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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.

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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.

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  • $\begingroup$ I guess I should have attempted this problem a little longer! Very concise and elegant, if I could upvote twice I would. $\endgroup$ – The Wheel is Before Descartes Dec 4 '19 at 3:29
  • $\begingroup$ @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. $\endgroup$ – John Omielan Dec 4 '19 at 3:30
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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, $\alpha \in F[\alpha] \setminus \{0\} = (F[\alpha])^{\times}$.

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