Which rational functions $\mathbb{P}^1\rightarrow k$ are regular at the point at infinity?

Question

I am trying to solve a problem in § 1.7 of Shafarevich's "Basic Algebraic Geometry 1":

"Let $k$ be an algebraically closed field. Which rational functions $\mathbb{P}^1\rightarrow k$ are regular at the point at infinity? What order of zero do they have there?"

Here's my attempt so far:

Let $u$ be a rational function on $\mathbb{P}^1$ that is regular at the point at infinity $\mathcal{O}:=[1:0]$. I have already proved in another exercise that if $u$ is regular on all of $\mathbb{P}^1$ then it is constant; hence let $u=p/q$ where $p$ and $q$ are homogeneous polynomials in $k[X,Y]$ of the same degree $n\geq 1$, say $p = \sum_{i+j = n} a_{i j} X^i Y^j$ and $q = \sum_{i+j = n} b_{i j} X^i Y^j$. Then since $u$ is regular at $\mathcal{O}$, $q(\mathcal{O}) = b_{n0} \neq 0$ and we see $u(\mathcal{O}) = a_{n 0}/b_{n 0}\in k$. From the question it looks like we ought to be able to show $a_{n 0} = 0$, but I don't really know how to proceed from here. One idea I had was to suppose $a_{n 0}\neq 0$ and look at what happens to $u$ on the affine part $\mathbb{A}_y^1 :=\left\{[a:b]\in\mathbb{P}^1 : b\neq 0\right\}$, but I can't seem to get anywhere with this.

Any tips would be appreciated!!

Answer 1

The required rational functions are the fractions of the form $\varphi(t)=\frac{p(t)}{q(t)}$ with $p(t),q(t)\in k[t] \:(q(t)\neq 0)$ polynomials satisfying $\operatorname {deg}q(t)\geq \operatorname {deg}p(t)$.
The order of the zero at infinity of $\varphi(t)$ is the difference $\operatorname {deg}q(t)-\operatorname {deg}p(t) \in \mathbb N$.
[This is valid for an arbitrary field $k$, algebraically closed or not.]

Answer 2

I assume from the context we are over the field of complex numbers.

A rational function $\frac uv$ with $u,v\in k[X,Y]$ is regular at $[1:0]$ if and only if $v(1,0)\ne 0$. Since we use projective coordinates, this is equivalent to $v(x,0)\ne 0$ for all $x\in k\setminus\{0\}$. This, again, is now equivalent to $v(X,0)\ne 0$ where capital $X$ denotes the actual variable.

Now think of $v(X,Y)$ as a polynomial in $Y$ over $k[X]$.

Does this help?