[Problem] (Fulton's Algebraic curves Problems 7.9)
Let $C=V(x^4+y^4-xyz^2)$. Write down equations for a nonsingular curve $X$ in some $\mathbb{P}^N$ that is birationally equivalent to $C$. (Use the Segre imbedding)
[My attempt]
First, I found the multiple points (singular points) by using gradient. The only singular point is $P=[0:0:1]$.
Next, I tried to blow up this curve by considering $[x:y:z] \times [u:t] \in \mathbb{P}^2 \times \mathbb{P}^1$ $$ \begin{cases} x^4+y^4-xyz^2=0 \\xt-yu=0 \end{cases}$$
Then by multiplying $u^4$ to the first equation, we obtain $x^4u^4+y^4u^4-xyz^2u^4= x^4u^4+(xt)^4-xz^2u^3(xt)=x^2(x^2u^4+x^2t^4-z^2u^3t) =0$
Therefore this $x^2u^4+x^2t^4-z^2u^3t=0$ is the first result of blowing up of the curve.
However, at $[x:y:z] \times [u:t]=[0:0:1] \times [0:1]$ , by observing affine case ($z=1, t=1$), $x^2u^4+x^2+u^3=0$ have singularity.
I'm stuck here. I don't know how can I proceed the blowing up this curve one more time and how to use the Segre imbedding..
Thank you for your attention.
I find a tiny but nice trick in this situation.
The curve $ x^2u^4+x^2t^4-z^2u^3t=0$ have an open subset $x^2u^4+x^2t^4-z^2u^3t=0 , u \neq 0$.
These two varieties are obviously birationally equivalent since it is just an open dense subset.
Now, $$ x^2u^4+x^2t^4-z^2u^3t=0 , u \neq 0 $$ $$ \Rightarrow x^2u^2+\frac{x^2t^4}{u^2}-z^2ut=0 \Leftrightarrow x^2u^2+y^2t^2-z^2ut=0$$ (in the last part, use $xt=yu$)
Therefore, we find the blowing up $x^2u^2+y^2t^2-z^2ut=0$ which has $(2,2)$ bi-degree.
Finally, we can use Segre embedding and the answer is $T_{1,1}^2+T_{2,2}^2-T_{3,1}T_{3,2}=0$ in $\mathbb{P^5}$.