3
$\begingroup$

Let $k[x_0,x_1,...,x_n]_d$ be a space of all forms (in other words, homogenous polynomials) of degree $d$ of variables $x_0, x_1,...,x_n$ over algebraically closed field $k$. Let's think of $k[x_0,x_1,...,x_n]_d$ as an affine space.

Of course, every such polynomial has a well defined zero-locus on a projective space $\mathbb P^n$, because it is homogenous.

Let $X\subset k[x_0,x_1,...,x_n]_d$ be a subset of polynomials, whose zero-loci are smooth subets of $\mathbb P^n$.

I would like to prove, that $X$ is open and dense subset of $k[x_0,x_1,...,x_n]_d$, but i have no idea how to do that. I guess it has something to do with the fact, that the set of smooth points of an algebraic set is dense and open, but i don't know how to use this fact in this setting.

$\endgroup$
  • 1
    $\begingroup$ what's a smooth subset of $\Bbb P^n$ ? $\endgroup$ – mercio Feb 7 '16 at 18:33
  • $\begingroup$ i meant a smooth subvariety of $\mathbb P^n$, which is a variety whose all points are smooth $\endgroup$ – adam Feb 7 '16 at 18:35
7
$\begingroup$

Many of these are proved using the universal hypersurface. Let $P$ the projective space of all degree $d$ forms (since the equation and any non-zero constant multiple give the same variety) and consider $Z\subset P\times\mathbb{P}^n$ the universal hypersurface defined in the obvious way - these are pairs $(f,p)$ with $f(p)=0$. Consider $T\subset Z$, defined as $(f,p)$ such that $f(p)=0$ and all the partial derivatives of $f$ are zero at $p$. Then, $Z$ is closed in $P\times\mathbb{P}^n$ and $T$ is closed in $Z$.If $f$ is not in the image of $T$ under the first projection, it is easy to check that $f$ is smooth. Since image of $T$ is closed, suffices to show that there is at least one smooth hypersurface. Though it can be shown for positive characteristics, easier for zero characteristic, by noting that the Fermat hypersurface, $\sum x_i^d=0$ is smooth.

$\endgroup$
  • $\begingroup$ that's a neat trick, thanks! $\endgroup$ – adam Feb 7 '16 at 20:07
  • $\begingroup$ How can it be shown in characteristic $p$ that for each $d$ there is a smooth hypersurface of degree $d$ in $ \mathbb P^n$ ? $\endgroup$ – Georges Elencwajg Feb 9 '16 at 10:37
  • $\begingroup$ @Georges Elencwajg Here is an ad-hoc example at least when $d\geq 3$. Let $p$ be the characteristic and $F_d$ the Fermat equation. If $p$ does not divide $d$, $F_d=0$ is smooth and so assume that $p$ divides $d$. Consider $G_d=\sum_{i=0}^{n-1} x_ix_{i+1}^{d-1}$. Then $F_d+G_d=0$ is smooth. This can be seen as follows. The partial derivative with respect to $x_0$ is $x_1^{d-1}$. Thus the singular point must have $x_1=0$. Next, derivative with respect to $x_1$ gives $(d-1)x_0x_1^{d-2}+x_2^{d-1}$. Since $d>2$, we see that $x_2=0$ at any singular point, since $x_1=0$. Rest, I hope is clear. $\endgroup$ – Mohan Feb 10 '16 at 16:47
  • $\begingroup$ Yes, the rest is clear: bravo and +1. $\endgroup$ – Georges Elencwajg Feb 10 '16 at 20:24
  • $\begingroup$ Dear Mohan, the only case apparently not covered by your analysis is $p=d=2$. But in that case your hypersurface $F_2+G_2=0 $ turns out to be smooth too. Actually the simpler hypersurface $G_2=0$ is already smooth. $\endgroup$ – Georges Elencwajg Feb 10 '16 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.