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I'm reading some notes in which following remark is given:

The Zariski topology is quite different from the usual ones. For example, on affine space $ \mathbb A^n$ a closed subset that is not equal to $ \mathbb A^n$ satisfies at least one non trivial polynomial equation and has therefore necessarily dimension less than $n$, so the closed subsets in Zariski topology are in a sense "very small".

My questions are the following:

  1. What is the meaning of dimension here?
  2. What is the meaning of 'so the closed subsets in Zariski topology are in a sense "very small"'?
  3. What are some other "weird" properties of the Zariski topology?
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    $\begingroup$ An example of "weird" property that might also help for your second question : in the affine space $\mathbb{A}^n$, any open set is dense. So closed sets have no interior point. $\endgroup$ – Arnaud D. Aug 30 '15 at 9:24
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If we consider the Zariski topology on the spectrum of a ring (and not only on its maximal ideals), a point is not necessarily closed. Actually the closed points correspond to maximal ideals. For instance if $A$ is an integral domain,the $0$ prime ideal is dense in $\operatorname{Spec}A$.

Zariski topology is not Hausdorff, but it is Kolmogorov, i.e., given two distinct points there is a neighbourhood of one of them which does not contain the other.

The idea behind Zariski topology is that to know an algebraic variety: we also must know all its subvarieties.

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  • $\begingroup$ Zariski topology on Spec(R) is sometimes Hausdorff $\endgroup$ – Prince M Nov 19 '17 at 2:29
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1) Closed sets are defined by some zero set of an ideal, say $I$, so dimension means the dimension of the ring $k[x_1,...,x_n]/I$.

2) For example, closed sets in $\mathbb{A}^1$ are finite set of points as the zero set of a polynomial is bounded by the degree.

3) Zariski topology is not Hausdorff.

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One way to explain "very small" is that if you think about $\mathbb A^n$ over $\mathbb C$ as $\mathbb C^n$ with the Euclidean topology (which strictly contains the Zariski topology in the sense that Zariski-closed implies Euclidean-closed), all proper Zariski-closed subsets have Lebesgue measure $0$.

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