Well-defined sets involving equivalence classes Let $k$ be a field and define $P^n(k)$ to be the n-dimensional projective space over $k$, meaning $$P^n(k):=(k^{n+1} \setminus \{0\})/{}\sim{}, \ x \sim y : \iff \exists \lambda \in k \setminus \{0\}: x=\lambda y.$$ If $F$ is now a polynomial in two variables, $F \in k[x,y]$, one can define the zero set of $F$ over $k^2$, $$Z(F):=\{x \in k^2 \ | \ F(x)=0\}.$$ My question is whether I can also define this for every polynomial over the projective space. I suspect that I can't, in general, do this over the projective $2$-space, since $F([x:y])=0$ would not be independent of the representing element for an arbitrary $F$.
This leads to my actual question, that is, does the condition of a set always have to be independent of the choice of the represening element? I suspect that indeed this has to be the case, because suppose $x,y$ are elements of $P^n(k)$ satisfying $x \sim y$ with $F(x)=0, F(y) \neq 0$. Then $$Z(F) \ni x=y,$$ despite $y$ not satisfying the condition to be in the set, which should be problematic. Is this correct?
I also wrote a question touching on a connected topic here: Is it possible for a set to contain an element that does not have the defining property?. In this post it has also been discussed that sets can contain elements that don't directly satisfy the defining property, as long as an equal object satisfies it. The difference to the situation now would be, though, that $y$ explicitly does not satisfy the condition, meaning the object $x$ satisfies the condition whereas $y$ does not, while they are the same object, which should be problematic. I hope I worded my questions well enough, since this is slightly confusing.
 A: 
Does the condition of a set always have to be independent of the choice of the represening element?

Yes, for a condition to be well defined. Otherwise, contradictions arise. For example, the polynomial $f(x,y) = x - y^2$ and the point $[1:2] \sim [4:2] \in P^1(k)$. Clearly, $f(4, 2) = 0$ but $f(2, 1) = 1 \neq 0$. Hence, the condition '$[4 : 2] \in Z(f)$' is not well defined since there is a representative of $[4 : 2]$ which is nonzero. This is a necessary condition; otherwise, there would be a contradiction as $[4 : 2] \in Z(f)$ and $[4 : 2] \notin Z(f)$ since $[4 : 2] \sim [2 : 1]$.
This corresponds to the following general fact. If $\sim$ is an equivalence relation on a set $S$, there is a projection function $S \to (S/\sim)$ given by $x \mapsto [x]$. Here I denote $(S/\sim)$ the set of eqiuvalence classes of this relation. This map is endowed with the following property. If $f: S \to T$ is a function and $f(x) = f(y)$ for every $x \sim y$, then $f$ descends to a unique map $(S/\sim) \to T$ given by $[x] \mapsto f(x)$.
In Algebraic Geometry, this arises in the following way. We consider polynomials $f(x_1, \dots, x_n)$ as functions on $k^n = \mathbb{A}^n_k$. Now, finding exactly which polynomials descend to well defined functions on $(\mathbb{A}^n_k \setminus \{0\} )/\sim = P^{n - 1}(k)$ reduces to finding exactly which polynomials satisfy $f(\lambda x_1, \dots, \lambda x_n) = f(x_1, \dots, x_n)$. As it turns out, this is very restrictive and leaves very little to work with. To remedy this, we consider the next best thing, polynomials $f \in k[x_1, \dots, x_n]$ so that $$f(\lambda x_1, \dots, \lambda x_n) = C f(x_1, \dots, x_n)$$ for all $[x_1: x_2: \cdots : x_n] \in P^{n - 1}(k)$ and some $C \in k$, mimicing the equivalence relation on projective space. While this doesn't yield well defined functions, it does yield a well defined way to define the zero set of $f$ in $P^{n - 1}(k)$.
Functions of the above form are called homogeneous and algebraic sets(or zariski closed sets) in $P^{n}(k)$ are by definition, zero sets of some family of homogeneous functions.
