Complete intersection in $\mathbb{P}^{n+2}$ as K3 surface

algebraic-geometrybirational-geometrycomplex-geometryk3-surfaces

I am reading D. Huybrechts' lecture "Lectures on K3 surfaces", http://www.math.uni-bonn.de/people/huybrech/K3Global.pdf
He claims that a smooth complete intersection of type $(d_1,\cdots, d_n)$ in $\mathbb P^{n+2}$ is a K3 surface if and only if $\sum {d_i} = n + 3$.

What confuses me is that he then says under the natural assumption that all $d_{i}>1$ there are in fact only three cases (up to permutation):

(a) $n=1, d_1=4$.

(b)$n=2, d_1=2, d_2=3$.

(c)$n=3, d_1=d_2=d_3=2.$

I wonder why we asssme $d_i>1$? What will happen if some $d_i=1$?

Any help would be appreciated. Thanks a lot!

Best Answer

If, say, $d_1 = 1$ this complete intersection is equal to a complete intersection in $\mathbb{P}^{n+1}$ of type $(d_2,\dots,d_n)$, so there is no sense in considering it separately.

Related Question