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How to prove that the element $1\otimes \arccos\frac{1}{3}\in\mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ isn't equal to zero?

I know why $$\arccos\frac{1}{3}\neq \frac{m}{n}\pi,$$ where $m\in\mathbb{Z}$ and $n\in\mathbb{N}$.

So, am I right that sufficiently state on $\mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ (on decomposable elements) equivalence relation $$x\otimes y \sim x\otimes z \Leftrightarrow (y-z)\in\mathbb{Q}?$$

Thanks.

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    Could you rephrase your last sentence? I don't understand the English grammar.2012-02-20
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    Oops, I mean the 2nd-to-last sentence. The sentence "Thanks." is fine. :)2012-02-20

2 Answers 2

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The hypothesis that $\arccos\frac{1}{3}\neq \frac{m}{n}\pi$ is irrelevant. What counts is that $\arccos\frac{1}{3}\notin \mathbb Q$.

Choose a basis $(r_i)_{i\in I}$ of $\mathbb R$ over $\mathbb Q$ with $0\in I$ and $x_0=1$.
Every $\xi \in \mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ can be written uniquely as $\xi=\Sigma x_i\otimes \bar y_i$ and in particular $$1\otimes \bar y_i=0 \iff \bar y_i=0 \in \mathbb R/\mathbb{Q}\iff y_i\in \mathbb Q$$
which proves that $1\otimes \overline {\arccos\frac{1}{3}}\neq 0$

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    $@$Georges: I was about to type out a similar answer, but I was stymied at giving a reference for the irrationality of $\arccos \frac{1}{3}$. Do you know of one?2012-02-20
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    @Pete: This has been asked and answered in [a separate question](http://math.stackexchange.com/questions/111288/why-arccos-frac13-is-an-irrational-number).2012-02-20
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    Thanks a lot to Pete: he was absolutely right to raise the issue. Indeed I didn't know a reference for the irrationality of $\arccos\frac{1}{3}$ and I am extremely grateful to @joriki for providing one.2012-02-20
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    Dear Georges: Thank you very much for your comment. I deleted my answer. I feel at the same time very embarrassed and very grateful, ... but more grateful than embarrassed, so positive feelings win! (By the way I was the first one of a hopefully very long list of users upvoting *your* answer. I liked it already when I believed mine was correct...)2012-02-20
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    @Pierre-Yves: If you don't mind my asking, what was wrong with your original post? I remember reading it and thinking to myself "yes, that makes sense", then running off to class.2012-02-20
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    Dear @Jason: I undeleted the answer. Thanks for you kind comment.2012-02-20
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    @joriki: thanks.2012-02-20
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WARNING!!! As explained in Georges's comment, the argument below is completely wrong. After having read this comment I deleted the answer, but Jason asked me very kindly what precisely was incorrect. So I decided to undelete it, hoping that it will serve as an example of what one should not do. (I hope the warning is conspicuous enough. If it isn't, please let me know.)

Hint. The tensor $1\otimes a\in\mathbb R\otimes_\mathbb Q\mathbb R/\mathbb Q$ is nonzero if and only if $a$ is irrational.

Proof: The natural $\mathbb Q$-bilinear map $f$ from $\mathbb R\times\mathbb R/\mathbb Q$ to $\mathbb R/\mathbb Q$ sends $(1,a)$ to $a$.

(The map $f$ is defined, with obvious notation, by $f(x,\overline y)=\overline{xy}$.)

Edit. More generally, the argument shows this. Let $A$ be a subring of $B$. Then $1\otimes\overline b$ is zero in $B\otimes_AB/A$ if an only if $b$ is in $A$. (We assume that $B$ is commutative.)

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    Dear Pierre-Yves, I'm afraid the map $f$ is not well defined. Indeed, it would lead to $f(\pi,\bar 0)=\bar 0=f(\pi,\bar 1)=\overline {\pi}\neq \bar 0$2012-02-20
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    That really is a subtle error! Thank you again for undeleting it.2012-02-20