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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 $n,m \in \mathbb{Z}$?
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It's hard to solve polynomial equations exactly even when we don't allow the roots, so we shouldn't expect these to be any easier. E.g we can't in general express the roots of a 5-th degree polynomial in terms of radical functions of the coefficients (Abel-Ruffini theorem) so we can't solve $$\sqrt[3]{ x^5+ax^4+bx^3+cx^2+dx+e }$$ either. –  Ragib Zaman Sep 2 '12 at 8:39
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Thanks, but I was aware of Abel-Ruffini. I only want to turn the special polynomial into a normal polynomial. Then, I hope that I can solve it. But my question is only about the transformation. –  Johannes Sep 2 '12 at 8:45
    
At least as soon as complex roots are considered, it makes littel sense to distinguish $\sqrt X$ and $-\sqrt X$, say. In that view, $\sqrt X+X+1=0$ and $\sqrt X-X-1=0$ have the same solutions, i.e. $z_{1,2}-\frac12\pm i\frac{\sqrt3}2$. Both of these have one square root that makes $\sqrt X + (X+1)=0$ and one that makes $\sqrt X - (X +1)=0$: The two square roots of $z_1$ are $\pm z_2$ and we have $+z_2+(z_1+1)=0$ and $-z_2-(z_1+1)=0$. –  Hagen von Eitzen Sep 2 '12 at 8:51
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The binomi trick is in fact a piece of Galois theory. If you make all combinations with differnt signs before square roots (and different $n$-th roots of unity before $n$-th roots), then all elementary symmetric expressions in these have no more roots in them. For $\sqrt X + X+1$ this means you are axtually solving $Y^2=X$ and $Y+X+1=0$, the symmetric expressions are $(Y+X+1)+(-Y+X+1)=2(X+1)$ and $(Y+X+1)(-Y+X+1)=(X+1)^2-Y^2$. The first is uninteresting, the other is known to be 0, leads to the solution for $X$ and then find $Y$ from $Y+X+1=0$. In general, however, degrees tend to grow fast. –  Hagen von Eitzen Sep 2 '12 at 9:07
    
Cool, I think that solves the first two questions. Indeed, for $3$rd roots, it should work, too! Now I am not completely sure about which X must be excluded from the domain (3rd point on my list...). For $(\sqrt{X}+(X+1))$, I only need to check that solutions are no roots of $(\sqrt{X}-(X+1))$. Which I could do by just putting them into $(\sqrt{X}-(X+1))$. Correct? –  Johannes Sep 2 '12 at 9:19
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