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I have a question about the computation of global regular function of $\mathbb{P}^2-V(x_0^2+x_1^2+x_2^2)$. I konw this is isomorphic to an affine variety, and i tried to compute the ring of global regular functions by Veronese map, but i'm not sure it is right. By Veronoese map $\mathbb{P}^2$ is mapped in $\mathbb{P}^5$ by $$[x_0,x_1,x_2] \rightarrow [x_0^2,x_1^2,x_2^2,x_0x_1,x_0x_2,x_1x_2].$$ In particular the conic is (say $[t_0,...,t_5]$ coordinates of $\mathbb{P}^5$) given by $t_0+t_1+t_2=0$ (intersect with the image of $\mathbb{P}^2$).

Now $\mathbb{P}^2-V(x_0^2+x_1^2+x_2^2)=X\cap \{t_0+t_1+t_2\not=0\}$ where X is the image of $\mathbb{P}^2$ by Veronese map. Moreover changing the variables of P^5 i can suppose that $$[x_0,x_1,x_2] \rightarrow [x_0^2,x_1^2,x_2^2+x_1^2+x_0^2,x_0x_1,x_0x_2,x_1x_2]$$ so my initial variety is given by points of the form $$(\frac{x_0^2}{x_0^2+x_1^2+x_2^2},\frac{x_1^2}{x_0^2+x_1^2+x_2^2},\frac{x_0x_1}{x_0^2+x_1^2+x_2^2},\frac{x_0x_2}{x_0^2+x_1^2+x_2^2},\frac{x_1x_2}{x_0^2+x_1^2+x_2^2})$$ in $\mathbb{A}^5$. I would to understand which variety is isomophic to it (it seems to be an hypersurface in $\mathbb{A}^3$) in such way to determine easily the ring of global regular functions. Maybe there is a faster way? or a different reasoning to conclude? Thanks for the help.

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    $\begingroup$ If $f$ is a homogeneous irreducible polynomial of degree $d$ in say $n+1$ variables, so that it defines a hypersurface in $\mathbb{P}^n$, it complement is affine and the regular functions on it can be described as the ring of functions of the form $F/f^m$ where $F$ is homogeneous of degree $md$ for all $m$. $\endgroup$ – Mohan Jul 7 at 15:39
  • $\begingroup$ How can say that it is in my case by hand? Can I conclude by my reasoning? $\endgroup$ – Matvey Tizovsky Jul 8 at 5:51
  • $\begingroup$ You can use your Veronese argument and use the fact I mentioned for projective spaces and hyperplanes. $\endgroup$ – Mohan Jul 8 at 14:32
  • $\begingroup$ Is it correct to say that in $\mathbb{P}^5-V(t_0+t_1+t_2)$ the global regular function are $k[t_0,...,t_5][\frac{1}{t_0+t_1+t_2}]$ and so imposing regularity on $X $ we obtain exactly the functions of the form $F/f^m$? thanks for attention and the help. $\endgroup$ – Matvey Tizovsky Jul 9 at 15:00
  • $\begingroup$ No, the ring you write down is the ring of regular functions on affine 3-space minus $V(t_0+t_1+t_2)=f.$. To get regular functions on projective plane minus the set you write, you must only take functions of the form $F/f^m$ where $\deg F=m$. For instance, $1/f$ is not allowed. $\endgroup$ – Mohan Jul 10 at 3:02

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