# Why are projective coordinate rings not isomorphic when the corresponding projective varieties are?

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!

• There isn't a transformation on the projective plane that carries a line to a conic. Feb 23, 2016 at 2:39
• @JohnBrevik: Thank you for your reply! But what if $\phi$ is just an isomorphism of varieties? Feb 23, 2016 at 9:53
• It's very subtle and interesting, isn't it? I can't give you a good complete answer here, but I'll give you a great example to think about: Look at $k[t^4, t^3u, tu^3, u^4]$, the projective coordinate ring of the line embedded in $\mathbb P^3$ by those degree-$4$ forms. The coordinate ring is missing" $t^2u^2$, and in fact it misses that form so much that it isn't normal (integrally closed), as $t^2u^2$ satisfies the integral equation $x^2-t^4u^4$. Feb 23, 2016 at 22:14
• How is $\alpha$ defined? Feb 23, 2016 at 22:22
• @Heinrich: Is $\alpha$ not well-defined? It has to be rational with homogeneous degree polynomials. So the proof does not even work for linear change of coordinates then. Is that right? Thank you! Feb 24, 2016 at 10:41

Thanks to @Heinrich's comments, I realized that we cannot define the regular map $\alpha$ as I stated in my proof. I think the following proof works.
An isomorphism of the $\mathbb{C}$-algebras would correspond to an isomorphism between the affine cones over the two projective varieties. 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. Suppose $V=V(f_1,…,f_r)$. Then $W=V(f_1\circ \psi,…,f_r\circ \psi)$. As affine varieties they are apparently isomorphic.