Exercise 7.12 from Fulton's Algebraic Curves
Find a quadratic transformation of $\; F = Y^2 Z^2 − X^4 −Y^4$ with only ordinary multiple points.
By checking the partial derivatives, I found that $P=[0,0,1]$ is a singular point. Checking the multiplicity of $P$ on $F$, I get $2$. Since the tangent lines aren't distinct and are $y=0$, an exceptional line, so $F$ is not in a good position and so I have to apply a change of coordinates before using the standard Cremona transformation.
Now, I did $T : x = X+Y, y=X-Y, z=Z$ as usual, which gives $$T(F) = 4x^2z^2+4y^2z^2 - 8xyz^2 - 8x^3y -8xy^3$$
that has (obviously) still got a double non ordinary point of multiplicity, but with $y=x$ as tangent (not an exceptional line). So now I should be able to apply Cremona and remove the non ordinary points. But when I calculate $T(F) (yz,xz,xy)$ and I calculate the proper transform I get $$xy^3 + x^3 y-8x^2y^2-8y^2z^2-8x^2z^2$$ that still has a singularity at $P=[0,0,1]$ (and that's okay), but that has $x+iy$ and $x-iy$ as tangent lines there, so $P$ is now ordinary.
So I solved it while writing the question, I'm not really sure why I thought this was wrong before.
@moderators can this be useful for future reference? If not, feel free to delete it
