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I am asked to prove that $\mathbb{V}(y-x^2)$ and $\mathbb{V}(y-x^3)$ are isomorphic, but I cannot find an inversible morphism from $\mathbb{V}(y-x^2)$ to $\mathbb{V}(y-x^3)$.

In order to make the morphism inversible, I think we can only consider linear map such as $$ (x,y)\mapsto(ax+by,cx+dy) $$

But this method doesn't work.

Can anyone help?

hxhxhx88
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    Hint: Both are isomorphic to $\mathbb{A}^1$. – Martin Brandenburg Oct 15 '13 at 20:53
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    Another intuitive reason, is that the coordinate ring of the first is $k[x,x^2]$ and the second is $k[x,x^3]$, both of which are $k[x]$, which explains Martin Brandenburg's answer. More generally the "graph" of a any polynomial function $f(x)\in k[x]$ is obviously going to have to coordinate ring $k[x]$ and so be isomorphic to $\mathbb{A}^1$. This matches our intuition, since all graphs should just be isomorphic to $\mathbb{A}^1$ by projection to the $x$-axis. – Alex Youcis Oct 15 '13 at 21:58
  • @MartinBrandenburg, got it, thank you! – hxhxhx88 Oct 16 '13 at 07:46
  • @AlexYoucis, inspiring, thanks! – hxhxhx88 Oct 16 '13 at 07:48

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You can map $(x,y)$ to $(x,x^3)$ from left to right; and $(x,y)$ to $(x,x^2)$ from right to left. This may look funny, but $(x,y) \in {\mathbb V}(y - x^2)$ is equal to $(x,x^2)$ and $(x,y) \in {\mathbb V}(y - x^3)$ is equal to $(x,x^3)$, making these maps each other's inverse.

Magdiragdag
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