How to prove every rational map from $\Bbb P^1 \to \Bbb P^n$ is regular. How to prove every rational map from $\Bbb P^1 \to \Bbb P^n$ is regular.
For $f=(\frac {f_1}{f'_1},...,\frac {f_n}{f'_n})$ where each $ f_i,f'_i$ are monomials of same degree. But now how to show this?
 A: Letting $\mathbb P^1=\{[x:y]|x,y\in k \text{ and } (x,y)\neq (0,0)\}$ with the usual identification. The polynomials $f_i$ and $f_i'$ in defining your rational functions can in fact be homogeneous polynomials in $x$ and $y$ of the same degree. 
We check the regularity of $f$ in the affine charts. So, first set $y=1$ to get the affine part of $f$ as $\tilde f=(\frac{g_0}{g_0'},\cdots, \frac{g_n}{g_n'})$ where $g_i(x,y)=g_i(x,1)$ and $g_i'(x,y)=g_i'(x,1)$. 
First off, we can assume that $g_i$ and $g_i'$ share no common roots. So, if $\tilde f$ is not regular, it can be because of two reasons:


*

*All the $\frac{g_i}{g_i'}$ vanish simultaneously at the same point $x_0$. This means $(x-x_0)^k$ divides all the $g_i$ maximally so can be factored out as part of the property of projective space and now not all the $\frac{g_i}{g_i'}$ vanish at $x_0$. One problem solved!

*One of the denominators $g_i'$ vanishes at say $x_0$ say with multiplicity $k$. Then multiply $(\frac{g_0}{g_0'},\cdots, \frac{g_n}{g_n'})$ through by $(x-x_0)^k$. Now $(x-x_0)^k\frac{g_i}{g_i'}$ does not vanish at $x_0$ any longer. Repeat this process with all the roots of the denominator.


Voila! $(\frac{g_0}{g_0'},\cdots, \frac{g_n}{g_n'})$ is now regular in the chart given by $y=1$ . Now repeat this process in the other chart $x=1$ to show that $f$ is regular in that chart too. So $f$ is a regular map on $\mathbb P^1$ because the two charts we checked regularity in cover it.
