Let $f\in k[x_1, \ldots, x_{n+1}]$ be an homogeneous degree $p$ polynomial and let $~^d\! f\in k[x_1, \ldots, x_n]$ be defined by:
$$~^d\! f(x_1, \ldots, x_n):=f(x_1, \ldots, x_n, 1).$$
Let $P_f(k)$ be the projective curve defined by $f$ and let $C_{~^d\!f}(k)$ be the affine curve defined by $~^d\!f$. Is it true that there is an identification:
$$P_f(k)\simeq C_{~^df}(k)\sqcup C_{~^d f_p}(k)$$
where $C_{~^d f_p}(k)$ is the algebraic curve defined by the degree $p$ component of the polynomial $~^df$?
For instance, consider the degree 2 homogeneous polynomial:
$$f(x, y, z)=xy-z^2$$
Then $$~^d f(x, y)=xy-1.$$ In this case the degree 2 part of $~^d f$ is: $$~^ df_2(x, y)=xy.$$
Notice:
$$P_f(k)=\{(a: b: c): c^2=ab\},$$ whereas $$C_f(k)=\{(a, b): ab=1\}$$ and $$C_{~^df_2}(k)=\{(a, b): ab=0\}.$$
Furthermore:
$$P_f(k)=\{(a: b: c): c^2=ab\}=\{(a: b: 0): ab=0\}\sqcup \{(a: b: 1): ab=1\},$$ from which follows the identification. Is it true in general?
Thanks.
I was able to prove it, so I'll share it here:
Notice that:
$$\begin{align*} P_f(k)&=\{(a_0: \ldots: a_n): f(a_0, \ldots, a_n)=0\}\\ &=\{(a_0: \ldots: 1): f(a_0, \ldots, a_{n-1}, 1)=0\}\sqcup \{(a_0: \ldots: a_{n-1}: 0): f(a_0, \ldots, a_{n-1}, 0)=0\}\\ &=\{(a_0: \ldots: a_{n-1}: 1): {}^d f(a_0, \ldots, a_{n-1})=0\}\sqcup \{(a_0: \ldots: a_{n-1}: 0): f(a_0, \ldots, a_{n-1}, 0)=0\} \end{align*}.$$
The first part is clearly in bijection with $C_{~^d\!f}(k)$, whereas the second part is in bijection with $C_{~^d\! f_p}(k)$ where $p$ is the degree of $f$.