Working out the normalization of $\mathbb C[X,Y]/(X^2-Y^3)$

I'm trying to identify the normalization of the ring $A := \mathbb C[X,Y]/\langle X^2-Y^3 \rangle$ with something more concrete.

First, $X^2-Y^3$ is irreducible in $\mathbb C[X,Y]$, making $\langle X^2-Y^3\rangle$ prime, so $A$ is a domain and it makes sense to talk about its normalisation, i.e., its integral closure in $\mathrm{Frac}(A)$. Then, we try to understand $\mathrm{Frac}(A)$: the composite arrow $$ \mathbb C[X,Y] \twoheadrightarrow A \hookrightarrow \mathrm{Frac}(A) $$ maps every element not in $\langle X^2-Y^3 \rangle$ to an invertible one in $\mathrm{Frac}(A)$. By the universal property of the localization, it defines an arrow $h : \mathbb C[X,Y]_{\langle X^2-Y^3\rangle} \to \mathrm{Frac}(A)$ making the following diagram commute : $$ \begin{matrix} \mathbb C[X,Y] & \twoheadrightarrow & A & \hookrightarrow & \mathrm{Frac}(A) \\ \downarrow &&&& \| \\ \mathbb C[X,Y]_{\langle X^2-Y^3\rangle} & &\stackrel h \longrightarrow & & \mathrm{Frac}(A). \end{matrix}$$

The arrow $h$ is onto: for $P,Q \in \mathbb C[X,Y], Q \notin \langle X^2-Y^3 \rangle$, $h(P/Q) = \pi(P) // \pi(Q)$ (denoting '/' the fraction of the left localization, '//' the one in $\mathrm{Frac}(A)$, and $\pi \colon \mathbb C[X,Y] \twoheadrightarrow A$). Then, we have a description of the fraction field of $A$ as $$\mathrm{Frac}(A) \simeq \mathbb C[X,Y]_{\langle X^2-Y^3\rangle} \,\big/\, \ker (h) \simeq \{P/Q \in \mathbb C(X,Y) \mid Q \notin \langle X^2-Y^3\rangle\} \,\big/\, \langle X^2 - Y^3 \rangle.$$

Am I correct so far ? If so, I'm having trouble to determine algebraic integers in this $A$-algebra. Any hint ?

Solution 1:

Hints:

• Show that $t=X/Y$ is integral over $A=\Bbb{C}[X,Y]/(X^2-Y^3)$.
• Show that $A[t]=\Bbb{C}[t]$ is integrally closed.

The general facts related to this:

• integral closure of the coordinate ring of an algebraic curve is broken only at singular points (here the cusp at the origin), so the coordinate ring of a non-singular curve is integrally closed.
• in the case of a curve, the integral closure of the coordinate ring (inside the function field) is the intersection of the DVRs containing $R$ - this is why integral closure can be studied locally

There is more to be said about the interplay between integral closure and local behavior of varieties, but I don't know/remember more. Waiting for somebody else to take over...