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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 ?

user26857
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Pece
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1 Answers1

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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...

Jyrki Lahtonen
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    So blowig up the cusp here removes the singularity, and also gives the integral closure. – Jyrki Lahtonen Sep 21 '13 at 15:21
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    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 ? – Pece Sep 21 '13 at 15:43
  • Correct, Pece. You may want to wait for an algebraic geometer to show up, and say more :-) – Jyrki Lahtonen Sep 21 '13 at 16:36
  • @JyrkiLahtonen Just out of curiosity, what would you want an algebraic geometer to say? Geometrically, the normalization of the cuspidal cubic (CC) is the parabola, which has coordinate ring $k[t]$. So, the normalization of the coordinate ring of the CC is isomorphic to $k[t]$. More explicitly, the normalization gives rise to the reverse maps of coordinate rings which finds that $X\mapsto t^3$ and $Y\mapsto t^2$. So, we see that the element of the coordinate ring of the CC that is missing is $t=\frac{X}{Y}$. – Alex Youcis Sep 22 '13 at 08:07
  • @Alex: Thanks for asking. I have this recollection that normalization is equivalent to finding a smooth model for curves, but with higher dimensional varieties there is more going on. May be answering that would take us too far afield? I guess I could ask that question myself, but the recollection is too vague to turn it into a useful question. – Jyrki Lahtonen Sep 22 '13 at 08:11
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    @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 – Alex Youcis Sep 22 '13 at 08:14
  • @JyrkiLahtonen you can show that its coordinate rings localization, say $A(C)_{(x-a,y-b)}$, does not have principally generated maximal ideal, and so not a DVR (which, in our case, is equivalent to integrally closed since A(C) is already dimension 1 and Noetherian). Does that at all answer anytyhing for you? – Alex Youcis Sep 22 '13 at 08:15
  • Sure, thanks @Alex. If there is an easy to describe 2-dimensional counterexample to integral closure being equivalent to non-singular, that would be a nice completion, but "may take us too far afield" :-) – Jyrki Lahtonen Sep 22 '13 at 08:17
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    @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? – Alex Youcis Sep 23 '13 at 05:38
  • Of course, I meant localization at every prime is regular local :) – Alex Youcis Sep 23 '13 at 05:45
  • @JyrkiLahtonen: But we also need to check whether there is a map $t \mapsto (f(t),g(t))=(x,y)$, or in the above case: $t\mapsto (t^3,t^2)$ ? What if we can't find such a map? – Dan Feb 05 '15 at 15:58
  • @Alex Thanks for your insights. Do you also have geometric intuition for the element that needs to be added to the ring to normalize it? In this example, we needed to add $t = X/Y$ to normalize the ring. How does this relate geometrically to the cusp on the curve? Is there some inuition that would lead us to consider this element instead of, say $Y/X$, or any other element (without checking algebraically)? – Anakhand Jan 27 '24 at 11:34
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    @Anakhand AlexYoucis knows this stuff better than I do, but I think a key point is that $t=X/Y$ is a local parameter for the relevant place of the function field. Or, the key point in a smooth model of the function field. Wait for a more detailed explanation from him. – Jyrki Lahtonen Jan 27 '24 at 12:15