If $S \subset \mathbb{P}^n$ is a closed set (in the Zariski sense) then $\mathcal{I}(S) \subset k[x_0,\ldots,x_n]$ denotes the homogeneous ideal of polynomials which vanish at $S$.

I want to find an example of two projective subvarieties $X \subseteq \mathbb{P}^n$ and $Y \subseteq \mathbb{P}^m$ such that $X \cong Y$ but $k[x_0,\ldots,x_n] / \mathcal{I}(X) \not\cong k[x_0,\ldots,x_m] / \mathcal{I}(Y)$. Here, a subvariety is irreducible, so $\mathcal{I}(Y),\mathcal{I}(Z)$ are necessarily prime (I think?)

I'm not looking for straight answers, but a hint is very welcome.

  • 2
    $\begingroup$ Take for $X$ and $Y$ a line and an irreducible conic in $\mathbb P^2$. $\endgroup$ – Georges Elencwajg Jun 4 '15 at 22:15
  • $\begingroup$ Thank you, that was exactly the hint I needed! $\endgroup$ – rwols Jun 5 '15 at 18:52

To turn Georges comment into an answer, the precise choices of homogeneous ideals in $S=k[x,y,z]$ could be $I = (xz-y^2)$ and $J=(x)$. Clearly, the two quotients are not the same because $S/I$ has dimension $3$ in degree $1$ and $S/J$ does not. However, you should be able to prove $\mathbb P^1\cong C:=\mathcal V(I)$ by mapping \begin{align*} [u:v] &\longmapsto [u^2:uv:v^2]. \end{align*} Note that on $u=1$, we have $[1:v]\mapsto [1:v:v^2]$, which is indeed injective. This is not a full proof to show that the two are isomorphic, but you should be able to go from here and glue functions on affine patches in both directions to get isomorphisms.

  • $\begingroup$ Yes, this is indeed the example I worked out, thank you very much! $\endgroup$ – rwols Jun 5 '15 at 18:52

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