TITLE: How to show that a hypersurface is a diagonal intersected with hyperplanes?
QUESTION [5 upvotes]: Suppose I have a hypersurface $V(F) = \{ \mathbf{x} \in k^n: F(\mathbf{x}) = 0 \}$, where $F$ is a homogeneous form of degree $d > 1$. I would like to show that there exists some diagonal form $D(y_1, \ldots, y_N) = A_1 y_1^d + \ldots + A_N y^d \in k[y_1, \ldots, y_N]$ and some linear forms $L_i(\mathbf{y}) \in k[y_1, \ldots, y_N]$, say $1 \leq i \leq T$, such that 
$$
V(F) = V(D) \cap V(L_1) \cap \ldots \cap V(L_T).
$$ 
Here $k = \mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$. I seem to recall someone telling me this follows fairly easily from some algebraic geometry (possibly using Veronese embedding) but I couldn't figure it out. Any comments would be appreciated.

REPLY [5 votes]: The map from $V(L_1) \cap \dots \cap V(L_T)$ to $\mathbb P^{N-1}$ is a linear map. So an equivalent way of stating this is that there are $N$ linear forms $y_1,\dots,y_N$ in $x_1,\dots,x_n$ such that $\sum_{i=1}^N A_i y_i^d = F$.
One way to think about this is to consider, for each $N$, the locus in the space of degree $d$ homogeneous forms that can be written as $\sum_{i=1}^N A_i y_i^d$. If the dimension of this locus for some $N$ is equal to the dimension for $N+1$, it follows that the sum of a generic element of this locus with a generic linear form is a generic element of the locus, so the sum of two generic elements of the locus is a generic element of the locus, so the locus is a dense subset of a linear subspace. However, the space of linear forms raised to the $d$th power is not contained in any linear subspace (this can be expressed as a statement of the Veronese embedding, but can also be proved directly algebraically), so for this $N$ the locus is in fact dense in the whole space. Hence for $2N$ the locus is the whole space. (And their must be such an $N$ because the dimension is bounded.)