I am reading morphism in Hartshorne and his comment confuses me. Firstly, he defines regular function on the affine space as follows:
Let $Y$ be a quasi-affine variety in $\mathbf{A}^{n}$. A function $f:Y\longrightarrow k$ is regular at a point $P\in Y$ if there is an open neighborhood $U$ with $P\in U\subseteq Y$ and polynomials $g,h\in A:=k[x_{1},\cdots,x_{n}]$ such that $h$ is nowhere zero on $U$, and $f=g/h$ on $U$.
Then, he commented as follows: "Here of course we interpret the polynomials as functions on $\mathbf{A}^{n}$, hence on $Y$. " Ok, I understand what he means here, because for each $f\in k[x_{1},\cdots, x_{n}]$, it can be considered as a function from $\mathbf{A}^{n}$ to $k$ by defining $f(P):=f(a_{1},\cdots, a_{n})$ for any $P:=(a_{1},\cdots, a_{n})\in\mathbf{A}^{n}$. So basically he wants to remind me that polynomials are also functions on $\mathbf{A}^{n}$.
Then, he defines regular function on the projective space as follows.
Let $Y$ be a quasi-projecitve variety in $\mathbf{P}^{n}$. A function $f:Y\longrightarrow k$ is regular at a point $P\in Y$ if there is an open neighborhood $U$ with $P\in U\subseteq Y$ and polynomials $g,h\in S:=k[x_{0},\cdots,x_{n}]$ such that $h$ is nowhere zero on $U$, and $f=g/h$ on $U$.
However, he comments as follows: "note that in this case, even though $g$ and $h$ are not functions on $\mathbf{P}^{n}$, their quotient is a well-defined function whenever $h\neq 0$, since they are homogeneous of the same degree."
What does he mean here? $g$ and $h$ are homogeneous polynomials right? Homogeneous polynomial $g$ gives a function from $\mathbf{P}^{n}$ to $\{0,1\}$ by $f(P):=0$ if $f(a_{0},\cdots, a_{n})=0$ and $f(P):=1$ if $f(a_{0},\cdots, a_{n})\neq 0$. Right?
Then, why wouldn't they be functions on $\mathbf{P}^{n}$? Is there any specific definition of "function" assumed here?
Thank you!
Best Answer
The function you've described makes perfect sense (and it's useful in that it allows us to define the vanishing locus of $f$), but in order to make it well-defined you had to mod out by scalar multiples in the codomain. Regular functions are by definition always valued in $k$, not in $k/k^{\times} = \{0, 1\}$--regardless of whether we're talking about affine or projective varieties.
An example might help. Consider $\mathbf P^2$ with projective coordinates $[x_0 : x_1 : x_2]$. Clearly, the homogeneous polynomial $x_0$ does not give a well-defined function $\mathbf P^2 \to k$, because for example the coordinates $[1 : 0 : 0]$ and $[2 : 0 : 0]$ represent the same point but have different values of $x_0$. But I claim that $f = \frac{x_1}{x_0}$ does give a well-defined function to $k$, at least on the locus $U \subset \mathbf P^2$ where $x_0 \neq 0$. The reason is simple: if you scale coordinates by a nonzero constant $c$, then $$ f([c x_0 : c x_1 : c x_2]) = \frac{c x_1}{c x_0} = \frac{x_1}{x_0} = f([x_0 : x_1 : x_2]). $$ The same is true for $\frac{g}{h}$ whenever $g$ and $h$ are homogeneous polynomials of the same degree.