How to show that $\mathbb{R}[x, y] / \langle x^2 + y^2 - 1 \rangle $ is a Dedekind domain?
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2I would start with the easy parts and see how far I come. If I were stuck then, I would ask for help here and present my previous work on the problem. – asdq Aug 17 '19 at 11:09
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@DietrichBurde you only gave the answer. I still do not know how to show that the ring is integrally closed. – Shri Aug 17 '19 at 11:20
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Vincenzo proves this (integrally closed) at the duplicate in (ii) ! – Dietrich Burde Aug 17 '19 at 11:22
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For such a smooth curve, for each $(a,b) \in \Bbb{C}^2, a^2+b^2=1$ there exists a polynomial $p(x,y)\in \Bbb{R}[x,y]$ having a simple zero at $(a,b)$ (for $a \ne \pm 1$ it means $P'(0)\ne 0$ where $P(t) = p(a+t,\sqrt{1-(a+t)^2})$), let $I(a,b) = { f \in \Bbb{R}[x,y], f(a,b)=0}$ then the localization $\Bbb{R}[x,y]{I(a,b)}$ is a DVR (because its only maximal ideal $(p{a,b}(x,y))$ is principal) which implies the ring is a Dedekind domain. To factorize an ideal $J$ say that $\Bbb{R}[x,y]J$ is in finitely such DVR so that $J = \prod (J,p{a,b})$. – reuns Aug 17 '19 at 12:07