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.
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.