Fix an algebraically closed **†** ground field $k$ of any characteristic. I want to use the classical definition of projective $n$-space $\mathbb{P}^n$ as set quotient of $\mathbb{A}^{n+1}\setminus 0$ by the action of $k^\*$, and for its Zariski topology the definition that closed sets are the "zero sets" of homogeneous polynomials in $n+1$ variables considered as functions from $\mathbb{P}^n$ to {$0,1$}.


I'm not satisfied with my understanding of why this is the same as the quotient topology from the map   
$\mathbb{A}^{n+1}\setminus 0\to \mathbb{P}^{n}$. This means proving the map is surjective, continuous, and that a set with closed pre-image is closed, the first two properties being clear. The last part is the least trivial, meaning that:



> 
> If a set $S\subseteq \mathbb{P}^n$ has closed preimage $\widehat{S}$ in $\mathbb{A}^{n+1}\setminus 0$, i.e. $\widehat{S}$ is cut out by some polynomials $f\_1,\ldots,f\_r$, then $S$ is closed in $\mathbb{P}^n$, i.e. $\widehat{S}$ is cut out by some *homogeneous* polynomials $g\_1,\ldots,g\_s$.
> 
> 
> 


I'm bad at algebra, so I can't seem to start doing anything here that doesn't look ugly/hard... I really want to "see" what's going on and avoid just quoting results about invariants without understanding how they work in this "toy" example.


Thanks for the help, anyone!


**† Follow up:** the algebraic closed hypothesis is not necessary, since David/Kevin's proof works for any infinite field, and for a finite field our spaces (as defined here) are discrete so the result is trivial.



Elementary proof that projective space is a quotient
Let $f$ be a polynomial which vanishes on $\hat{S}$. Write $f=\sum f\_i$, where $f\_i$ is homogenous of degree $i$. The set $\hat{S}$ is homogenous so, for any $\lambda \in k^\*$, the polynomial $f(\lambda \cdot x) = \sum \lambda^i f\_i$ also vanishes on $f$.


Since $k$ is infinite, we can find more equations of the form $\sum \lambda^i f\_i=0$ than there are terms in $f$. Taking linear combinations, we deduce that each $f\_i$ individually vanishes on $\hat{S}$.


