TITLE: Proof that all rational maps $\mathbb{P}^1\to\mathbb{P}^N$ are regular without using codimension?
QUESTION [7 upvotes]: There is a theorem in Shafarevich's Basic Algebraic Geometry (Theorem 3, pg. 109) which states that if $X$ is a nonsingular variety, and $\varphi\colon X\to\mathbb{P}^N$ a rational map to projective space, then the set of points at which $\varphi$ is not regular as codimension $\geq 2$.
So if $X=\mathbb{P}^1$, this immediately implies that every rational map $\mathbb{P}^1\to\mathbb{P}^N$ is also regular. 
Is there a way to see this in a more "down-to-earth" way, I feel like this theorem is a bit overkill for the result. Thanks.

REPLY [8 votes]: Suppose $f:P^1\to P^n$ is a rational map. On a nonempty open subset of the domain, then, it can be written as $(x:y)\mapsto(f_0(x,y):\cdots:f_n(x,y))$ with the $f_i$ homogeneous of the same degree. 
If all the $f_i$ vanish at a point, they all have a factor in common, and you can remove it. That way, you can extend the map to the whole of $P^1$.