Show that $F_{\theta}=\mathbb{Q}(\sin\theta); \theta\in\mathbb{R}$ is a field. Moreover, Show that $E_{\theta}=\mathbb{Q}(\sin\frac{\theta}{3}); \theta\in\mathbb{R}$ is a field extension of $F_{\theta}$ and find the maximum value of [$E_{\theta}:F_{\theta}$]

I am unable to show that for every $x\in F_{\theta}$, it has an inverse.
Also, I have shown that for some specific values of $\theta$, $E_{\theta}$ is a field extension of $F_{\theta}$.
But I can't prove for a general $\theta$.

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    $\begingroup$ What does $\Bbb Q(\sin\theta)$ mean to you? Because to me, it is by definition a field. The literal meaning of those symbols in my book is "The smallest subfield of $\Bbb R$ which contains $\Bbb Q$ and $\sin\theta$". $\endgroup$ – Arthur Aug 19 at 9:29
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    $\begingroup$ @Kumar Yeah, that's not going to be a field most of the time. And it's not the standard definition of $\Bbb Q(\sin\theta)$. Are you certain that that's your definition? I mean, $\Bbb Q(\sqrt2)$ looks like that, for instance, but it's not usually defined that way. $\endgroup$ – Arthur Aug 19 at 9:41
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    $\begingroup$ Sorry, but what you defined will be a field only if $\sin\theta$ is algebraic over $\mathbb{Q}$ with degree at most $2$. This is not even close to being true for all $\theta$. $\endgroup$ – Mark Aug 19 at 9:45
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    $\begingroup$ Maybe you defined it specifically for elements like $\sqrt{2}$. $\endgroup$ – Mark Aug 19 at 9:48
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    $\begingroup$ @Kumar Arthur and Mark are right. I gave an explanation of what they said by editing my answer. $\endgroup$ – Scientifica Aug 19 at 9:49

Hint: $\forall x\in\mathbb R, \sin(3x)=3\sin(x)-4\sin^3(x)$.

Edit: Didn't see your first question concerning inverses in $F_\theta$, my bad. As Arthur said in comment, $\mathbb Q(\sin(\theta))$ should be by definition the smallest field containing $\mathbb Q$ and $\sin(\theta)$. Also for some values of $\theta$, $\sin(\theta)$ is not algebraic over $\mathbb Q$. Hence for such a value $\theta$, you can't express $\dfrac{1}{\sin(\theta)}$ as a $\mathbb Q$-linear combinaision of powers of $\sin(\theta)$.

Edit 2: (See Edit 3 for mistakes) The definition you gave in comments is the one for Q[sin(theta)]. It is only normal that you struggle to show that it is a field, because in fact it is a field if and only if $\sin(\theta)$ is algebraic over $\mathbb Q$, otherwise it's merely an integral domain. For example, if you take $\theta\in\mathbb R$ such that $\sin(\theta)=e^{-1}$, then knowing that $e^{-1}$ is not algebraic over $\mathbb Q$, then $\dfrac{1}{\sin(\theta)}=e\notin\mathbb Q[e^{-1}]$, because otherwise there exists a polynomial $P(X)$ with coefficients in $\mathbb Q$ such that $P(e^{-1})=e$, and by multiplying in both sides of the expression with a high enough power of $e$, you get $Q(e)=1$ for some polynomial $Q(X)\in\mathbb Q[X]$, which contradicts the transcendence of $e$ over $\mathbb Q$.

Edit 3: As Mark said in comment, the definition you gave doesn't even make the structure a ring, and the smallest ring containing it is $\mathbb Q[\sin(\theta)]:=\{P(\sin\theta)\mid P(X)\in\mathbb Q[X]\}$ where $\mathbb Q[X]$ is the ring of polynomials with coefficients in $\mathbb Q$. As for $\{a+b\sin\theta\mid a,b\in\mathbb Q\}$, all that can be said in a general matter about it ($\theta\in\mathbb R$ arbitrary) is that it is a $\mathbb Q$-vector space, of dimension $1$ if $\sin\theta\in\mathbb Q$, and $2$ otherwise.

  • $\begingroup$ Using the above identity only, I was successful in showing the extension, for specific values of $\theta$. $\endgroup$ – Kumar Aug 19 at 9:34
  • $\begingroup$ @Kumar I think it depends on how you understand what $\mathbb Q(\alpha)$ is for $\alpha\in\mathbb R$. Normally $\mathbb Q(\alpha)$ is by definition the smallest field containing $\mathbb Q$ and $\alpha$. Please see the edit above as to why you may struggle for some values of $\theta$. $\endgroup$ – Scientifica Aug 19 at 9:39
  • $\begingroup$ Actually this is not even a definition for $\mathbb{Q}[\sin(\theta)]$. Let's assume it is a ring, then $\sin^2(\theta)$ has to be there. But then we can write $\sin^2(\theta)=a+b\sin(\theta)$, so $\sin(\theta)$ must be algebraic over $\mathbb{Q}$ with degree at most $2$. So if $\sin\theta$ is transcendental or even algebraic of a higher degree then $\{a+b\sin\theta:a,b\in\mathbb{Q}\}$ will not be closed to multiplication. $\endgroup$ – Mark Aug 19 at 9:59
  • $\begingroup$ @Mark Woops you're right. Thank you for pointing that out! Gonna fix it. $\endgroup$ – Scientifica Aug 19 at 10:09

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