# The set of polynomials which “cut out” smooth subsets of projective space is open and dense

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.

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

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.
• How can it be shown in characteristic $p$ that for each $d$ there is a smooth hypersurface of degree $d$ in $\mathbb P^n$ ? – Georges Elencwajg Feb 9 '16 at 10:37
• @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. – Mohan Feb 10 '16 at 16:47
• 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. – Georges Elencwajg Feb 10 '16 at 20:36