$ are the complete set of the roots of unity of order $n$. (This leads to the computation of $F(X;0,0,\dots,0,\pm 1)$.)

  So we are exactly in the setting of the related question:

  MSE 5019227

  More exactly, let $z_1,z_2,\dots,z_n=-1$ in cyclic order be the first $n$ roots of the one of order $2n$. Here $z_j=\exp(2\pi ij/(2n))$. These are the notations used in my answer to MSE 5019227. We specialize these values in the present setting. When plugging in this system, we formally write $z$ for the system $z=(z_1,z_2,\dots,z_n)$, and $z^2$ for the system $(z_1^2,z_2^2,\dots,z_n^2)$. With these notations observe that:$$ \begin{aligned} 0 &= e_1(z^2) = E_1(z)\ ,\\ 0 &= e_2(z^2) = E_2(z)\ ,\\ &\vdots\\ 0 &= e_{n-1}(z^2)=E_{n-1}(z)\ ,\\ \pm1 &= e_n(z^2) = E_n(z)\ ,\\ P(X,z) &= \prod_{\mu\in J(P)}(X-\mu\cdot z) \\ &=U(P)(X;E_1(z),E_2(z),\dots,E_{n-1}(z),E_n(z)) \\ &=U(P)(X;0,0,\dots,0,\pm 1) \ ,\\ Q(X,z) &= \prod_{\mu\in J(Q)}(X-\mu\cdot z) \\ &=U(Q)(X;E_1(z),E_2(z),\dots,E_{n-1}(z),E_n(z)) \\ &=U(Q)(X;0,0,\dots,0,\pm 1) \ ,\\ f(X,z) &=\frac 12(P(X,z)+Q(X,z)) \ . \end{aligned} $$

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• In my answer to MSE 5019227, the above polynomial was explained, it is a polynomial of the shape $$ X^d + cX^{d-2n}+\text{(lower terms)}\ ,$$ and $c$ (from there, which is $\pm c$ from here) satisfies a refined divisibility condition w.r.t. powers of $2$.

$\square$