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I would like to compute the normalization of the ring $A=\mathbb C[x,y,t]/(t^3-x^3y)$, but I do not know how to proceed.

I am not an expert in normalizations, and the only examples I saw were from one-dimensional rings (normalization of curves). I would really appreciate if someone could explain how one can attack such a problem, and guide me to a solution.

Just to let you know why I find this problem difficult, I will include roughly my attempts.

So, I look at the equation $t^3=x^3y$ of the surface and I see that what essentially is missing in $A$ is a third root of $y$. This suggests to make a "change of coordinates $\mathbb C[x,y,t]\to \mathbb C[u,v,z]$, where now we send $y\mapsto u^3$, say. In the (still ghost) normalization this third root will have the form $(\textrm{image of }t)/(\textrm{image of }x)$, so if I send $x\mapsto v$, I should then send $t\mapsto uv$. Well, but I completely ignored $z$ and this sounds like a problem doesn't it? Also, the function $t^3-x^3y$ gets sent to $(uv)^3-v^3u^3=0$, which is silly because what I was planning in order to get $\tilde A$ was to divide out $\mathbb C[u,v,z]$ by the image of $t^3-x^3y$. Probably none of this makes sense, but at least you can see where my difficulties are.

Thank you!

Community
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Jack
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  • Why do you need the $z$? Your ring has dimension 2, so that you only need two variables. Also, the fact that the image of $t^3-x^3 y$ is zero is a good thing. It means that you have a homomorphism from your ring to $\mathbb{C}[u,v]$. – Ofir Mar 18 '14 at 14:14
  • I wanted the $z$ in order to get the normalization as a quotient of a polynomial ring in $3$ variables, which would give me its affine equation. Maybe it is wrong, I do not know the best procedure. – Jack Mar 18 '14 at 14:38
  • Maybe you can find some inspiration here: http://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings – user26857 Mar 18 '14 at 17:30

1 Answers1

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The element $q=\frac{t}{x}$ is the most obvious element in the integral closure of $A$ which is not contained in $A$. Notice that $qx=t$ and hence $t^3=x^3 y$ becomes $q^3 = y$ (using that $x^3 \neq 0$). It follows that $A[q]=\mathbb{C}[x,q]$. Try to show that $x$ and $q$ are algebraically independent here, so that $A[q]$ is really a polynomial ring and hence normal. But then it must be the normalization.

The corresponding morphism of varieties is $\mathbb{A}^2_{\mathbb{C}} \to \mathrm{Spec}(A)$, $(a,b) \mapsto (a,b^3,ab)$.

The real picture is as follows:

enter image description here

Notice that the singularity at the origin is blown away by the normalization.

Martin Brandenburg
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