Proof that $\frac{\mathbb{C}[x,y]}{\langle x^2+y^2-1\rangle}$ is a Dedekind Domain I want to prove that $S=\frac{\mathbb{C}[x,y]}{\langle x^2+y^2-1\rangle}$ Is a Dedekind Domain. Hence I want to prove that
$1$. $S$ is Noetherian,
$2$. $S$ is integrally closed, and
$3$. in $S$ every non-zero prime ideal is a maximal ideal.
I was able to show that $S$ is Noetherian, because of Hilbert's Basis Theorem.
Further with few substitutions, I was also able to show that $S \cong \mathbb{C}[u,\frac{1}{u}]$ and hence it being a localisation of the principal ideal domain $\mathbb{C}[x]$. It is also a PID. Hence it is integrally closed.
Now I want to prove that every nonzero prime Ideal of $S \cong \mathbb{C}[u,\frac{1}{u}]$ is maximal. I thought I will use going up theorem, but it looks like $ \mathbb{C}[u,\frac{1}{u}]$ is not integral over $\mathbb{C}[u]$. I particular the element $\frac{1}{u}$ is not integral. Hence I cannot proceed further. Can anyone help me how to solve this problem?
 A: You already showed that $S$ is a PID. But then it already follows that every non-zero prime ideal is maximal (in some sense this is a very characteristic property of PIDs; for example a UFD where this holds is already a PID)!
To see this, take some non-zero prime ideal $\mathfrak p<S$. Since $S$ is a PID we find a prime element $p\in S\setminus\{0\}$ such that $\mathfrak p=\langle p\rangle$. Now, take some ideal $I$ such that $\mathfrak p\subseteq I\subseteq S$. We find $x\in S$ such that $I=\langle x\rangle$ again since $S$ is a PID. Now $p=xy$ for some $y\in S$. This means that $x\mid p$ and $y\mid p$. But $p$ is a prime element implying that $p\mid x$ or $p\mid y$. Hence either $y$ is a unit or $x$ is. But this means that either $\mathfrak p=I$ or $R=I$. Thus, we have shown that $\mathfrak p$ is maximal.
A: Question: "Hence I cannot proceed further. Can anyone help me how to solve this problem?"
Answer: A Noetherian integral domain $A$ of dimension one is a Dedekind domain iff every localization $A_{\mathfrak{m}}$ is a discrete valuation ring (this is Atiyah-Macdonald Theorem 9.3). In your case for $S$ - since $C:=Spec(S)$ is a regular curve, it follows $dim_k(\mathfrak{m}/\mathfrak{m}^2)=1$, hence $(A_{\mathfrak{m}},\mathfrak{m}A_{\mathfrak{m}})$ is a discrete valuation ring for any maximal ideal $\mathfrak{m}$ by Atiyah-Macdonald Prop. 9.2. Hence $S$ is a Dedekind domain. A similar result holds for any ring $A$ where $C':=Spec(A)$ is a regular curve over an algebraically colsed field $k$.
