# 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?

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 :

$$P(X)=a_kX^k+...+a_0$$

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

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