1

Suppose that the minimal polynomial of $x$ in $\mathbb{Q}[X]$ is $$\sum_{n=0}^k a_n X^n.$$ How can I find the minimal polynomial of $ix$ (where $i$ is the imaginary unit) in $\mathbb{Q}[X]$ from the coefficients $a_n$?

I ran into a problem because some powers of $i$ (that is, $i$ and $-i$) are not in $\mathbb{Q}$.

Vestoo
  • 409
  • 1
    You can't, in general: you can have $x$ and $y$ which have the same minimal polynomial over $\mathbb{Q}$ such that $ix$ and $iy$ have different minimal polynomials over $\mathbb{Q}$. Is there some more specific context you are interested in? – Eric Wofsey Mar 13 '23 at 20:32
  • @EricWofsey A part of a proof that $\pi$ is transcendental is the following implication: "if $\pi$ is algebraic, then $i\pi$ is algebraic". (Where "algebraic" means a root of a polynomial equation with rational coefficients). I found this implication to be non-trivial because some powers of $i$ are not rational and this led me to the question above. I don't know how to prove it. – Vestoo Mar 13 '23 at 20:40
  • to give two examples of differing behavior: try $x=1$. Then, try $x =\sqrt[4] 2$. – peter a g Mar 13 '23 at 20:42
  • 2
    Note that showing $ix$ is algebraic is quite different from finding its minimal polynomial: you just have to find some polynomial with rational coefficients which it is a root of, which is not necessarily the minimal polynomial. – Eric Wofsey Mar 13 '23 at 20:47
  • 1
    In any case, see https://math.stackexchange.com/a/141540/86856 for the standard way to prove statements like this (which does not actually explicitly construct a polynomial at all!). – Eric Wofsey Mar 13 '23 at 20:52
  • 2
    @Vestoo "some powers of i are not rational" $-$ $z = ix$ is a root of $P(-iz)$. Collect all odd powers of $i$ and write it as $P(-iz) = A(z) + i B(z)$ with $A, B$ rational polynomials, then $z$ is a root of $A^2(z)+B^2(z)$ which is a rational polynomial. It may not be the minimal one, but it doesn't have to be if all you are after is proving $z$ is algebraic. – dxiv Mar 13 '23 at 21:24

1 Answers1

1

Suppose you have $y=ix$ you can express that in rational coefficients as $x^2+y^2=0$. Now that's a polynomial in $x$ and your original minimal polynomial is another polynomial in $x$. You can compute the resultant of both these polynomials; it has to be zero for both polynomials to have a common zero, i.e. for there to be an $x$ which satisfies them both. The resultant is a determinant in the coefficients of the individual polynomials, but since one of these coefficients is $y^2$ that gives you a polynomial in $y$ which can be used to characterized $y$. Whether it's minimal or not is yet another question, so you best factor the result and see.

MvG
  • 42,596