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A problem on an algebra qual reads

Show that the ring $R = \mathbb{C}[x,y]/(y^2 - x^3 +1)$ is a Dedekind domain. (Hint: compare $R$ with the subring $\mathbb{C}[x]$.)

$R$ is clearly Noetherian. It is an integral extension of $\mathbb{C}[x]$, so inherits its dimension, which is one. How do I know it is normal?

Eric Auld
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    It's normal because the curve $y^2=x^3-1$ is smooth. But somehow that sounds like cheating. It shouldn't be too hard to show that $R$ is the integral closure of $\Bbb{C}[x]$ in the field of fractions $\Bbb{C}(x,y)$ of $R$, though. – Jyrki Lahtonen Aug 27 '17 at 20:54

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Say $A$ is an UFD containing $\frac{1}{2}$ and $d\in A$ is a square-free element, not zero, not a unit. Then the ring $B=A[y]/(y^2 - d)$ is a normal domain. Indeed, the polynomial $y^2-d$ is irreducible. The field of fractions of $B$, denoted by $L$, is $K[y]/(y^2-d)$. Consider an element $l =k_1 + k_2 \sqrt{d} \in L$. Assume $l$ is an integer. Then the trace and the norm are in $A$, that is $2k_1$ and $k_1^2 - d k_2^2 \in A$, and this implies right away $k_1$, $k_2 \in A$.

orangeskid
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    After coming across this answer again and having to run through this argument twice now: since $A$ is a UFD, we can write $k_2 = a/b$ in lowest terms, i.e., $a$ and $b$ have no prime factor in common. Suppose $p$ is a prime factor of $b$, so $p^2 \mid b^2$. Writing $d k_2^2 = c \in A$, then $p^2 \mid c b^2 = d a^2$. Since $a$ and $b$ have no common prime factor, then $p^2 \mid d$. But $d$ is assumed squarefree, contradiction. Thus $b$ has no prime factors, hence must be a unit in $A$, so $k_2 = a/b \in A$. +1, though; actually +1 from 2017. :) – Viktor Vaughn Dec 28 '18 at 02:58