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If $g=x_1+2x_2+3x_3, s_1=x_1+x_2+x_2, s_2=x_1x_2+x_1x_3+x_2x_3$ and $s_3=x_1x_2x_3$ , write $x_1, x_2$ and $x_3$ in function of $g, s_1, s_2$ and $s_3$.

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  • $\begingroup$ I tried solve the system with 4 equations, but I no had success! $\endgroup$ – Carlos Gomes May 12 '15 at 3:20
  • $\begingroup$ can you write the system and tell us where you get stuck? $\endgroup$ – hjhjhj57 May 12 '15 at 3:22
  • $\begingroup$ The equations are complicated. I can not isolate any of the variables $x_1, x_2$ or $x_3$. $\endgroup$ – Carlos Gomes May 12 '15 at 3:25
  • $\begingroup$ $s_1$, $s_2$ ,$s_3$ can naturally give you a cubic equation whose roots are $x_i$ $\endgroup$ – Yimin May 12 '15 at 3:45
  • $\begingroup$ This seems a hard problem; if it is asking for polynomial expressions, I cannot see easily why it should be solvable at all. Can you provide more context of where this problem comes from? Is this an exercise from some book/course, and if so on which subject? Really, "this question is missing context and details" (which is a possible motive to close it, though I won't vote to do so) applies here. $\endgroup$ – Marc van Leeuwen May 12 '15 at 3:52
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I'm pretty sure a complete answer is too long for this format, but I will give a sketch of my approach.

First off, while we could try to find relations using powers of $g$, the bookkeeping will be awful. To try and ameliorate this, instead consider $$ (g-2\sigma_1)^2 = (x_3-x_1)^2 = x_3^2+x_1^2 - 2x_1x_3 $$ This element is fixed by the Galois action interchanging $x_1$ and $x_3$. It follows that this element generates the fixed points of this action, and thus we can in theory write $x_2$ in terms of this element. The minimal polynomial for this element is degree $3$ over the base field $\Bbb Q(\sigma_1, \sigma_2, \sigma_3)$, so there are symmetric polynomials $F_1, F_2, F_3, F_4$ such that $$ x_2 = \frac{F_1 + (g-2\sigma_1)^2F_2 + (g-2\sigma_1)^4F_3}{F_4} $$ If we rewrite the desired relation as $$ F_1 + (g-2\sigma_1)^2F_2 + (g-2\sigma_1)^4F_3- x_2F_2 = 0 $$ we see that we might as well assume that the $F_i$ are homogeneous. Some counting lets us see that we must have a solution in total degree $4$, since we can take arbitrary linear combinations of the following $10$ elements: $$ \begin{array}{lll} (g-2\sigma_1)^4 & (g-2\sigma_1)^2\sigma_2 & (g-2\sigma_1)^2\sigma_1^2 \\ x_2\sigma_3 & x_2\sigma_2\sigma_1 & x_2\sigma_1^3 \\ \sigma_3\sigma_1 & \sigma_2^2 & \sigma_2\sigma_1^2\\ \sigma_1^4, \end{array} $$ and the space of degree-4 polynomials symmetric in $x_1$ and $x_3$ is $9$-dimensional, spanned by $$ \begin{array}{llll} x_2^4 & x_1^4+x_3^4 & x_2^3(x_1+x_3) \\ x_2(x_1^3+x_3^3) & x_1x_3(x_1^2+x_3^2) & x_2^2(x_1^2+x_3^2) \\ x_1^3x_3^3 & x_2^2x_1x_3 & x_2x_1x_3(x_1+x_3) \end{array} $$ At this point, finding suitable $F_i$ is "just" linear algebra. After obtaining an expression for $x_2$, you can get $x_3 = \frac{g- \sigma_1 - x_2 }{2}$, $x_1 = g-2x_2-3x_3$.

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