Every variety is isomorphic to an intersection of a linear space and a Veronese surface

Question

"Deduce that any projective variety is isomorphic to an intersection of a Veronese variety with a linear space"

I've been trying to solve this exercise from Joe Harris book.

I can see that if a variety $X\subset \mathbb{P}^n$ has only polynomials of degree $d$, then in the coordinates of $\nu_d(\mathbb{P}^n)$ it is, indeed, a linear space.

The problem is when I have different degrees. If I take a family of polinomial of maximum degree $d$, and I have a polynomial $f_0$ in this family, with $m=deg ~f_0<d$, i can multiplicate $f_0$ by $X_0^{d-m}$, so I get $X_0^{d-m}f_0$ with degree d, and it is a linear space in $\nu_d(\mathbb{P}^n)$.

The problem is that it may not be the same variety $X$ after this proceeding.

There is a way to fix that? Or should I try in a different way?

Thanks.

Clearly stating the problem before talking about your work would be helpfil — Stella Biderman, Feb 01 '16 at 15:24
I put the problem on the title of the question, sorry if it was not clear. I edited, and its on the question now. — ett, Feb 01 '16 at 15:28
Sorry, I must have missed that. It's generally considered good form to fully state the problem in the post, even if it's in the title. — Stella Biderman, Feb 01 '16 at 15:37

Eric Wofsey · Accepted Answer · 2016-02-01T15:51:36.370

4

Instead of just taking $X_0^{d-m}f_0$, take $X_i^{d-m}f_0$ for all $i$ at once. If you have a point $p=[a_0,\dots,a_n]\in\mathbb{P}^n$ such that $X_i^{d-m}f_0$ vanishes at $p$ for all $i$, then $a_i\neq 0$ for some $i$, and then the condition that $X_i^{d-m}f_0$ vanishes forces $f_0$ to vanish at $p$.

edited Feb 01 '16 at 15:51

answered Feb 01 '16 at 15:33

Eric Wofsey

330,363

Every variety is isomorphic to an intersection of a linear space and a Veronese surface

1 Answers1