0
$\begingroup$

Suppose a nonzero complex number $\alpha$ is a root of a polynomial of degree $n$ with rational coefficients. Show that $\frac{1}{\alpha}$ is also a root of a polynomial of degree $n$ with rational coefficients.

I have no idea how to solve this question. I tried to write that suppose $\alpha$ is a root of $f(x)$, and f(x) can be written in $(1-\alpha)g(x)$, but I am not sure. Could you please help me to solve this question? Thank you.

$\endgroup$
  • 1
    $\begingroup$ Suppose $p(X)$ annihilates $\alpha$, and has degree $m$. Then $X^m p(X^{-1})$ is a polynomial that annihilates $\alpha^{-1}$. $\endgroup$ – Pedro Tamaroff Mar 24 '16 at 0:24
5
$\begingroup$

Suppose $a_0+a_1\alpha+\dots+a_n\alpha^n=0$, with $a_0\ne0$, which is possible because $\alpha\ne0$.

Consider the polynomial $a_n+a_{n-1}X+\dots+a_1X^{n-1}+a_0X^n$.

$\endgroup$
0
$\begingroup$

The smallest subfield of $\mathbb{C}$ that contains $\alpha$ must also contain $\alpha^{-1}$, thus the degrees of the minimal polynomials must be the same.

$\endgroup$
  • $\begingroup$ you are using that $\alpha$ is algebraic, thus so is $\alpha^{-1}$, so that the minimal polynomial of $\alpha^{-1}$ exists, but to prove its degree is the same as the minimal polynomial for $\alpha$, you will have to prove first that if $\alpha$ is a root of $f(x)$, then $\alpha^{-1}$ is a root of $x^n f(x^{-1})$ ? $\endgroup$ – reuns Mar 24 '16 at 0:28
  • $\begingroup$ If the minimal polynomial for $\alpha$ has degree $n$, then $\alpha^{-1} \in \mathbb{Q}(\alpha)$ and so it must have degree smaller than $n$. A similar argument applied to $\alpha^{-1}$ shows that the degrees are the same. $\endgroup$ – sqtrat Mar 24 '16 at 0:31
  • $\begingroup$ yes but what I mean is that for proving all this, you will probably use that $f(\alpha) = 0 \implies \alpha^{-1}$ is a root of $x^n f(x^{-1})$ $\endgroup$ – reuns Mar 24 '16 at 0:32
  • $\begingroup$ Ok, true that, I will delete it. $\endgroup$ – sqtrat Mar 24 '16 at 0:33
  • $\begingroup$ no, don't delete it, that was only a comment (and a question) on how to prove all this $\endgroup$ – reuns Mar 24 '16 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.