Complement of a point in $\mathbb{P}^{2}$ This is question $5$ from Shafarevich's book page $66$. Let $X=\mathbb{P}^{2} \setminus x$ where $x$ is a point. Prove that $X$ is not isomorphic to affine nor a projective variety. How to prove this?
 A: Okay yeah yeah I think the stuff in the comments works. Here's how to see the only regular functions on $\mathbb{P}^2 - pt$ are $k$, where $k$ is your algebraically closed ground field (your argument for why that would imply $\mathbb{P}^2 - pt$ is just a single point is legit, I think). 
(Global) regular functions on $\mathbb{P}^2 - pt$, by definition, are regular functions on $\mathbb{A}^3 - line$ which are $k^*$ invariant.
Let $f$ be such a function, introduce coordinates $x,y,z$, where the line is $\{x = 0\} \cap \{y = 0\}$ (the $z$ axis). On $\{x \neq 0 \}$, $f$ must be of the form $p(x,y,z,x^{-1})$, and on $\{y \neq 0\}$, $f$ must be of the form $p(x,y,z,y^{-1})$, where $p$ just means ``a  polynomial in". 
Suppose for a contradiction $f = \sum_k p_k(y,z)x^k$ on $\{x \neq 0\}$ had terms with $x$ to a negative power. 
Now on $\{ y \neq 0\}$, we have an equality of functions $$x^{-k}p_{-k}(y,z) + \ldots  = p_0(y, y^{-1},z) + p_1(y,y^{-1},z)x + \ldots$$Now multiply by through by $x^{k}$, now we have something of the form $$p_{-k}(y,z) + \ldots  = p_0(y^{-1}, y,z)x^k + \ldots$$Now just evaluate at any point of the form $(0,y \neq 0,z)$, we get 0 on the RHS every time, which tells us the LHS must be the 0 polynomial in $y,z$, as desired.
So this tells us $\mathbb{P}^2 - pt$ ain't affine, and as to why it ain't projective, as in the comments, in general if we have $X \rightarrow Y$, $X$ proper, and $Y$ separated, then $X \rightarrow Y$ has closed image.
Were $\mathbb{P}^2 - pt$ isomorphic by $i$ to some projective variety $X$, consider $$X \xrightarrow{i} \mathbb{P}^2 - pt \rightarrow \mathbb{P}^2$$so $\mathbb{P}^2 - pt$ is closed, a contradiction. 
Well, there's something annoying more to be said. We already know $\mathbb{P}^2 - pt$ is open, why can't it be clopen? Well, $\mathbb{P}^2$ has the quotient topology, so we're asking, is $\mathbb{A}^3$ connected? You can see this in various ways, one would be that if it weren't we could write $\mathbb{A}^3 = V(I) \coprod V(J)$, both nonempty, the $LHS$ has (Krull) dimension 3, the RHS has dim $\leqslant 2$. 
A: You can use the fact that if a polynomial $F \in k[X_{1}, X_{2}, \ldots, X_{n}], \#k = \infty$, vanishes in the complementary of a finite subset of $k^{n}$, then $F = 0$ (proof by induction on $n$). Therefore, you cannot have $\mathbb{P}^{2} \setminus \{ x \}$ as a set of common zeroes of polynomials.
I think it is true that the only regular functions on $\mathbb{P}^{2} \setminus \{ x \}$  are the constant ones, and this might be proved using the above fact about polynomials.
A: I am not sure about the projective variety but you should be able to show that a line without a point is not an affine variety, and similarly an affine space without a point is not an affine variety. The view that a projective space is 'glued' from affine space, and projective variety is "glued" by affine varieties may help but I cannot write a proof explicitly. 
A: You can use topological generalities to solve the problem (at least showing non-projectiveness), assuming for instance we are over the complex numbers.
$\mathbb A^2\setminus\{x\}$ is not compact in the classical topology (take the open cover $(\mathbb R\times (-i,i))_i$) so not a projective variety (i.e. not a closed subset of some projective space). Working with the Zariski topology for compactness is not so simple. Rather, using the Zariski topology, it is not a complete variety, since a canonical embedding in $\mathbb P^2$ is open (and not all of $\mathbb P^2$) and therefore not closed since $\mathbb P^2$ is irreducible.
$\mathbb P^2\setminus\{(0,0)\}$ is not compact in the classical topology (take the open cover $(((\infty\cup\mathbb R\setminus\{0\})\times\mathbb R)\cup((\infty\cup\mathbb R)\times (\infty,-1/i)\cup(1/i,\infty))\cup\{\infty\times\infty\})_i$), and using the Zariski topology, it is not complete, taking for instance the canonical embedding into $\mathbb P^2$, which is irreducible. You can do the same with any point (adding some notation) or take an isomorphism to $\mathbb P^2\setminus\{x\}$.
Showing that they are not affine is more difficult, since for instance $\mathbb A^1\setminus\{x\}$ is affine. You must show that you can only remove at most codimension 1 subsets from affine varieties to obtain varieties that are still affine. But there are other arguments, I think that of uncookedfalcon could be ok, but I did not really read it.
EDIT: I tried to correct the example of open cover for $\mathbb P^2\setminus\{(0,0)\}$, that is tricky.
