Proof that a Zariski closed subset of a projective space is the common zeros of homogeneous polynomials

Let $K$ be an algebraically closed field. Let $n \ge 0$ be an integer. We consider $K^{n+1}$ as a topological space with Zariski topology. Let $G = K^*$ be the multiplicative group of $K$. Let $X = K^{n+1} - {0}$. Then $G$ acts on $X$. Then the projective space $P^n$ over $K$ is, by definition, the orbit space $X/G$. We regard $P^n$ as a topological space with the quotient topology. Let $Z$ be a closed subset of $P^n$. Then how do you prove that $Z$ is the common zeros of homogeneous polynomials?

A start: Take a $Z'\subset K^{n+1}$ which is a closed and such that $\forall k\in K^*, z\in Z': kz\in Z'$.

Let $I\subset K[x_1,...,x_{n+1}]$ be the ideal of polynomials $f$ such that $f(z)=0$ for all $z\in Z'$.

Take any polynomial $f\in I$ and write it as the sum $\sum f_i$ where the $f_i$ are homogeneous of degree $i$.

Then for any particular $z_0\in Z'$, define $p(x)=\sum f_i(z_0)x^i$. So $p(x)\in K[x]$. If $f_i(z_0)\neq 0$ for some $i$, then $p(x)$ is a non-zero polynomial, so there exists a $k\in K^*$ such that $p(k)\neq 0$. (You don't need $K$ algebraically closed, just $K$ infinite for this step.)

But $p(k)=\sum f_i(z_0)k^i = \sum f_i(kz_0) = f(kz_0)$. So that would contradict the condition that if $z\in Z'$ then $kz\in Z'$.

So that means that $p(x)$ has to be the zero polynomial, which means that $f_i(z_0)=0$ for all $z_0\in Z'$, which means that $f$ is the sum of homogeneous polynomials $f_i$ such tat $f_i(z)=0$ for all $z\in Z'$.

Any closed $Z\subseteq P^n$ must be the image of such a $Z'\subset K^{n+1}$, so it seems like you are done.