I have to prove that the surface $Z(xy^2-z^2)$ in the affine space is a rational one, where the base field is the complex one. Is it right to use the morphism $t \rightarrow (t,t,t^{3/2})$ with the projection over the first component for the inverse? Thank you.
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3Hi user52342, I think you will get better/faster/more answers if you make sure to use proper grammar and capitalization in your questions. You can also enclose your math in dollar signs to make it render nicely. Click the 'edit' button to see how. Best regards, – Potato Jun 17 '13 at 16:45
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Recall the definitions. Algebraic geometry deals with polynomials, so $t^{3/2}$ isn't well defined here (actually also complex root functions are multivalued and therefore aren't really functions).
Hint: Consider the open subvariety $Z(xy^2-z^2) \setminus Z(y)$. Show that it is isomorphic to an open subvariety of $\mathbb{A}^2$.
Martin Brandenburg
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Ok thank you. I'm thinking about the morphism that goes from the open set of $\mathbb{A}^2$ $s; \ne0$ to $(\frac{t^2}{s^2},t,s)$ with inverse the projection over last two components. Is this right ? – user52342 Jun 18 '13 at 07:55
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Yes it maps $(s,t)$ to $(t^2/s^2,s,t)$. But I think the best way for this isomorphism is to manipulate the coordinate rings. – Martin Brandenburg Jun 18 '13 at 09:33
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Sorry, I don't understand what you have said to me. What do you mean with the best way for this isomorphism? – user52342 Jun 18 '13 at 09:42
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Two affine varieties are isomorphic iff their coordinate rings are isomorphic. – Martin Brandenburg Jun 18 '13 at 13:30
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OK. If we call $U$ the open subvariety of our surface like before and we consider the morphism from $U$ to the open set $W_t$ of $\mathbb{A}^2$ (where the the second affine cordinate is different from zero) given by $(x,y,z)$ that goes to $(s^2,t,st)$, than $K[U]$ is isomorphisc to $K[W_t]$ and so the conclusion. Is this time right ? – user52342 Jun 18 '13 at 14:59