Prove that $\mathbb{A}^1 - \{p_1, \dots, p_n\}$ and $\mathbb{A}^1 - \{q_1, \dots, q_m\}$ are not isomorphic for $n \neq m$ I want to prove that $\mathbb{A}^1 - \{p_1, \dots, p_n\}$ and $\mathbb{A}^1 - \{q_1, \dots, q_m\}$ are not isomorphic for $n \neq m$, where the $p$'s and $q$'s are points. This is one of those theorems that seems intuitively obvious, but I don't know how to prove. Perhaps by induction? 

I suppose $k$ is algebraically closed or at least the points $p_i, q_j$ have coordinates in $k$. Otherwise the proof is more complicate.
If they are isomorphic, then there exists an isomorphism of $k$-algebras ($k$ is the ground field): 
$$\phi: k[t, 1/(t-p_1), \dots, 1/(t-p_n)] \simeq k[t, 1/(t-q_1), \dots, 1/(t-q_m)].$$
This gives an isomorphism of their groups of units
$$ k^* (t-p_1)^{\mathbb Z}...(t-p_n)^{\mathbb Z} \simeq k^*(t-q_1)^{\mathbb Z}...(t-q_m)^{\mathbb Z}$$ 
which is identity on $k^*$ because $\phi$ is identity on $k$. Taking the quotient by $k^*$ we obtain an isomorphism of groups
$$ (t-p_1)^{\mathbb Z}...(t-p_n)^{\mathbb Z} \simeq (t-q_1)^{\mathbb Z}...(t-q_m)^{\mathbb Z},$$
hence $\mathbb Z^n\simeq \mathbb Z^m$ and $n=m$. 
Note that in general (when $n\ge 3$), $n=m$ is not sufficient to insure the existence of an isomorphism between the two curves.
