# Join and Zariski closed sets

A set in $\mathbb{C}^n$ is called Zariski-closed if it can be written as the set of zeroes of some set of polynomial equations $$V(f_1,...,f_m) = \left\{ z \in \mathbb{C}^n \mid f_1(z)=...=f_m(z)=0 \right\}$$ and in general as $V(I)$ where $I \subset \mathbb{C}[x_1,...,x_n]$ is an ideal. The Zariski-closed sets form a topology called "Zariski topology". A Zariski-closed set embedded in the affine space $\mathbb{C}^n$ is called affine variety. An open set of an affine variety is called a quasi-affine variety.

It is easy to see that in the case of 1-dimensional affine space $\mathbb{C}$ the Zariski-closed sets are $\emptyset, \mathbb{C}$ and all the finite sets $\{ z_1,...,z_k \}$ (since a polynomial has only a finite number of roots). Then $$[0,1] = \{ z \in \mathbb{R} \subset \mathbb{C} \mid 0 \le z \le 1 \}$$ is clearly not Zariski-closed, but it isn't evident it is Zariski open. So I ask: is it Zariski open? I think it is niether, since its complement is also infinite and hence not Zariski-closed.

Second question, suppose $W$ is an affine variety (and hence Zariski closed), is $W \times [0,1]$ (embedded in higher dimensional affine space) is Zariski closed or Zariski open? Or quasi-affine variety? A typical example is a band or a strip, let $\mathbb{C}^2$ and $W = \{ (x,y) \mid x = 0 \} \in \mathbb{C}^2$ and then $$W \times [0,1] \cong \{ (x,y,z) \mid x = 0 \mbox{ and } 0 \le z \le 1 \}$$ but there are more complicated examples.

The next generalization is as follows. Suppose we have two affine varieties in $\mathbb{C}^n$, call them $X$ and $Y$. The join $J(X,Y)$ of $X$ and $Y$ is defined as follows: $$J(X,Y) = \bigcup_{p \in X, q \in Y} \overline{pq}$$ where $p \in X$ and $q \in Y$ are points in $\mathbb{C}^n$ and $\overline{pq}$ is the line segment joining them, $$\overline{pq} = \left\{ \lambda p + (1-\lambda)q \mid \lambda \in [0,1] \right\} \ .$$ I want to determine if the Zariski closed, Zariski open, a quasi-affine variety or niether of these?

Again, a (relatively) simple example is the join of $y=0$ and $y=1$ in $\mathbb{C}^2$.

Remark: it is hard to visualize the varieties involved since we work over the algebraically closed field $\mathbb{C}$ and not over $\mathbb{R}$ so the Euclidean geometry may give wrong intuition.

References for definitions from Wikipedia: https://en.wikipedia.org/wiki/Zariski_topology and https://en.wikipedia.org/wiki/Join_(topology)

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If a polynomial $f\in\mathbb{C}[x]$ satisfies $f(z)=0$ for all $z\in[0,1]$, then we must have that $f=0$ (this is true of any analytic function).

Let's say that $W\subseteq\mathbb{C}^n$. Applying the above observation, if $f\in \mathbb{C}[x_1,\ldots,x_n,y]$ vanishes on $W\times[0,1]\subseteq\mathbb{C}^{n+1}$, then for any $(z_1,\ldots,z_n)\in W$, the polynomial $f(z_1,\ldots,z_n,y)\in\mathbb{C}[y]$ must be zero everywhere, so that $f$ vanishes on $W\times\mathbb{C}$. Thus, the closure of $W\times[0,1]$ in the Zariski topology on $\mathbb{C}^{n+1}$ is $W\times\mathbb{C}$.

The same reasoning applies to any line segment, and hence in particular to any $\overline{pq}$; thus, $J(X,Y)$ will never be Zariski-closed except in degenerate cases (e.g. $X$ or $Y$ being $\varnothing$ or $\mathbb{C}$, or $X=Y$). The same reasoning applies to the complement of any line segment, so that $J(X,Y)$ will never be Zariski-open.

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Thanks, I thought so. When you talk about the degenerate cases you mentioned $X = \mathbb{C}$. What about the case when $X = \mathbb{C}^3$ and $Y = \mathbb{C}$ and they are algebraic affine varieties embedded in $\mathbb{C}^4$? I thik that $J(X,Y)$ won't be closed or open by the above arguments (this is a sort of the strip example I mentioned in my question. –  LinAlgMan Mar 10 '13 at 16:26
To put it bluntly, the real segment $[0,1]$ is persona non grata in algebraic geometry circles. We algebraic geometers can be quite an elitist, snobbish, unwelcoming crowd :-) –  Georges Elencwajg Mar 10 '13 at 20:03