2
$\begingroup$

Let $A$ denote the set of algebraic elements in $\mathbb{C}$ over $\mathbb{Q}$. Then if $a\in A$, prove that $a^{-1} \in A$.

Since $a\in A$, then there is a polynomial $f(x)=\sum_{i=0}^n \lambda_i x^i$ such that $f(a)=0$. Then $f(a^{-1})=\sum_{i=0}^n \lambda_i a^{-i}$ which is not $0$. I wonder how to modify $f$ so that we can have a new polynomial $\tilde{f}$ such that $\tilde{f}(a^{-1})=0$.

  • 1
    For broader perspectives, see [Enlightening proof that the algebraic numbers form a field](http://math.stackexchange.com/q/331017). – 2017-02-18

1 Answers 1

4

The number $a^{-1}$ is a root of

$$\tilde f:=\sum_{i=0}^n \lambda_{n-i} x^i$$

so it is algebraic.

  • 1
    You can see this by dividing you polynomial by $x^n$. – 2017-02-18