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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.

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  • $\begingroup$ Are you willing to use Hilbert's Nullstellensatz, or looking for something more elementary? $\endgroup$
    – stewbasic
    Jun 22, 2016 at 0:27
  • $\begingroup$ 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). $\endgroup$ Jun 22, 2016 at 0:32

1 Answer 1

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Let $I=\mathbb R[x,y](1-x^2-y^2)$, and suppose $f\in\ker(\phi)$.

Proof 1

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 $\mathbb R$, $f\in I$.

Proof 2

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 $f\in I$.

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  • $\begingroup$ What exactly is meant by the first definition? Is I an ideal or a quotient ring? $\endgroup$ Jun 22, 2016 at 1:37
  • $\begingroup$ I meant for I to be the ideal generated by (1-x^2-y^2). In your question you denote this by just (1-x^2-y^2), but I thought that might be ambiguous. $\endgroup$
    – stewbasic
    Jun 22, 2016 at 1:44

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