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I have an homogeneous quartic in $(u,v,w)\:$: $Q(u,v,w)=u^4+au^2w^2+bw^4-v^2$

For $\lambda$ in the base field $K$, I wish to calculate: $\lambda^4Q(u,v,w)=Q(u',v',w')$.

I 'm not sure but I find $u'=\lambda u$, $v'=\lambda^2v$, $w'=\lambda w$.

Do you think it's correct ?

I thank you in advance...

beginarray
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  • I'm not really sure what the purpose of the exercise is here, but yes your $u',v',w'$ fulfill that $Q(u',v',w') = \lambda^4Q(u,v,w)$. – 0CT0 Aug 29 '19 at 10:53
  • Bonjour @Dario Cavallaro. The purpose of the exercice is the following: I have to establish an isomorphism between the above quartic and a cubic of the form $E: y^2=x^3+a'x^2+b'x$. In order to do that, I need to homogenize bi-rational transformations formulas between the two projective curves. – beginarray Aug 29 '19 at 11:24

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