This question is asked in MSE but no effective answer appeared.
Suppose $C$ is a irreducible closed curve in $\mathbb P^3$(projective space over an algebraically closed field), I need to prove there are three homogeneous functions $f_1,f_2,f_3$ such that $C=V_+(f_1,f_2,f_3)$. How to get it?
EDIT: this answer has some issues. See the comments. I think a similar line of reasoning shows you can do it with four equations, though.
For reasons of dimension your curve is contained in a hypersurface defined by a single equation $f_1 = 0$. Again for dimension reasons there must be another independent equation $f_2 = 0$ which holds on your curve. Now $f_1 = 0, f_2 = 0$ defines a one-dimensional algebraic set $X$ which may have several irreducible components, one of which is your curve. Each irreducible component $X_i$ corresponds to a distinct homogeneous prime ideal $\mathfrak{p}_i$ of codimension 1 in the coordinate ring of projective space. Without loss of generality take $X_1$ to be your curve. It is a general fact of commutative algebra that if an ideal $\mathfrak{q}$ of a ring is contained in the union $\bigcup_j \mathfrak{q}_j$ of some finite set of prime ideals, then there exists $k$ with $\mathfrak{q} \subseteq \mathfrak{q}_k$. Since all the $\mathfrak{p}_i$ are distinct prime ideals of the same height, no $\mathfrak{p}_i$ contains $\mathfrak{p}_1$ other than $\mathfrak{p}_1$ itself. Thus by the commutative algebra lemma there exists $f_3 \in \mathfrak{p}_1 - \bigcup_{i \neq 1} \mathfrak{p}_i$. By construction $f_3$ only vanishes on $X_1$ and no other $X_i$, so $f_1, f_2, f_3$ have vanishing locus equal to $X_1$, as desired.
