1
$\begingroup$

Let's define special polynomials as polynomials in $\mathbb{Q}[X]$, where we allow to make roots, too. Examples:

$\sqrt{X^4+1}$, $\sqrt[3]{X}+\sqrt{X+1}$, $\sqrt{X+\sqrt{X+1}}$

How can I transform a special polynomial equation into a usual polynomial equation? Let's consider $\sqrt{X}+(X+1) = 0$. For "most" $X$, we can just use binomi:

$(\sqrt{X}+(X+1))(\sqrt{X}-(X+1))=X-(X+1)^2=0$.

Now, we have a nice polynomial and in this case, we can solve it :). Here are my questions, and it's okay if you can already answer one point:

  • Can I solve something like this in general if I have only roots with index 2?
  • What if the index is, let's say, 3?
  • In my above equation, I used the word "most". I think it is okay for all $X$ that do not solve $(\sqrt{X}-(X+1))=0$. But how do I get those $X$?
  • Bonus question. What if I use $K:=\mathbb{Q}(\sqrt[n]{m})$ for all $n,m \in \mathbb{Z}$?
  • 0
    Oh, also: May I assume that instead of calculating $a-b=0$, calculating $a^n-b^n$ for arbitrary $n \in \mathbb{N}$ is equivalent? That way I could finally transform every sum of $n$th roots into a root-less sum?2012-09-02

0 Answers 0