The twisted cubic, $C$, is given by the image of the map $v_3: \mathbb{P}^1 \rightarrow \mathbb{P}^3$ defined as $$[x_0: x_1] \mapsto [x_0^3: x_0^2 x_1: x_0 x_1^2: x_1^3]$$
We can also see that $C$ lies in the image of the segre map $s: \mathbb{P}^1 \times \mathbb{P}^1\rightarrow \mathbb{P}^3$, given by
$$([x_0: x_1], [y_0: y_1]) \mapsto [x_0y_0: x_0y_1: x_1y_0: x_1y_1]$$
Thus, instead of the usual homogenous polynomials (quadratic or not) defining $C$, we should be able to find a set of bihomogenous polynomials that define $C$, using the segre map.
The homework exercise is to actually find a single bihomogenous polynomial that does this. I see that we can achieve this with $x_0 = y_0^2$ and $x_1 = y_1^2$, so my natural guess is to try $x_0y_1^2 – x_1y_0^2$. Applying this gives \begin{equation*}\begin{split}[x_0y_0: x_0y_1: x_1y_0: x_1y_1] &= [x_0y_0y_1^2: x_0y_1^3: x_1y_0y_1^2: x_1y_1^3] \\&= [x_1y_0^3: x_1y_0^2y_1: x_1y_0y_1^2: x_1y_1^3]\\ &= [y_0^3: y_0^2y_1: y_0y_1^2: y_1^3]\end{split}\end{equation*}
Which is exactly what we want. However, my concern is when multiplying by $y_1^2$, we could have $y_1 = 0$, and similarly when dividing out by $x_1$. Are there different polynomials that work, or is there some open cover type justification behind this?
Best Answer
I think I got it. The first thing that we want to check is if $y_1 =0$, then the starting point is $[1:0:0:0]$, which is the same as the point we would like to see. Because $y_1 = 0$, we must have $y_0 \neq 0$, and so the polynomial condition gives us that $x_1 = 0$, and so $x_0 \neq 0$. This gives us the point $[x_0y_0: 0: 0: 0] = [1: 0: 0: 0]$, as we wanted.
Next, we want to check that if $y_1 \neq 0$, then $x_1 \neq 0$, which is easier via contrapositive. If $x_1 = 0$, then $x_0 \neq 0$, and our polynomial condition tells us that $y_1 = 0$.
Thus this map is well defined on the whole space.