Algebraic number field has multiplicative inverse I am struggling with showing that for algebraic number $\alpha$, the ring generated by $\mathbb{Q}[\alpha]$ is a field. I understand that to do this, I will have to show that any $r+s\alpha, r,s\in \mathbb{Q}$ has an inverse in $\mathbb{Q}[\alpha]$. I'm lost on how to go about doing this, though. Help? 
 A: The ring $\mathbb{Q} [\alpha]$ is the image of $f: \mathbb{Q}[x] \to \mathbb{C}$ where $f: p(x) \mapsto p(\alpha)$ ($f$ is evaluation at $\alpha$.) Since $\alpha$ is algebraic it has a minimal polynomial $q_{\alpha}(x)$. Show the kernel of $f$ is the principal ideal generated by $q_{\alpha}(x)$. Since $q_{\alpha}(x)$ is irreducible its principal ideal is maximal. Thus the image of $f$, the subring $\mathbb{Q}[\alpha]$ generated by $\alpha$, is a field.
A: *

*From the very definition of    $\mathbb{Q}[\alpha]$ (look in your course if you don't understand), this is a ring ($\alpha$ could be transcendental here).


*From point $1$, the only thing you need to do is to find an inverse for each element of $\alpha$. Use the fact that $\alpha$ is algebraic to show that $\alpha^{-1}\in \mathbb{Q}[\alpha]$.

Hint $\alpha$ is algebraic, then there is a minimal polynomial (irreducible) $P\neq 0$ so that $P(\alpha)=0$. Write :
$$P(X)=a_kX^k+...+a_0$$
It follows that :
$$\alpha(a_k\alpha^{k-1}+...+a_2\alpha+a_1)=-a_0 $$
After you justified that  $a_0\neq 0$, conclude that the inverse of $\alpha$ is indeed in $\mathbb{Q}[\alpha]$.



*Using $2$, show that any element in $\mathbb{Q}[\alpha]$ is invertible in $\mathbb{Q}[\alpha]$.

A: If $\alpha$ is an algebraic number, then $\mathbb{Q}[\alpha]$ is a finite-dimensional vector space over $\mathbb{Q}$.
The map $x \mapsto \alpha x$ is an injective linear transformation and so is surjective.
This means that $1$ is in the image and so $\alpha$ has an inverse in $\mathbb{Q}[\alpha]$.
(Injectiveness follows from $\mathbb{Q}[\alpha]\subseteq \mathbb{C}$.)
The main advantage of this approach is that it works for finding the inverse of any $\beta \in \mathbb{Q}[\alpha]$ simply by solving a linear system, without any tricks or flashes of insight.
