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 $s_3$.

  • $\begingroup$ 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

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 $x_3 = \frac{g- \sigma_1 - x_2 }{2}$, $x_1 = g-2x_2-3x_3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.