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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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  • $\begingroup$ For the record: the unusually large number of votes on the (many times) duplicate answers below is likely because of the extra exposure the question got due to being tweeted. There are surely many good dupe targets but I don't have time to search at the moment. $\endgroup$ Commented Sep 6 at 13:45

2 Answers 2

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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

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

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

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

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

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

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. $\endgroup$ Commented Dec 4, 2019 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. $\endgroup$ Commented Dec 4, 2019 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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