I don't remember where but I've read that the Picard number of a complete intersection of dimension at least 3 is $1$. Why?
In Dimension $2$ case the Picard number of $P^1\times P^1$ is $2$, by the Segre Embedding it is a hypersurface in $P^3$. But anything else indicated that the picard number of a complete intersection of dimension at least 3 is $1$ ? Thanks!
The Picard number of projective space is $1$. Moreover, there's a version of the Lefschetz hyperplane theorem for Picard groups: if $X$ is a smooth projective variety of dimension $\geq 4$ over a field of characteristic zero and $D \subset X$ is an ample effective divisor, then the restriction map $\operatorname{Pic}(X) \to \operatorname{Pic}(D)$ is an isomorphism. (For reference, see example 3.1.25 in Lazarsfeld's Positivity in Algebraic Geometry I.)
So, for any smooth complete intersection of dimension $\geq 3$ over a field of characteristic zero, you can apply the above argument inductively to conclude that the Picard group is isomorphic to that of the ambient projective space, and thus is isomorphic to $\mathbb{Z}$. I'm not sure what happens in the case of singular complete intersections (actually, I suppose you can handle some singularities with the above inductive method, but all the singularities have to come "at the last step", i.e., it has to be given by an ample divisor in a 4-dimensional smooth complete intersection) or over a field of positive characteristic.