Why is $k[x,y,z,u]/(x^2-y^2z, uy-x) \cong k[y,u]$?

Question

In the second paragraph of the accepted answer to this question, the answerer mentions that the normalization of the Whitney umbrella $k[x,y,z]/(x^2-y^2z)$ is $k[y,u]$, where $u = x/y$. I am trying to prove this claim, which I believe is equivalent to showing the isomorphism in the title, i.e. $k[x,y,z,u]/(x^2-y^2z, uy-x) \cong k[y,u]$.

From the second relation we get that the left-hand ring is isomorphic to $k[y,z,u]/(y^2u^2 - y^2z)$, but I'm not sure why this is isomorphic to $k[y,u]$. Factoring out the $y^2$, it seems to me that the relation is saying that $y^2(u^2 - z) = 0$, but I don't think I can "cancel out" the $y^2$ to conclude that $z = u^2$.

Is this misgiving correct? If so, it seems that I would need additional conditions on my ring (such as another relation) -- what might these be? If not, why is it allowed?

Answer 1

If you attach $u=x/y$ in the fraction field of $R=k[x,y,z]/x^2-y^2z$ to $R$, you get $R[u]=S$. Since $u=x/y$, you can write $x=uy$ and then since $R$ is a domain and so is $S$ (both have the same fraction field) you can write $S=k[y.z,u]$ with the relation $u^2-z$. Thus $S=k[u,y]$.

Answer 2

Since $R = k[x,y,z] / (x^2 - y^2z)$ is an integral domain, it's safe to talk about fractions. The element $x/y$ is integral over $R$ because $(x/y)^2 = z$, so the integral closure includes $x/y$.

Your argument shows that this is not the same as the ring $k[x,y,z,u] / (x^2 - y^2z, uy-x)$ since the latter is not a domain. Whereas adjoining elements of the fraction field to $R$ obviously leaves it a domain.

The missing relation from that presentation is just $u^2 = z$ though, and the relations $uy=x, u^2=z$ imply $x^2 = y^2z$. You can check that $k[x,y,z,u] / (uy-x, u^2-z) \cong k[u,y] \cong R[\frac{x}{y}] \subset \mathrm{Frac}(R)$.