I have two surfaces $A,B \subset \mathbb{P}^3$ generated by two homogenous polynomials $f, g$. I want to know when their intersection $A \cap B$ is a complete intersection. $V \subseteq \mathbb{P}^n$ with $\dim V = r$ is a complete intersection, if its ideal $I$ has $n - r$ generators.
In my case as $A = Z(f)$ and $B = Z(g)$, we also get $ C: = A \cap B = Z(f) \cap Z(g) = Z(f, g)$. That means if $I(Z(f, g)) = \sqrt{\langle f, g \rangle} = \langle F, G \rangle$ for some homogenous $F, G$ and $\dim C = 1$ (meaning $C$ is a curve), then $C$ is a complete intersection, right? Wouldn't that imply that if $C$ happen to be irreducible, then $C$ is automatically a complete intersection?
But I found in Hartshorne (I, Ex. 2.17) that a twisted cubic is not a complete intersection, but a set theoretic intersection, meaning that twisted cubic is exactly $Z(f) \cap Z(g)$. Where does my reasoning fail exactly? Maybe $ Z(f) \cap Z(g) = Z(f, g)$ is suspicious, but I don't see how.