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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 $$x\otimes y \sim x\otimes z \Leftrightarrow (y-z)\in\mathbb{Q}?$$


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Could you rephrase your last sentence? I don't understand the English grammar. – William DeMeo Feb 20 '12 at 11:25
Oops, I mean the 2nd-to-last sentence. The sentence "Thanks." is fine. :) – William DeMeo Feb 20 '12 at 11:27
up vote 8 down vote accepted

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? – Pete L. Clark Feb 20 '12 at 13:09
@Pete: This has been asked and answered in a separate question. – joriki Feb 20 '12 at 14:30
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. – Georges Elencwajg Feb 20 '12 at 14:38
@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. – Jason DeVito Feb 20 '12 at 15:03
@joriki: thanks. – Pete L. Clark Feb 20 '12 at 15:19

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 $f(\pi,\bar 0)=\bar 0=f(\pi,\bar 1)=\overline {\pi}\neq \bar 0$ – Georges Elencwajg Feb 20 '12 at 14:25
That really is a subtle error! Thank you again for undeleting it. – Jason DeVito Feb 20 '12 at 15:22

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