# Show that $\mathbb{Q}(\sin\theta)$ is a field

QUESTION:

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 $$\theta$$.

• 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 $\sin\theta$". – Arthur Aug 19 at 9:29
• @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. – Arthur Aug 19 at 9:41
• 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 $\theta$. – Mark Aug 19 at 9:45
• Maybe you defined it specifically for elements like $\sqrt{2}$. – Mark Aug 19 at 9:48
• @Kumar Arthur and Mark are right. I gave an explanation of what they said by editing my answer. – 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.
• Using the above identity only, I was successful in showing the extension, for specific values of $\theta$. – Kumar Aug 19 at 9:34
• @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 $\theta$. – Scientifica Aug 19 at 9:39
• 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. – Mark Aug 19 at 9:59