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I'm working on some computations...

The underlying field has characteristic $p$ and $q$ is a power of $p$.

Let $X$ be the affine variety defined by $\{(a,b,c,\xi_1,\xi_2) \in \mathbb{A}^5\ | (\det \begin{pmatrix} a&a^q&a^{q^2} \\ b&b^q&b^{q^2} \\ c&c^q&c^{q^2} \\ \end{pmatrix})^{q-1} = (\xi_1^{q+1} \xi_2)^{(q-1)(q^2+q+1)} \}.$

Moreover I can define an action of $\mathbb{G}_m^2$ on it through : $(z_1,z_2)*(a,b,c,\xi_1,\xi_2) = (z_1^{q+1} z_2\ a,z_1^{q+1} z_2\ b,z_1^{q+1} z_2\ c, z_1\ \xi_1, z_2\ \xi_2).$

Then I define the variety $Y$ by $\{[a:b:c:t] \in \mathbb{P}^3\ | (\det \begin{pmatrix} a&a^q&a^{q^2} \\ b&b^q&b^{q^2} \\ c&c^q&c^{q^2} \\ \end{pmatrix})^{q-1} = t^{(q-1)(q^2+q+1)} \}.$

I would like to compare the varieties $Y$ and $X/\mathbb{G}_m^2$ (this quotient variety exists), especially I would like to know if those varieties are isomorphic.

I've already shown that I have an obvious morphism $\varphi : X \to Y$, $(a,b,c,\xi_1,\xi_2) \mapsto [a:b:c:\xi_1^{q+1}\xi_2]$ and that it factors through $X/\mathbb{G}_m^2$, giving a bijective morphism $\overline{\varphi} : X/\mathbb{G}_m^2 \to Y$. This is not enough though... I tried using the universal property of the quotient variety $X/\mathbb{G}_m^2$ but this is not working very well.

If anyone has any insight about this question I'd be happy to read it, even if you don't make the computations!

Edit : ok the bold/italic part is false, this induced morphism isn't bijective since $(a,b,c,1,0)$ and $(a,b,c,0,1)$ are sent by $\varphi$ on $[a:b:c:0]$ but aren't in the same $\mathbb{G}_m^2$-orbit.

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What means the quotient $X/G$ ? –  user18119 Mar 28 '12 at 20:55
    
Oops... I meant $X/\mathbb{G}_m^2$... I'll edit it. Anyway, I just realized my morphism $\overline{\varphi}$ isn't even bijective... so it's not even relevant trying to prove it is an isomorphism. However, I'd like to see at least, if those varieties are isomorphic or not. –  ng_th Mar 29 '12 at 8:21

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