Let $\mathbb A^n_k$ be affine $n$-space and let $\mathbb P^n_k$ be the projective $n$-space.
An affine curve is a set of the form $$C_f(k):=\{(a_1, \ldots, a_n)\in\mathbb A^n_k: f(a_1, \ldots, a_n)=0\}$$ where $f\in k[x_1, \ldots, x_n]$ is non-constant.
A projective curve is a set of the form $$P_f(k):=\{(a_1:....: a_{n+1}): f(a_1, \ldots, a_{n+1})=0\}$$ where $f\in k[x_1, \ldots, x_{n+1}]$ is non-constant.
Every $f\in k[x_1, \ldots, x_n]$ produces an homogeneous polynomial ${}^hf\in k[x_1, \ldots, x_{n+1}]$ via $${}^h f(x_1, \ldots, x_{n+1}):=x_{n+1}^{\textrm{deg}(f)}\cdot f(x_1/x_{n+1}, \ldots, x_n/x_{n+1}).$$ Conversely, every homogeneous $f\in k[x_1, \ldots, x_{n+1}]$ induces a polynomial ${}^d f\in k[x_1, \ldots, x_n]$ via $${}^d f(x_1, \ldots, x_n):=f(x_1, \ldots, x_n, 1).$$
There are two natural questions:
1) Given $f\in k[x_1, \ldots, x_{n}]$, how does $C_f(k)$ relate to $P_{{}^h f}(k)$?
2) Given $f\in k[x_1, \ldots, x_{n+1}]$ homogeneous, how does $P_f(k)$ relate to $C_{{}^d f}(k)$?
I presume it works as follows:
There is an inclusion $$\imath: \mathbb A^n_k\hookrightarrow \mathbb P^n_{k}$$ given by $$\imath(a_1, \ldots, a_n):=(a_1: \ldots: a_n: 1).$$
Then:
Answer to 1) $\imath(C_f(k))=\imath(\mathbb A^n_k)\cap P_{{}^h f}(k)$. In particular, there is an identification $$C_f(k)\simeq \imath(\mathbb A^n_k)\cap P_{{}^h f}(k).$$
Answer to 2) I conjecture that $$P_f(k)\simeq C_{{}^d f}(k)\sqcup \{\mathcal{O}\}$$ where $\mathcal{O}\not\in C_{{}^d f}(k)$.
Are my answers to 1 and 2 correct?
Maybe my answer to 2) doesn't work always. I guess there must be a condition we must impose on $z_{n+1}$. In that case, what would it be the correct answer to 2)?
Thanks.
Subtle issues arise when you do this kind of homogenizing and dehomogenizing. It's not so straightforward. In my opinion the best introductory explanation of all this can be found in the brilliant book Ideals, Varieties, and Algorithms. Go through chapter $8$ section $2$.