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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 ?

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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...

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    $\begingroup$ So blowig up the cusp here removes the singularity, and also gives the integral closure. $\endgroup$ – Jyrki Lahtonen Sep 21 '13 at 15:21
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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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