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.
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ It doesn't seem clear what you are asking. Are $n$ and $N$ the same? If so then $V$ has codimension one in $\mathbb{P}_{\mathbb{C}}^{n-1}$, and for the desired inequality to occur then $F$ has to have the same vanishing locus of the diagonal form. If $n < N$ then one would have to specify some embedding of $\mathbb{P}_{\mathbb{C}}^{n-1}$ into $\mathbb{P}_{\mathbb{C}}^{N-1}$. This is possibly where you might use Vernoese embedding. You can also look at this paper (projecteuclid.org/euclid.dmj/1143936000) $\endgroup$– Stanley Yao XiaoCommented Aug 19, 2018 at 12:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.)
answered Aug 19, 2018 at 13:06
-
2$\begingroup$ I think the essential point of the problem is exactly to prove that this subvariety is not contained in a linear subspace, which is not at all immediately clear from the statement. One can either interpret it as the image of the Veronese embedding, after which it becomes immediately clear (it's amazing the psychological effect dualizing has) or do something explicit as in mathoverflow.net/questions/98714/…. $\endgroup$– dhyCommented Aug 19, 2018 at 13:14
-
2$\begingroup$ @dhy Here's how I would express it: If you write $y_i \sum_{i=1}^n z_i x_i$, you can take any linear form on this space and write it as a degree $d$ homogeneous polynomial in the $z_i$. If the form is nonzero on some monomial in the $x_i$, the corresponding polynomial will contain a nonzero coefficient of that monomial in the $z_i$ and hence be nonzero. (this uses characteristic $>d$, it is false without that) But every nonzero polynomial is nonzero on some point. $\endgroup$ Commented Aug 19, 2018 at 13:42