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Let $k$ be an algebraically closed field and $\mathbb A_k^n$ be the affine $n$-space. We can give the Zariski topology on it. Then consider $X=\mathbb A_k^m \times \mathbb A_k^n$ under product topology, where each of $\mathbb A_k^m$ and $\mathbb A_k^n$ are given the Zariski topology. And also consider $X=\mathbb A_k^{m+n}$ under the Zariski topology. My question is : For which $m,n\ge 1$, are these two topological spaces homeomorphic ?

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    $\begingroup$ Never for $m,n\geq 1$. $\endgroup$ – Mohan Aug 23 '18 at 0:01
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    $\begingroup$ @Mohan: why ? Note here that I am not asking when are the two topologies same ... I am asking when they are homeomorphic ... so ruling out only the identity map as a homeomorphism is not enough ... $\endgroup$ – user521337 Aug 24 '18 at 5:42
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To elaborate on Mohan's answer, consider the special case of $\mathbb{A} \times \mathbb{A}$ vs $\mathbb{A}^2$. Suppose here that the base field is infinite. Then we know that $\mathbb{A}$ is not Hausdorff so the diagonal is not closed.

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  • $\begingroup$ the diagonal of what is not closed ... ? And so what ... ? $\endgroup$ – user521337 Aug 23 '18 at 23:19
  • $\begingroup$ The diagonal of $\mathbb{A} \times \mathbb{A}$ is not closed. But the diagonal corresponds to the variety $x=y$. $\endgroup$ – Sheel Stueber Aug 23 '18 at 23:44
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    $\begingroup$ that only shows that the identity map is not a homeomorphism ... some other homeomorphism might be there ... $\endgroup$ – user521337 Aug 24 '18 at 5:37
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I'm posting this answer not because the other answer isn't good (in fact I think it's better than this one), but because I'm a fan of the ``do both sides of algebraic geometry to see how the geometry and algebra work together'' point of view, especially when you learn the subject, and thus we should see how the algebra works to fill out both sides of the picture. As a quick piece of notation, I'll write $\mathbb{A}_k^1 \times_{\mathbf{Top}} \mathbb{A}_k^1$ for the topological product of the affine line with itself and $\mathbb{A}^1_k \times_{\mathbf{Sch}} \mathbb{A}_k^1$ for the scheme-theoretic product of the affine line with itself. Consider that since you've assumed $k$ is algebraically closed, the underlying space of the affine line is $\mathbb{A}_k^1 = \lbrace (x - a) \; | \; a \in k \rbrace \cup \lbrace (0) \rbrace$ and so $$ \mathbb{A}_k^1 \times_{\mathbf{Top}} \mathbb{A}_k^1 = \bigg( \lbrace (0) \rbrace \cup \lbrace (x-a) \; | \; a \in k \rbrace\bigg) \times \bigg( \lbrace (0) \rbrace \cup \lbrace (x-a) \; | \; a \in k \rbrace \bigg). $$ However, note that $\mathbb{A}_k^1 \times_{\mathbf{Sch}} \mathbb{A}_k^1$ can be computed from the isomoprhism $$ \mathbb{A}_k^1 \times_{\mathbf{Sch}} \mathbb{A}_k^1 = \operatorname{Spec} k[x] \times_{\mathbf{Sch}} \operatorname{Spec} k[x] \cong \operatorname{Spec}\left(k[x] \otimes_k k[x]\right) \cong \operatorname{Spec} k[x,y], $$ so we can hence describe the points $\mathfrak{p}$ in the underlying space of the scheme-theoreetic product in terms of the prime ideals in $k[x,y]$. Note that each ideal $(x-a,y-b)$ is maximal in $k[x,y]$ for all $a,b \in k$, so there is an embedding $$ \mathbb{A}_k^1 \times_{\mathbf{Top}} \mathbb{A}_k^1 \hookrightarrow \mathbb{A}_k^1 \times_{\mathbf{Sch}} \mathbb{A}_k^1. $$ Thus in order to show these are not homeomorphic we need only exhibit extra points in the scheme theoretic-product. However, since there are irreducible curves in $k[x,y]$, say take $p(x,y) = x^3 + a_2x^2 + a_4x + a_6 - y^2 - a_1xy -a_3y$ to be an elliptic curve in $k$ (so $p$ in particular defines a nonsingular variety), then $(p)$ is irreducible in $k[x,y]$ and hence defines a prime ideal in $k[x,y]$. Then $(p) \in \mathbb{A}_k^1 \times_{\mathbf{Sch}} \mathbb{A}_k^1$ but $(p) \notin \mathbb{A}_k^1 \times_{\mathbf{Top}} \mathbb{A}_k^1$. You can generalize this by noting that you can always get new prime ideals in $k[x_1,\cdots,x_{n+m}]$ that require all $n+m$ variables to be defined.

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    $\begingroup$ please see my comment to the other answer ... $\endgroup$ – user521337 Aug 24 '18 at 5:41

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