While learning commutative algebra and basic algebraic geometry and trying to understand the structure of results (i.e. what should be proven first and what next) I came to the following question:
Is it possible to prove that $\mathbb A^2-point$ is not an affine variety, if you don't know that the polynomial ring is a unique factorisation domain?
It seems to me, that this question has some meaning, since when we define affine variety, we don't need to use the fact that the polynomial ring is an UFD. Don't we?
Best Answer
Yes, you can do it over any field.
First, it is enough to show $\mathcal O(Y) = k[x,y]$ ($Y=A^2-0$). If that is true and $Y$ is affine, then the embedding $Y \to A^2$ must correspond to some $k$-algebra map $k[x,y] \to k[x,y]$, which is absurd.
The key point now, as in Guillermo's post, is to show that $R_{(x)} \cap R_{(y)}= R$ ($R=k[x,y]$). It will follow from the
Indeed, if $f/x^m =g/y^n$, then $fy^n=0$ modulo $x^m$, so $f=hx^m$ and we are done.
The above Fact is elementary. For example you can induct on $m$. Clearly $m=1$ is OK. Now if $m>1$, use the short exact sequence:
$$0 \to R/{(x^{m-1})} \to R/(x^m) \to R/(x) \to 0$$
and Snake Lemma to conclude that $y^n$ is regular on the middle term as well.
Note that I used $x,y$ abstractly and all you need is that the elements $(x,y)$ form a regular sequence on a commutative Noetherian ring $R$ to start with. Then the proof shows that $\mathcal O(Y) =R$ if $Y=\text{Spec}(R) - V(x,y)$. In fact, more general results are true for ideals of depth at least $2$. If you are interested, it will be a good motivation to learn about depth and regular sequences.