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 ?



  • 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...

  • 1
    $\begingroup$ So blowig up the cusp here removes the singularity, and also gives the integral closure. $\endgroup$ – Jyrki Lahtonen Sep 21 '13 at 15:21
  • 1
    $\begingroup$ Thanks. So first hint is obvious : $T^2-Y$ admits $t$ as root. As $X=t^3$ and $Y=t^2$ in $\mathrm{Frac}(A)$, it gives $A[t] = \mathbb C[t]$. Now, $t$ is not algebraic over $\mathbb C$, which make $\mathbb C[t]$ isomorphic to the $\mathbb C$-algebra of polynomial in one variable $\mathbb C[T]$. As $\mathbb C$ is factorial, so is $\mathbb C[T]$. Factorial implies integrally closed, which concludes (integers over $A$ are integers over $A[t]$, so in $A[t]$ ; conversely, $A[t]$ is integral over $A$). Is that what you had in mind with those 2 hints ? $\endgroup$ – Pece Sep 21 '13 at 15:43
  • $\begingroup$ Correct, Pece. You may want to wait for an algebraic geometer to show up, and say more :-) $\endgroup$ – Jyrki Lahtonen Sep 21 '13 at 16:36
  • 1
    $\begingroup$ @JyrkiLahtonen The normalization, at least for one-dimensional varieties over alg. closed fields, corresponds merely to resolution of singularities. Indeed, in dimension 1, the obstruction to a variety being non-singular is normality. Namely, it's not hard to show that a curve is non-singular if and only if its coordinate ring is a Dedekind domain, which since the coordinate ring are already one-dimensional and Noetherian, is equivalent to integrally closed. Of course, integral closedness is local, and the basic problem is that at the point $(x-a,y-b)$ where the curve has a singularity $\endgroup$ – Alex Youcis Sep 22 '13 at 8:14
  • 1
    $\begingroup$ @JyrkiLahtonen The classic example is the cone $z^2=xy$ in $\mathbb{A}^3$. The basic thing is that, for an affine variety, being non-singular is the same as having your coordinate ring be regular (this just means your localizations at maximals are all regular local). In dimension $1$, being regular is equivalent to normal, but for higher dimensions this fails to be true. So, if you'd rather think in algebra land, find a ring $R$ which is normal, but is not regular and take its spectrum. Of course, as we've said, we will necessarily have that $\dim R>1$. Does that help at all? $\endgroup$ – Alex Youcis Sep 23 '13 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.