# Prove that $\Bbb{R}[\cos(\theta),\sin(\theta)]\cong\Bbb{R}[x,y]/(1-x^2-y^2)$ [duplicate]

More precisely, given the ring homomorphism $\phi:\Bbb{R}[x,y]\to\Bbb{R}^\Bbb{R}$, with $\phi(f(x,y)):\Bbb{R}\to\Bbb{R},\,\,\phi(f(x,y))(\theta)=f(\cos(\theta),\sin(\theta))$, where $\Bbb{R}[x,y]$ is the ring of formal polynomials in two variables on $\Bbb{R}$, I wish to show that $\mathrm{ker}(\phi)=(1-x^2-y^2)$. So far the best I can do is to show that $\mathrm{ker}(\phi)\supseteq(1-x^2-y^2)$, which is pretty trivial anyway. In particular, I'm interested in a proof of this statement from the perspective of algebraic geometry.

• Are you willing to use Hilbert's Nullstellensatz, or looking for something more elementary? Jun 22, 2016 at 0:27
• I'm familiar with the Nullstellensatz, so a proof using it would still be useful. However, a more elementary proof would be ideal (provided such a proof exists). Jun 22, 2016 at 0:32

Let $I=\mathbb R[x,y](1-x^2-y^2)$, and suppose $f\in\ker(\phi)$.
Let $p:\mathbb C\rightarrow\mathbb C^2$ be the map $\theta\mapsto(\cos(\theta),\sin(\theta))$. By assumption, $f\circ p$ vanishes on $\mathbb R$. It's analytic so it vanishes on $\mathbb C$ by the identity theorem. That is, $f$ vanishes on $\mathrm{im}\;p=Z(I)$. By Hilbert's Nullstellensatz, $f^r\in I$ for some $r$. Since $\mathbb R[x,y]$ is a UFD and $(1-x^2-y^2)$ is irreducible over $\mathbb R$, $f\in I$.
By induction on degree wrt $y$, there exist $g(x),\;h(x)\in\mathbb R[x]$ such that $f(x,y)\in g(x)+h(x)y+I$. For any $\theta\in\mathbb R$, $$0=f(\cos\theta,\sin\theta)=g(\cos\theta)+h(\cos\theta)\sin\theta.$$ Substituting $\theta\mapsto-\theta$, $$0=g(\cos\theta)-h(\cos\theta)\sin\theta.$$ Thus $g(\cos\theta)=h(\cos\theta)\sin\theta=0$. Allowing $\theta$ to vary, $g(x)$ and $h(x)$ both vanish on $(0,1)$. Since they are polynomials, they are zero, so $f\in I$.