Affine varieties isomorphic to projective line

Suppose $V$ is an affine subvariety of $\mathbb{C}^m$, i.e. $V$ is an affine variety (algebraic set of some ideal) and $V\subset \mathbb{C}^m$. I am thinking about whether
$1)$ $V$ can be isomorphic to the projective line $\mathbb{C}\mathbb{P}^1$,
$2)$ $V$ can be isomorphic to $\mathbb{C}\mathbb{P}^1 \setminus\{p\},$ and
$3)$ $V$ can be isomorphic to the projective line $\mathbb{C}\mathbb{P}^1 \setminus\{p,q\}$ where $p,q\in \mathbb{C}\mathbb{P}^1$ .

I know that coordinate ring of $V$ is $\mathbb{C}[x_1,\cdots, x_m]/I(V)$ and that of $\mathbb{C}\mathbb{P}^1$ is $\mathbb{C}$. I don't see how this could be useful though.

1. No. If $V\simeq \Bbb C\Bbb P^1$, then $\Bbb C[x_1,\dots,x_m]/I(V)\simeq \Bbb C$. Now, since $\Bbb C$ is a field, $I(V)$ is a maximal ideal of $\Bbb C[x_1,\dots,x_m]$ and, by the Nullstellensatz, $\exists (a_1,\dots,a_m)\in\Bbb C^m$ s.t. $I(V)=(x_1-a_1,x_2-a_2,\dots,x_m-a_m)$. Thus, $V=\{(a_1,\dots,a_m)\}$ which is clearly non-isomorphic to $\Bbb C\Bbb P^1$. This is absurd so there is no affine subvariety of $\Bbb C^m$ isomorphic to $\Bbb C\Bbb P^1$.

Edit: Another idea is the following: if $\varphi: \Bbb C\Bbb P^1\to V$ is an isomorphism, then the pullback of any regular function $f:V\to\Bbb C$ is a regular function on $\Bbb C\Bbb P^1$, hence it's constant. By taking for $f$ any of the restrictions to $V$ of the functions $x_i:\Bbb C^m\to\Bbb C$, we see that $\forall i,\,x_i\circ \varphi$ is constant, hence $\varphi$ is constant (its image is a point).

1. Yes, of course. Simply take a straight line in $\Bbb C^m$: it's isomorphic to $\Bbb C\simeq \Bbb C\Bbb P^1\setminus \{\infty\}$.

2. Yes. Note that, up to change of coordinates, one can choose $p=0$ and $q=\infty$ in $\Bbb C\Bbb P^1$. So $\Bbb C\Bbb P^1\setminus\{p,q\}\simeq \Bbb C^*$ and its ring of global regular functions (*) is $\Bbb C\left[x,\dfrac 1x\right]\simeq \Bbb C[x,y]/(xy-1) \simeq \Bbb C[x_1,\dots,x_m]/(x_1x_2-1,x_3,\dots,x_m)$ (where $m\geq 2$): this corresponds to a plane hyperbola of $\Bbb C^m$.

(*) Here we use a general fact: if $f$ is a global regular function on a complex affine variety $W$ (here, it's $\Bbb C$) whose ring of global regular functions is $\Bbb C[W]$, then the open subset $D(f):= W\setminus V(f)$ is an affine variety and its ring of global regular functions is $\Bbb C[W]_{f}=\Bbb C[W]\left[\dfrac 1f\right]$.

• Thanks! In $1)$ you say that co-ordinate rings are isomorphic by viewing both $V$ and $\mathbb{P}^1$ as quasi-projective varieties, right? – John Oliver Aug 16 '16 at 21:17
• You're perfectly right. – paf Aug 16 '16 at 21:38
• See also my edit above. – paf Aug 16 '16 at 21:45
• Yes, this is what I meant. Note also that you can see $\Bbb C\Bbb P^1$ as $\Bbb C\cup\{\infty\}$: hence, in this case, 0 and $\infty$ make sense. – paf Aug 17 '16 at 1:10
• @monomorphic: I think $V(y - x^2) \subseteq \mathbb{A}^2$ is an affine variety that is isomorphic to $\mathbb{A}^1 = \mathbb{P}^1 - \{\infty\}$ but isn't a line. Algebraically $k[x,y]/(y - x^2) \cong k[x]$ – Jay Aug 17 '16 at 19:20