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Given an arbitrary radical expression $a$ whose minimal polynomial over $\mathbb{Q}$ is known to be $P(x)$, $P(x^n)$ is an annulling polynomial for $\sqrt[n]{a}$, where $n\in\mathbb{N}$. When $P(x^n)$ is not irreducible, is there an algebraic way to determine which factor of $P(x^n)$ is the minimal polynomial of $\sqrt[n]{a}$ over $\mathbb{Q}$? Or is it in general necessary to do numerical approximations of each factor evaluated at $\sqrt[n]{a}$ and see which result is closest to zero?

In some sufficiently simple cases it would of course be possible to evaluate the factors algebraically rather than numerically, but in general I'm not aware of a set of algebraic simplifications that are guaranteed to determine whether an expression equals zero.

More context: I'm writing a program into which the user can enter any radical expression. The program returns the simplest form of the expression it can find. During the course of the computation, it's sometimes necessary to find minimal polynomials of radical expressions, the specifics of which depend on what the user entered. When the program has to solve the problem this question is about, it has the value of $n$, it has a representation of $a$ as essentially a binary tree with integers for leaves and arithmetic operations at the nodes, and it has the coefficients of $P(x)$.

  • Are you working with a concrete case? –  May 06 '18 at 02:18
  • @totoro Depends on what you mean. In context, the algebraic number will always be specified explicitly as a radical expression, but I need to be able to do handle any such expression. – Alex Kindel May 06 '18 at 02:21
  • Many algebraic numbers are not radical expressions. –  May 06 '18 at 02:22
  • @totoro I know, but that happens to be the case I'm working with. I suppose I should update the question to reflect that. – Alex Kindel May 06 '18 at 02:23
  • There you go. There is a case that you are working on. Then write it there. Focus the problem, make the specification as concrete as possible and then the solution as general as possible. –  May 06 '18 at 02:25

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