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Let $f(x)$ be a polynomial of degree $n$ with rational coefficients whose zeroes are nonnegative real numbers: $x_1, \dots, x_n\geq 0$.

General question. Does there exist a simple expression for the number $$r=\sqrt{x_1} + \sqrt{x_2} + \dots + \sqrt{x_n}$$ in terms of coefficients of $f(x)$ ?

Remark. Let $s_k = x_1^k + x_2^k + \dots + x_n^k$ be the sum of $k$-th powers of zeros of $f(x)$. It is clear that for nonnegative integer $k$, $s_k$ can be expressed as a polynomial of the coefficients of $f(x)$ (cf. Newton-Girard formulae). The number $r$ (and even its square $r^2$) represent a zero of a polynomial whose coefficients are polynomials in $s_k$:

$n=1:\qquad r^2 - s_1 = 0$

$n=2:\qquad r^4 - 2 s_1 r^2 + (2 s_2 - s_1^2) = 0$

$n=3:\qquad 3 r^8 - 12 s_1 r^6 + (6 s_1^2 + 12 s_2) r^4 + (- 20 s_1^3 + 72 s_1 s_2 - 64 s_3) r^2 + (3 s_1^4 - 12 s_1^2 s_2 + 12 s_2^2) = 0$

$n=4:\qquad 9 r^{16} - 72 s_1 r^{14} + (180 s_1^2 + 72 s_2) r^{12} + \dots = 0$

Unfortunately the degree of these polynomials grows as $2^n$, and in general it seems it cannot be made much smaller.

Specific question. Now, let us consider a special case of $f(x)$ satisfying the identity $f(y^2)=g(y)\cdot g(-y)$ where $g(y)$ is a polynomial with rational coefficients. Then each $\sqrt{x_i}$ is a zero of either $g(y)$ or $g(-y)$. Can we find a simple expression for $r$ in this case?

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Hi Max, any progress on this problem of yours? (I have a related one) – Ziofil Aug 24 '14 at 17:20
@Filippo: unfortunately, no progress. – Max Alekseyev Aug 26 '14 at 1:50

I don't think that anything better than what you figured out already can be done. From the algebraic point of view, you cannot distinguish the $2^n$ siblings of $r$, corresponding to the different choices of the signs of the square roots. So you get a degree $2^n$ polynomial where $r$ is one of the roots. I believe it is not possible to take positivity into account.

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In general case, I agree. But I have a hope that something better can be done in the special case when $f(y^2) = g(y)\cdot g(-y)$. – Max Alekseyev Feb 27 '13 at 15:49
I'm skeptical that anything can be done under this added assumption - I doubt that the hypothesis that $g$ has rational coefficients can be combined with the positivity requirement. Of course, this is not a proof, just a feeling. – Peter Mueller Feb 27 '13 at 16:14

Let us consider this in the rational function field of $n$ variables $x_1,\cdots, x_n$, Let $f(x)=(x-x_1)\cdots (x-x_n)=x^n+c_1x^{n-1}+\cdots+c_{n-1}x+c_n$ Then the coefficients are elementary symmetric functions in $x_1,\cdots, x_n$. We see that $K=\mathbb{Q}(x_1,\cdots,x_n)$ is a Galois extension of $k=\mathbb{Q}(c_1,\cdots,c_n)$ with Galois group $S_n$.

Now we consider $r=\sqrt{x_1}+\cdots +\sqrt{x_n}$. Then $r$ has degree $2^n$ over $K$, so there is a polynomial $p$ of degree $2^n$ which is satisfied by $r$. Let $p(x)=x^{2^n}+f_1 x^{2^n-1} +\cdots + f_{2^n}$, where $f_i\in K$. Further, we have $f_i\in k$, indeed each $f_i\in\mathbb{Z}[c_1,\cdots,c_n]$.

Thus, it follows that $F=k(r)$ also has degree $2^n$ over $k$. Note that $F$ is not a normal extension of $k$. However, $KF=\mathbb{Q}(x_1,\cdots,x_n,r)$ is a normal extension of $k$. Indeed, it is the normal closure of $k(r)$ over $k$. So, the Galois group $Gal(KF/k)$ has $S_n$ as a quotient. $S_n$ is not solvable if $n\geq 5$. Therefore, if $n\geq 5$, then we cannot have "a root formula" for $r$ in terms of $c_1,\cdots c_n$.

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