$\mathbb{R}[x,y]/(x^2+y^2-1)$ is integrally closed.

Asked Aug 17 '19 at 15:35

Active Aug 19 '19 at 06:51

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I have been thinking about how to prove that the integral domain $\mathbb{R}[x,y]/(x^2+y^2-1)$ is integrally closed and I am writing this post to ask for proof verification.

Attempt.

Let $R=\mathbb{R}[x,y]/(x^2+y^2-1)$. Then $Q(R)=\frac{R}{\{nonzero\:elements\: of\:R\}}$. But {nonzero elements of $R$} is the complement of $(x^2+y^2-1)$ which is the prime ideal of $\mathbb{R}[x,y]$. So letting $p=(x^2+y^2-1)$, $Q(R)=(\mathbb{R}[x,y]/(x^2+y^2-1))_p=R_p.$ Let $\overline{R}$ be the integral closure of $R$ in $Q(R)$. Then since $(\overline{R})_p$ is the integral closure of $R_p(=Q(R))$ in $Q(R)_p(=Q(R))$, $\overline{R}_p=R_p$. And for all other primes $p'$ in $\mathbb{R}[x,y]$, $p'^c\cap p\neq\emptyset$. So $R_{p'}=0={\overline{R}}_{p'}$, which concludes $R=\overline{R}$.

edited Aug 17 '19 at 21:02

asked Aug 17 '19 at 15:35

Compare with this post, and Vicenzo's answer. – Dietrich Burde Aug 17 '19 at 15:57
I think this answer provides an alternative approach. – Viktor Vaughn Aug 17 '19 at 17:32
TBH I don't understand a word from what you are doing here, starting with the description of Q(R). – user26857 Aug 17 '19 at 20:42
@user26857 by $Q(R)$, I mean the field of fractions of an integral domain. – Aug 17 '19 at 21:04
I have no doubt about it! – user26857 Aug 18 '19 at 10:22
1

Reopened. Reason: the OP has asked for a proof verification. – user26857 Aug 19 '19 at 06:52
$Q(R)=R_p$ is wrong. – user26857 Aug 19 '19 at 15:35

$\mathbb{R}[x,y]/(x^2+y^2-1)$ is integrally closed.

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