I was trying to prove the following question from An Invitation to Algebraic Geometry by Karen Smith:
Show that the homogeneous coordinate rings of projectively equivalent varieties are isomorphic.
Here is my attempt:
Let $V,W$ be projective varieties in $\mathbb{P}^n$. Let $\phi:V\rightarrow W$ be the linear change of coordinates map and $\psi:W\rightarrow V$ be its inverse map.
Define a map $\alpha:\mathbb{C}[V]→\mathbb{C}[W]$ by $\alpha(f)=f\circ\psi$. Let $f\circ\psi=0$ on $\mathbb{C}[W]$. Since $\psi$ is a bijection between $W$ and $V$, $f$ must be zero on $\mathbb{C}[V]$. This shows that $\alpha$ is injective.
To show that it is surjective, we see that for any $g\in\mathbb{C}[W]$, $f=g\circ \phi$ is sent to $g\circ \phi\circ\psi=g$ by $\alpha$.
Now we show that $\alpha$ is a ring homomorphism. $$\alpha(f+g)=(f+g)\circ\psi=f\circ\psi+g\circ\psi=\alpha(f)+\alpha(g)\\ \alpha(fg)=(fg)\circ\psi=(f\circ\psi)(g\circ\psi)=\alpha(f)\alpha(g)$$
My question is, why does this proof not work for any isomorphic varieties? I understand that homogeneous coordinate rings are not necessarily isomorphic even if the underlying varieties are isomorphic. The book provides an example:
$$\mathbb{C}[x,y,x]/(xz-y^2) \text{ and } \mathbb{C}[s,t]$$
The two varieties are isomorphic but the coordinate rings are not, since as affine cones one of them has a singularity at the origin, whereas the other is nonsingular. This example is very clear to me, but I still don't understand why the above proof does not work for this case. I examined each step with isomorphism instead of linear map, but couldn't see where it breaks down. Could anyone explain?
Thank you for your help!