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The homogeneous coordinate ring of a projective variety is not invariant under isomorphism.

For example I was taking let $X = \Bbb P_1$, and let $Y$ be the $2$-uple embedding of $\Bbb P_1$ in $\Bbb P_2$. Then $X\cong Y$. But show that $S(X)\ncong S(Y)$.

Now I can embed $\Bbb P^1$ in $\Bbb P_2$ s.t $[x_0,x_1] \to [x_0^2,x_0x_1,x_1^2]$. Now what I see that $V([x_0,x_1])=\{(0,0)\}$ and $V([x_0^2,x_0x_1,x_1^2])=\{(0,0,0)\}$.

So am I going in a right way? Because I am not getting how to get $X$ and $Y$ and how to arise in $S(X)\ncong S(Y)$?

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    $\begingroup$ The $2$-uple embedding of $\mathbb{P}^1$ into $\mathbb{P}^2$ is a closed embedding, so an isomorphism onto its image. But, the natural presentation of the image is $k[x,y,z]/(xz-y^2)$ . These graded rings are not isomorphic. In fact, the latter is isomorphic to $\displaystyle \bigoplus_{n\geqslant 0}R_{2n}$ where $\displaystyle R=\bigoplus_{n\geqslant 0}R_n$ is the usual presentation of $k[x,y]$ as a graded ring. $\endgroup$
    – Alex Youcis
    Feb 21, 2016 at 12:21
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    $\begingroup$ $S(X)$ and $S(Y)$ are not even isomorphic as ungraded rings. $\endgroup$
    – Georges Elencwajg
    Feb 21, 2016 at 17:01
  • $\begingroup$ @GeorgesElencwajg Hmm, also a good point! :) $\endgroup$
    – Alex Youcis
    Feb 21, 2016 at 17:16

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