What is the homotopy type of the affine space in the Zariski topology..?
Solution 1:
Here is a proof that for every $n\ge 2$, the affine space ${\mathbb C}^n$ (with Zariski topology) is contractible. (Since you already know how to do it for $n=1$, I am treating only the case $n\ge 2$.) Sadly, this proof reveals nothing interesting in algebra/algebraic geometry. The same proof works for any variety over ${\mathbb C}$.
Let $\Delta_R$ denote the open disk of radius $R$ in ${\mathbb C}$, centered at $0$.
I will need:
Lemma. For every $m\ge 1$ there exists a holomorphic function $F: \Delta_1\to {\mathbb C}^{m}$ whose graph $\Gamma=\Gamma_F$ satisfies the property that for every proper affine subvariety $V\subset {\mathbb C}^{m+1}$, the intersection $V\cap \Gamma$ is finite.
Proof. Let $f_1,...,f_m: \Delta_2\to {\mathbb C}$ be algebraically independent holomorphic functions. Let $F$ denote the restriction of the function $f=(f_1,...,f_m)$ to the disk $\Delta_1$. The graph of $F$ satisfies the required property. Indeed, algebraic independence of the functions $f_1,...,f_n$ implies that the intersection $V\cap \Gamma_f$ has to be zero-dimensional. But this intersection is an analytic subvariety; hence, its intersection with the tube $\overline{\Delta_1} \times {\mathbb C}^n$ is finite. qed
Now, consider the following map $H: {\mathbb C}^n\times {\mathbb C}\to {\mathbb C}^n$:
$H(z_1,...,z_n,1)=(z_1,...,z_n)$.
$H(z_1,...,z_n, 0)=(0,...,0)$.
Consider the subset $E\subset {\mathbb C}^{n+1}$ consisting of tuples $(z_1,...,z_n,w)$ such that $w\notin \{0, 1\}$. Since the graph $\Gamma$ as above has cardinality continuum, there exists a bijection $H|_E: E\to \Gamma\subset {\mathbb C}^n$.
I claim that the function $H$ thus obtained is continuous in Zariski topology (on domain and the range). Let $V\subset {\mathbb C}^n$ be a Zariski closed subset (an affine subvariety). Suppose first that $V$ does not contain $0\in {\mathbb C}^n$. Since the intersection $V\cap \Gamma$ is finite (and $H$ restricted to the complement of ${\mathbb C}^n \times 0$ is the identity map), $H^{-1}(V)$ is the union of a finite set and $V\times 0$. Such set is clearly Zariski closed. The case when $V$ contains $0$ is similar, you just have to add the subvariety ${\mathbb C}^n \times 0$ to the above inverse image. Therefore, $H$ is continuous. qed
Solution 2:
Here is a more elementary version of Moishe Cohen's argument, which also works over more general fields. Let $K$ be any field of cardinality $\geq 2^{\aleph_0}$; we will show that $K^n$ is contractible in the Zariski topology. Let $B\subset K$ be an algebraically independent set of cardinality $|K|$ and partition $B$ into sets of $n$ elements. Use these sets of $n$ elements as the coordinates of points of $K^n$, giving a set $S\subset K^n$ of cardinality $|K|$. I claim that if $f\in K[x_1,\dots,x_n]$ is a nonzero polynomial, it has only finitely many zeroes in $S$. Indeed, if $f(s)=0$, that means the coordinates of $s$ are algebraically dependent over the field generated by the coefficients of $f$, and this can only happen for finitely many $s\in S$.
Now, as in Moishe Cohen's answer, we can define a contraction $H:K^n\times[0,1]\to K^n$ by $H(x,0)=x$, $H(x,1)=0$, and on $K^n\times (0,1)$, $H$ is given by some bijection $K^n\times(0,1)\to S$ (here is where we use that $|K|\geq 2^{\aleph_0}$).
In fact, we can go even further. Assume $K$ is algebraically closed, and let $X$ be any irreducible variety over $K$. Let $Y\subseteq X$ be a dense affine open subvariety, and by Noether normalization let $f:Y\to K^n$ be a finite surjective morphism. The set $f^{-1}(S)\subset X$ then has the property that its intersection with any Zariski-closed proper subset of $X$ is finite (here is where we use irreducibility of $X$, to guarantee that there is no closed proper subset of $Y$ on which $f$ is surjective). We can then again construct a contraction $H:X\times [0,1]\to X$ as above, with $f^{-1}(S)$ in place of $S$. So any irreducible variety over an algebraically closed field of cardinality $\geq 2^{\aleph_0}$ is contractible in the Zariski topology.