# Why do rational functions in $\mathbb{P}^n$ have to be a ratio of homogeneous polynomials of the same degree?

To have a well defined homogeneous polynomial $$f:\mathbb{P}^n\rightarrow \mathbb{P}^n$$, we require that it is homogeneous so that $$f(x)=f(\lambda x)=\lambda^{\deg f} f(x)=f(x)$$ (so that it is well defined on projective space).

In various texts (e.g. Sutherland's lecture notes on Elliptic Curves, Harris Geometry of Algebraic Curves) a rational function $$g:\mathbb{P}^n\rightarrow \mathbb{P}^n$$ is defined to be the ratio of homogeneous polynomials $$f$$ of the same degree. Intuitively I assume this is to make sure $$g$$ is well defined.

However, let $$g=f/h,$$ where $$f,g$$ are homogeneous polynomials on projective space of some arbitrary degrees $$F,G$$. Then, $$g(x)=g(\lambda x)=f(\lambda x)/h(\lambda x)=\lambda^{F-G}f(x)/h(x)=g(x)$$. This does not require that $$F=G$$.

Question: Why then is it required that $$f$$ and $$h$$ have the same degree for $$g=f/h$$? Am I wrong in my analysis somewhere, or is there a different reason for the definition, or am I wrong in interpretation of the definition?

Edit: or do we consider $$g$$ as a function to affine space?

• $g(\lambda x)=g(x)$ seems like a nice property to have when working in a projective space. Commented Mar 21, 2022 at 18:15
• But it retains this property even if $F\neq G$. Commented Mar 21, 2022 at 18:20
• The two definitions are equivalent right. Multiply $h(x)$ by $x^{F-G}$ and you have rewritten $g$ as the quotient of two polynomials of the same degree. So probably convinience. Commented Mar 21, 2022 at 18:24
• I thought it was something like that, but there seem to be consequences of this; see Example 3.6 of Harris' curves linked above: let $F,F'$ be homogeneous polynomials of degree $d,d'$ and consider the divisors $V(F),V(F')$, then as divisors $V(F)-V(F')=(F/F')$. Then it says V(F), V(F') are linearly equivalent iff F, F' have the same degree (iff F/F' is rational). So the notion has some consequences that seem to supersede convenience. Commented Mar 21, 2022 at 18:41
• @TejasRao you have it by definition but $g(x)$ has it as a computational property. It's self-normalizing. Commented Mar 21, 2022 at 18:52

First of all, a ratio of homogeneous polynomials of the same degree defines a function $$\mathbb P^n \to \mathbb P^1$$, not $$\mathbb P^n \to \mathbb P^n$$. For the latter, the usual description is as an $$(n+1)$$-tuple of coordinate maps, each of which is a homogeneous polynomial of the same degree, not all vanishing at any common point.
For example, consider $$\mathbb P^1$$ with projective coordinates $$[x : y]$$. If $$f, g \in k[x, y]$$ are homogeneous of degrees $$F, G$$ and $$h := [f : g]$$, then we have: \begin{align*} h([\lambda x : \lambda y]) & = [f(\lambda x, \lambda y) : g(\lambda x, \lambda y)] \\ & = [\lambda^F f(x, y) : \lambda^G g(x, y)], \end{align*} which agrees with $$[f(x, y) : g(x, y)]$$ if and only if $$F = G$$.
I think your mistake is in passing between affine and projective coordinates. If $$[f : g]$$ is a map to $$\mathbb P^1$$ as above, we can view it in affine coordinates as a ratio $$f/g$$ of homogeneous polynomials, where the zero locus of $$g$$ is sent to the point at infinity. But affine coordinates are not defined up to scalar multiplication, so $$\lambda^{F-G} f(x)/g(x)$$ is not the same thing as $$f(x)/g(x)$$.