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?


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.


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.

  1. 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).

  2. 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 :


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 $\mathbb{Q}[\alpha]$.

  1. Using $2$, show that any element in $\mathbb{Q}[\alpha]$ is invertible in $\mathbb{Q}[\alpha]$.
  • $\begingroup$ I understand 1, it's precisely 2 that I don't understand how to do :(. $\endgroup$
    – nargles324
    Dec 17 '15 at 9:56
  • $\begingroup$ @nargles324, ok I'll give a more precise hint. $\endgroup$ Dec 17 '15 at 9:57
  • $\begingroup$ ahh, does the above work because the ring is closed under multiplication and addition? thank you so much, that was silly of me! $\endgroup$
    – nargles324
    Dec 17 '15 at 10:03
  • 1
    $\begingroup$ You only have to justify $a_0 \ne 0$, right? Then $a_k\alpha^{k-1}+...+a_2\alpha+a_1\neq 0$ follows automatically from the equation $\alpha(a_k\alpha^{k-1}+...+a_2\alpha+a_1)=-a_0$. $\endgroup$
    – TonyK
    Dec 17 '15 at 10:16
  • 2
    $\begingroup$ This only gives an inverse for $\alpha$. What about the other elements of $\mathbb{Q}[\alpha]$? $\endgroup$
    – lhf
    Dec 17 '15 at 11:19

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