Given $\alpha,\,\beta,\,\gamma$ three roots of $g(x)\in\mathbb Q[x]$, a monic polynomial of degree $3$. We know that $\alpha+\beta+\gamma=0$, $\alpha^2+\beta^2+\gamma^2=2009$ and $\alpha\,\beta\,\gamma=456$. Is it possible to find the polynomial $g(x)$ only from these?

I've been working with the degree of the extension $\mathbb Q \subseteq \mathbb Q(\alpha,\,\beta,\,\gamma)$. I've found that it must be $3$ because $g(x)$ is the irreducible polynomial of $\gamma$ over $\mathbb Q(\alpha,\,\beta)$. But there is something that doesn't hold, there must be some of these roots that are not algebraic or something. May be this approach is totally wrong. Is there anyone who knows how to deal with this problem?

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    $\begingroup$ Are you asking if the equations in $\alpha,\beta,\gamma$ have a unique (modulo permutations) solution in $\mathbb{Q}$? Surely $g(x) = (x-\alpha)(x-\beta)(x-\gamma)$? Perhaps I am missing something? $\endgroup$ – copper.hat Nov 8 '12 at 18:34

The polynomial is $(x-a)(x-b)(x-c)$ with the roots being $a,b,c$. By saying "three roots" you imply all these are different. Note that when multiplied out and coefficients are collected you have three symmetric functions in the roots. For example the constant term is $-abc$, while the degree 2 coefficient is $-(a+b+c)$. The degree 1 coefficient is $ab+ac+bc$, which can be written as $$\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}.$$ So it looks like you can get all the coefficients of the monic from the givens you have.

Note: Just saw copper.hat's remark, essentially saying what's in this answer. I'll leave it up for now in case the poser of the question needs it (or can even use it...)


Recall $\rm\:g(x) = (x-a)(x-b)(x-c) = x^3\! - s_1 x^2 + s_2 x - s_3\:$ where the $\rm\,s_i\:$ are the elementary symmetric polynomials. You are given $\rm\,s_1,\,s_3\,$ and you seek $\rm\,s_2,\,$ given also the symmetric polynomial $\rm\,p_2 = a^2+b^2+c^2.\:$ By the Fundamental Theorem of Symmetric Polynomials, every symmetric polynomial can be written uniquely as a polynomial in the elementary symmetric polynomials, using a very simple algorithm due to Gauss. Let's recall how this works.

If $\rm\ a^i\ b^j\ c^k\ $ is the highest monomial in the symmetric polynomial $\rm\:p(a,b,c)\:$ (w.r.t. dictionary lex order where $\rm\ a > b > c),\: $ then subtract $\rm\ s_1^{i-j}\ s_2^{j-k}\ s_3^k\:.\:$ The result is a symmetric polynomial with smaller lex-degree, so iterating this reduction yields a representation of $\rm\:p(a,b,c)\:$ as a polynomial in the elementary symmetric polynomials $\rm\:s_i\:.\:$

Let's apply Gauss's algorithm to your symmetric polynomial $\rm\,p_2.\:$ Here since $\rm\:p_2\:$ has highest term $\rm\ a^{\color{#C00}2} b^0 c^{\color{#0A0}0} $ in lex-order, we subtract from $\rm\,p_2$ the term $\rm s_1^{\color{#C00}2-0}\, s_2^{0-\color{#0A0}0}\ s_3^{\color{#0A0}0} =\, s_1^2 = (a+b+c)^2$ yielding

$$\rm a^2 + b^2 + c^2 - (a + b + c)^2 =\, 2\,(ab + bc + ca)$$

This has leading monomial $\rm\: 2\, a^{\color{#C00}1} b^1 c^\color{#0A0}0,$ so to kill it we subtract $\rm\: 2\, s_1^{\color{#C00}1-1} s_2^{1-\color{#0A0}0} s_3^{\color{#C00}0} = 2\, s_2\ $ yielding

$$\rm 2\,(ab + bc + ca) - 2\, s_2 = 0 $$

Thus we've proved $\rm\:p_2 - s_1^2 = 2\,s_2,\:$ which allows us to solve for $\rm\:s_2\:$ since we know $\rm\:p_2,\,s_1.\:$

More generally, this leads to a simple recursion that yields closed-form Newton's Identities expressing the power sums in terms of elementary symmetric functions. Of course I could have simply referred you to this Wikipedia page to look up said formula. But that would be little help when you need to do the same for other symmetric polynomials - as you probably will, since such symmetries are ubiquitous.


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