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