Suppose we have a polynomial $p \in \mathbb{Q}[x]$ and we consider the roots $\alpha_0,\dots,\alpha_d \in \mathbb{C}$ of $p$. Then, $p$ imposes some polynomial relations on the $\alpha_i$. Let us denote this ideal of the relations by $I \trianglelefteq \mathbb{Q}[\alpha_0,\dots,\alpha_n]$. How can we compute a basis of $I$?

Example: If we have $p=x^2-x-1$, then the roots are $\phi=\frac{1+\sqrt{5}}{2}$ (the golden ratio) and $\psi=\frac{1-\sqrt{5}}{2}$. It can be easily checked that $\phi=-1/\psi$. Hence, they satisfy, e.g., $\phi \psi +1 = 0$ or $\phi^2 \psi^2-1=0$ (which also follows from the previous relation). So, $\langle \phi \psi +1 \rangle \subseteq I$. How can we find a basis for $I$ and how can we do this in general for other polynomials?

What kind of theory is behind this? Can Galois theory or algebraic number theory answer this?

  • 2
    $\begingroup$ Interesting question. You definitely mean $\mathbb{Q}[\alpha_0, \ldots, \alpha_d]$ instead of $\mathbb{C}[\alpha_0, \ldots, \alpha_n]$. A few initial thoughts: the quotient is finite-dimensional since $\alpha_i^d$ can always be reduced to lower powers. Additional relations come from Vieta's formulas. This is effectively computable in principle, though the naive brute force linear algebra approach would be a slow mess. $\endgroup$ Apr 8 '21 at 8:13
  • 1
    $\begingroup$ The field generated by all the roots of $p(x)$ is isomorphic to the splitting field $K$. By the primitive element theorem $K$ is generated by a single element $\beta$ which you can take to be $\beta = \sum \lambda_i \alpha_i$ for generic $\lambda_i \in \mathbf{Q}$. Let $\beta$ have minimal polynomial $B(x)$. Then $K = \mathbf{Q}[x]/B(x)$. But then each $\alpha_i$ will be equal to $A_i(\beta)$ for some polynomial of degree less then $[K:\mathbf{Q}]$. So then $I$ is generated by $A_i(\beta) - \alpha_i$ and $B(\beta) = 0$. $\endgroup$
    – user902362
    Apr 8 '21 at 11:57
  • $\begingroup$ What do you mean with a basis of $I$? Did you mean some elements generating the ideal? If so then we do it by factoring $p=\prod_j p_j^{e_j}$ in $\Bbb{Q}[\alpha_1][x]$, $p_2$ is the $\Bbb{Q}[\alpha_1]$ minimal of $\alpha_1$, then repeating factoring $p_2=\prod_j q_j^{e_j}$ in $\Bbb{Q}[\alpha_1,\alpha_2][x]$, $q_3$ is the minimal polynomial of $\alpha_3$, and so on. This is how we find the splitting field of a polynomial. $\endgroup$
    – reuns
    Apr 8 '21 at 13:07
  • $\begingroup$ For your interest, search for Marden theorem for cubic and Lin McMullin’s Theorem for quartic. $\endgroup$ Apr 9 '21 at 0:46
  • $\begingroup$ @user902362 Thank you very much! This looks very good to me. I checked with the example and it seems to give the ideal which can be expected $I=\langle (\phi-\psi)^2-5,\frac{1-\psi-\phi}{2} \rangle = \langle \psi^2-\psi-1, \phi + \psi -1\rangle$ (the latter being the Groebner basis). Is it obvious why the polynomials from your method really generate the entire ideal? $\endgroup$
    – blablablup
    Apr 9 '21 at 7:23

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