Let $k$ be an algebraically closed field. For every positive integer $n$ we can consider the affine $n$-space $\mathbb A^n_k$ with Zariski topology (the topology of zero sets of family of polynomials). As a topological space, $\mathbb A^n_k$ has krull dimension (maximal length of chain of closed irreducible subsets) $n$ , so for $m \ne n$ , $\mathbb A^m_k$ and $\mathbb A^n_k$ are not homeomorphic. My question is : Does there exist positive integers $m\ne n$ such that some open subset of $\mathbb A^m_k$ is homeomorphic with some open subset of $\mathbb A^n_k$ ?
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No, any open subset of $\mathbb{A}^n_k$ is of dimension $n$. Use that the topology on $\mathbb{A}^n_k$ is generated by sets of the form $D(f)=\mathrm{Spec}(k[x_1,\ldots,x_n]_f)$. Now any maximal ideal in $k[x_1,\ldots,x_n]$ not containing $f$ gives rise to a chain of closed subsets of length $n$ in $D(f)$.
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$\begingroup$ What do you mean by "Spec" ? Do you mean prime spectrum or maximal spectrum ? (note that the Zariski topology on $\mathbb A^n_k$ is actually the maximal spectrum of $k[X_1,...,X_n]$ ... $\endgroup$– userJun 8, 2018 at 10:52
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$\begingroup$ I generally work with the spectrum (and not the maximal spectrum) - but I don't think it makes a difference in this simple case. $\endgroup$– LouisJun 8, 2018 at 10:59
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$\begingroup$ Well, there are some subtle differences ... for example the maximal spectrum (hence $\mathbb A^n_k$ ) is always $T_1$ whereas the prime spectrum is $T_1$ iff the ring is zero dimensional ... and by the way , the Zariski topology on $\mathbb A^n_k$ is INDEED actually the maximal spectrum of $\mathbb k[X_1,..,X_n] $ ... so it would be really helpful if you elaborate on your answer $\endgroup$– userJun 8, 2018 at 11:02
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$\begingroup$ are you sure every open set has dimension $n$ ?In proposition 1.10 of Chapter 1 of Hartshorne's Algebraic Geometry, it is only stated and proved that $\dim Y=\dim \overline Y$ for any quasi affine variety $Y$ ... if such a strong claim as yours were to hold I think it would be mentioned there ... can you please provide a detailed proof of your claim ? $\endgroup$– userJun 8, 2018 at 11:15
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$\begingroup$ For any open subset $U \subseteq \mathbb{A}^n_k$, we have $\bar{U}= \mathbb{A}^n_k$, because $\mathbb{A}^n_k$ is irreducible. $\endgroup$– LouisJun 8, 2018 at 11:16