# Is $\frac{1}{\alpha} \in \mathbb{Q}[\alpha]$ for irrational $\alpha$?

I have been trying to pick up abstract algebra and just attempted an exercise from Landin's An Introduction to Algebraic Structures which asks to prove whether $$\frac{1}{\pi} \in \mathbb{Q}[\pi]$$, and would like to ask a (slightly) more general question.

Let $$\alpha$$ be an irrational number. Is $$\frac1\alpha \in \mathbb{Q}[\alpha]$$?

$$\mathbb{Q}[\alpha]$$ being the ring of numbers of the form $$a_0 +a_1\alpha+a_2\alpha^2\dots a_n\alpha^n$$ with $$a_i\in\mathbb{Q}$$. I'm really not sure where to begin, and haven't found a similar question on MSE (perhaps because the solution is simpler than I realize.) I am stuck on what we can conclude about $$\frac{1}{\alpha}$$ other than that it is an irrational number greater than $$1$$. Clearly, $$\alpha^k n \in \mathbb{Q}[\alpha]$$, but I'm reluctant to make any claims about $$\frac{1}{\alpha}$$ and $$\alpha^k n$$, as I'm sure a solution would use other simpler methods.

*edited title and parts of the body because $$\alpha$$ need not be less than one.

Not, it's not true for all irrational numbers $$0 \lt \alpha \lt 1$$. To see this, assume it's true so you get

$$\frac{1}{\alpha} = \sum_{i=0}^{n}a_i \alpha^n \tag{1}\label{eq1A}$$

for some set of $$a_i \in \mathbb{Q}$$. Multiply by $$\alpha$$ on both sides and then subtract $$1$$ from both sides to get

$$\sum_{i=0}^{n}a_i \alpha^{n+1} - 1 = 0 \tag{2}\label{eq2A}$$

This means $$\alpha$$ is a root of the polynomial

$$p(x) = \sum_{i=0}^{n}a_i x^{n+1} - 1 \tag{3}\label{eq3A}$$

However, since all of the coefficients of the terms in $$p(x)$$ are rational, this can only happen with $$\alpha$$ being an algebraic number, so it's not true for all irrational, i.e., it doesn't hold for cases where $$\alpha$$ is a transcendental number.

• I guess I should have attempted this problem a little longer! Very concise and elegant, if I could upvote twice I would. – Descartes Before the Horse Dec 4 '19 at 3:29
• @heepo Thanks for the compliment. There are many cases where I've tried to solve something for a while and, when I saw how to solve it, wondered why I didn't see it earlier myself. – John Omielan Dec 4 '19 at 3:30

John's answer is nice, but I thought I might write an answer which emphasizes that this has nothing to do with $$\mathbb{Q}$$ (or irrational numbers, for that matter).

Proposition: Let $$F$$ be a field, and let $$R$$ be an $$F$$-algebra which is an integral domain. Fix an element $$\alpha \in R^{\times}$$. Then $$\alpha \in (F[\alpha])^{\times}$$ if and only if $$\alpha$$ is algebraic (integral) over $$F$$.

Proof: This is mostly a matter of analyzing what $$F[\alpha]$$ means: it is the image of the unique (universal) $$F$$-algebra homomorphism $$\varphi \colon F[X] \to R$$ sending $$X$$ to $$\alpha$$. Note, then, that $$\alpha$$ is algebraic over $$F$$ if and only if $$\ker(\varphi) \neq \langle 0 \rangle$$. Moreover, since $$R$$ is a domain, the image of $$F$$ is an integral domain. Hence, we see the following:

If $$\ker(\varphi) = \langle 0 \rangle$$, then $$\alpha$$ is transcendental over $$F$$, and the $$F$$-subalgebra $$F[\alpha] \subset R$$ is isomorphic to $$F[X]$$. In particular, $$\alpha \notin (F[\alpha])^{\times}$$, since $$X$$ is not a unit of $$F[X]$$. (It generates a maximal ideal of $$F[X]$$!)

If $$\ker(\varphi) \neq \langle 0 \rangle$$, then $$\ker(\varphi)$$ is a nonzero prime ideal of $$F[X]$$, since the image of $$\varphi$$ is a domain. Since $$F[X]$$ is a principal ideal domain, it follows that $$\ker(\varphi)$$ must be maximal, and so by the first isomorphism theorem, $$F[\alpha]$$ is a field. In particular, $$\alpha \in F[\alpha] \setminus \{0\} = (F[\alpha])^{\times}$$.