I am trying to find the dimension and the necessary and sufficient conditions under which $A[X,Y]/(Y^2 - aX)$ is regular, that is, the localizations of $A[X,Y]/(Y^2 - aX)$ at all maximal ideals are regular, where $A$ are the Gaussian integers and $a \in A$. Any hints?
I can't even see what the max ideals are. Thanks.
Let $M$ be a maximal ideal of $A[X,Y]/(Y^2-aX)$. Then $M=\mathfrak m/(Y^2-aX)$, where $\mathfrak m$ is a maximal ideal in $A[X,Y]$ containing $Y^2-aX$. We have $$\left(\dfrac{A[X,Y]}{(Y^2-aX)}\right)_M=\dfrac{A[X,Y]_{\mathfrak m}}{(Y^2-aX)}.$$ (I) If $a=0$, then $Y\in \mathfrak m$, so $Y^2\in \mathfrak m^2$, hence $A[X,Y]_{\mathfrak m}/(Y^2)$ isn't regular.
(II) If $a\ne 0$ and there is $p\in A$ irreducible such that $p\mid a$, then consider $\mathfrak m=(p,X,Y)$, and note that $p\in \mathfrak m-\mathfrak m^2$. If $A[X,Y]_{\mathfrak m}/(Y^2-aX)$ is regular, then its factor ring by the ideal $(p)$ is also regular, that is, $(A/pA)[X,Y]_{\mathfrak m}/(Y^2-\bar aX)$ is regular. But $\bar a=\bar 0$ in $A/pA$ and we are in case (I).
(III) Suppose $a\ne0$ and there is no irreducible $p\in A$ such that $p\mid a$, that is, $a$ is invertible. In this case $A[X,Y]/(Y^2-aX)\simeq A[Y]$ is regular.
Conclusion: $A[X,Y]/(Y^2-aX)$ is regular if and only if $a$ is invertible.
Remark. The proof works for $A$ a PID.
