Why is the polynomial $F \in K[Z_0, \dots, Z_n]$ on $K^{n+1}$ not a well defined function on $\mathbb{P}^n$ in general? I'm reading Joe Harris' Algebraic Geometry and he says 
"A polynomial $F \in K[Z_0, \dots, Z_n]$ on the vector space $K^{n+1}$ does not define a function on $\mathbb{P}^n$" 
where $K$ is a algebraically closed field, $\mathbb{P}^n$ is the projective space over $K$.
I'm confused why the function $F$ is not well defined. Given $X \in \mathbb{P}^n$, a one dimensional subspace of $K^{n+1}$, isn't $F(X)$ simply the image of $F$ under $X$? Why would that not be well defined? Is it because we require the image to also be an element of $\mathbb{P}^n$?
 A: "Well defined" is a term generally used for a function that may or may not be "well defined" on a quotient set; that is, on the set of equivalence classes of an underlying set modulo an equivalence relation.
For example, take $\mathbb Z$ and $\mathbb Z_2$. The latter is a quotient set of the former, where the equivalence relation is $a \equiv b$ if and only if $a-b$ is even. Then, the function $\mathbb Z \to \mathbb R, \, n \mapsto (-1)^n$, gives a "well defined map" from $\mathbb Z_2 \to \mathbb R$. The condition to check, in all such cases, is that under the original map, on the original set, elements of the original set that are equivalent under the equivalence relation have the same image under the function.
In this regard, a polynomial $f \in K[Z_0, \dots, Z_n]$ defines a function from $K^{n+1}$ to $K$ in the usual way. $\mathbb P^n$ is a quotient set of $K^{n+1} \setminus \{0\}$, modulo multiplication of vectors by nonzero scalars.
The point in Harris's statement is that, given a polynomial $f$ of $n+1$ variables, the value of the corresponding polynomial function at $a_0,
\dots, a_n$ is different from the value of the same polynomial at $(ta_0, \dots, ta_n$) $(t\in K \setminus \{0\})$. This is true even for homogeneous polynomials. The only polynomials that do, in fact, work are the constants.
