An exercise from the Harris's book I need a hint on an exercise:

Let $K$ be an algebraically closed field. Prove that any finite set $\Gamma \subset KP^n$ such that not all of its points lie on the same line can be given by polynomials of degree less than $\operatorname{card}\Gamma$.

 A: Given a set of $r+1 (r\geq 2)$ points $p_0,p_1,...,p_r \in \mathbb P^n(K) $ , I' ll show by induction on $r$ that  if the points  are not colinear the set is an intersection of hypersurfaces of degree $r$.    
Suppose $r=2$ (Initialization of induction)
 Recall our points are not aligned and suppose that $p_0$ is not on the line determined by $p_1,p_2$ .
Let $L_j \; (j=0,1,2)$ be  a linear form such that its hyperplane of zeros $V(L_j)$ does not contain $p_j$ but contains the other two $p_i$'s.
Then we can, as required,  describe our three-element set as the intersection of three  degenerate quadrics (=union of two hyperplanes):
$$ V(L_0.L_1)\cap V(L_1.L_2) \cap V(L_2.L_0)= \lbrace p_0,p_1,p_2 \rbrace        $$
Suppose $r\geq 3$.
 Choose (by induction hypothesis) a family $(F_i)_{i\in I}$ of homogeneous polynomials of degree $r-1$ whose zero sets $V(F_i)$ intersect in exactly the set $\lbrace p_1,...,p_r\rbrace $.
 Also,  choose $n$ linear forms $L_1, ...,L_n$ whose zero sets $V(L_j)$ (hyperplanes) intersect exactly in the singleton set $\lbrace  p_0 \rbrace$.
Then the homogeneous polynomials $F_i.L_j \; (i\in I,\; j=1,...,r)$ of degree $r$ have zero sets $V(F_i.G_j)$  which exactly define the required initial set:
$$\bigcap_{i\in I \; j=1,...,r} V(F_i.Lj)= \lbrace p_0,p_1,...,p_r \rbrace  $$
Remark
 The above works for all fields; algebraic closedness is irrelevant.
Why "not colinear" ?
If our set of $r+1$ points were included in a line, any hypersurface of degree $r$ containing the points  would contain the line: this is essentially the elementary fact that a  polynomial  of degree $r$ with $r+1$ zeros must be the zero polynomial.
But then obviously we can keep intersecting such hypersurfaces till we get blue in the face: they will never cut out our $r+1$ points, but at best the line they lie on.
