$\alpha$ is a root of a polynomial. Show that $\frac{1}{\alpha}$ is also a root of a polynomial.

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.

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

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 $a_n+a_{n-1}X+\dots+a_1X^{n-1}+a_0X^n$.
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.
• 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 $x^n f(x^{-1})$ ? – reuns Mar 24 '16 at 0:28
• 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. – sqtrat Mar 24 '16 at 0:31
• 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 $x^n f(x^{-1})$ – reuns Mar 24 '16 at 0:32