Interior points of Affine Variety in $\mathbb{C}^2$ I've just read that no non-trivial affine variety in $\mathbb{C}^2$ has an interior point. I can't see why this is obviously true. Could somebody give me a hint?
In particular, is this something special about $\mathbb{C^2}$, or is it true generally in $\mathbb{A}^n$?
Many thanks!
 A: This is true in $\mathbb C^n$ generally, but does not make sense in $\mathbb A^n$ for all algebraically closed fields as not all alg. closed fields have a canonical metric topology. To see this, note that if a variety $V$ were to have an interior point, then it would contain some open set $U$, and so any holomorphic function which vanishes on $V$ vanishes on $U$ so is zero everywhere. Thus the only polynomials which vanish on $V$ are those which vanish everywhere, which is just the zero polynomial, hence $V$ is trivial.
A: I do believe it to be true in general, assuming the field is algebraically closed (i am not familiar with AG over nonclosed fields so i won't say anything about that).
All statements concern Zariski topology unless stated otherwise.
Do you know the algebraic dimension of a variety? If no, see Hartshorne, the definition just after corollary 1.6.
First, any open $U \subset \mathbb{A}^n$ is a quasi affine variety and has the same dimension: $n$ (Hartshorne 1.10).
So pick a nontrival variety $X \subset \mathbb{A}^n$, it has dimension strictly lower then $n$. If it would contain an interior point, it would contain an open, which has dimension $n> \text{dim}(X)$, which is impossible. Hence $X$ contains no interior points.
Note that exactly the same argument works in $\mathbb{P}^n$. We can even generalize to nontrival subvarieties of another variety: $X \subsetneq Y$, where of course we call points of $X$ interior if they are contained in some open in $Y$ that is contained in $X$. Then also $X$ contains no interior points.
Now if we are in the case that the field is in fact $\mathbb{C}$, the same holds in the metric topology. If $X$ would contain a metric interior point, it would contain some metric open ball. However $X$ is Zariski closed, so contains the Zariski closure of the open ball, which is the whole of $\mathbb{A}^n$. But you assumed X to be a proper subvariety, contradiction.
(to see that a Zariski closure of a metric open ball is the whole space, you can easily prove that a polynomial that vanishes on a metric open ball is zero everywhere. Just fill in values for all but one of the variables of the polynomial, and use the fundamental theorem of algebra. So the polynomial has a finite amount of zeroes on a line, which is a contradiction if you pick a line through the ball)
Hope this helps.
Joachim
