I was investigating relationships between commutative algebra and real analysis when the following problem came into mind.
Problem. Let $ P \in \mathbb{R}[x,y] $. If $ P(\sin(\theta),\cos(\theta)) = 0 $ for all $ \theta \in \mathbb{R} $, then is it true that $ P $ lies in the principal ideal $ \langle x^{2} + y^{2} - 1 \rangle $?
This problem does not appear difficult, but I have yet to find a solution. Thanks!
Let $ A \stackrel{\text{df}}{=} \mathbb{R}[y] $, so that $ P \in A[x] $. As $ P $ is a monic polynomial, there exist polynomials $ Q,R \in A[x] $ such that $$ P = (x^{2} + y^{2} - 1) Q + R, $$ where the $ x $-degree of $ R $ is less than $ 2 $. We can thus write $$ R(x,y) = f(y) x + g(y). $$
Now, suppose that $ f(y) $ and $ g(y) $ are not zero. Then as \begin{align} \forall \theta \in \mathbb{R}: \quad 0 & = P(\sin(\theta),\cos(\theta)) \\ & = R(\sin(\theta),\cos(\theta)) \\ & = f(\cos(\theta)) \cdot \sin(\theta) + g(\cos(\theta)), \end{align} we can deduce that $$ \forall t \in [0,1]: \quad f(t) \sqrt{1 - t^{2}} + g(t) = 0, $$ which implies that $ [f(t)]^{2} (1 - t^{2}) \equiv [g(t)]^{2} $ as polynomials. This is seen to be a contradiction, just by looking at the order of vanishing of both polynomials at $ t = 1 $. Therefore, $ P \in \langle x^{2} + y^{2} - 1 \rangle $.