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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!

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    $\begingroup$ There isn't a transformation on the projective plane that carries a line to a conic. $\endgroup$
    – John Brevik
    Feb 23, 2016 at 2:39
  • $\begingroup$ @JohnBrevik: Thank you for your reply! But what if $\phi$ is just an isomorphism of varieties? $\endgroup$
    – KittyL
    Feb 23, 2016 at 9:53
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    $\begingroup$ 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 $x^2-t^4u^4$. $\endgroup$
    – John Brevik
    Feb 23, 2016 at 22:14
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    $\begingroup$ How is $\alpha$ defined? $\endgroup$
    – SlavaM
    Feb 23, 2016 at 22:22
  • $\begingroup$ @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! $\endgroup$
    – KittyL
    Feb 24, 2016 at 10:41

1 Answer 1

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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.

(Notice that this proof does not work for general isomorphic projective varieties.)

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