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In Smith's An Invitation to Algebraic Geometry, following the definition of the projective closure of an affine variety, it was remarked that "the closure may be computed in either the Zariski topology on $\mathbb{P}^n$, or in the Euclidean topology on $\mathbb{P}^n$; the result is the same, and both correspond to our intuitive idea of a closure.'' (Varieties in this book are taken to be over $\mathbb{C}$.)

I was wondering why this is true, since the Zariski topology is coarser than the Euclidean topology. Can someone sketch a proof of this fact? Smith offers no explanation for this.

Partly I think I'm confused about the notion of "Euclidean topology" on projective space. There are at least two topologies that could be considered the "Euclidean topology", and I hope they're the same:

  1. The standard affine cover of $\mathbb{P}^n$ gives rise to charts where the open sets are affine $n$-space $\mathbb{C}^n$. If $\mathbb{C}^n$ is equipped with the Euclidean topology, this makes $\mathbb{P}^n$ a complex manifold.

  2. There is a surjective map from $\pi: \mathbb{C}^{n+1} \setminus \{0\} \to \mathbb{P}^n$ that identifies lines given by $\pi(z_0,\ldots,z_n) = [z_0:\cdots:z_n]$. If $\mathbb{C}^{n+1}$ is given the Euclidean topology, then $\mathbb{P}^n$ can be given the quotient topology. This should be the same as declaring that a set $V$ in $\mathbb{P}^n$ is closed iff its affine cone $\pi^{-1}(V) \cup \{0\}$ is closed in $\mathbb{C}^{n+1}$ with the Euclidean topology. (A related question: If $\mathbb{C}^{n+1}$ is given the Zariski topology instead, is the quotient topology the Zariski topology on $\mathbb{P}^n$?)

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  • $\begingroup$ Zariski open sets are still open in the euclidean topology, so it shouldn't seem too far off that closing them up in Zariski closes them up in euclidean as well and vice versa. $\endgroup$
    – dezign
    Apr 20, 2013 at 1:15

2 Answers 2

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This has little to do with the projective space.

Let $X$ be an algebraic variety over $\mathbb C$. It can be endowed with the natural topology induced by the complex absolute value. This topology is usually called the complex topology.

Lemma Let $Z$ be a complex algebraic variety. Let $T$ be a Zariski closed subset of $Z$, Zariski nowhere dense in $Z$. Then $T$ is complex nowhere dense in $Z$.

Let $Z^0$ be an open subset of a closed subset of $X$ (all in the sense of Zariski topology). We want to show

Claim: the Zariski closure of $Z^0$ coincides with its complex closure.

In your question, $X$ is a projective space, and $Z^0$ is an affine subvariety of $X$.

Proof: Let $Z$ be the Zariski closure of $Z^0$ and let $Z^c$ be the complex closure of $Z^0$. As the Zariski topology is coarse than the complex one, we have $Z^c\subseteq Z$. As $Z\setminus Z^0$ is Zariski closed and Zariski nowhere dense in $Z$, by the previous lemma, $Z\setminus Z^0$ is complex nowhere dense in $Z$, hence $Z^0$ is complex dens in $Z$. This implies that $Z^c=Z$.

It remains to prove the lemma. It is well-known, but I don't have a reference, so let me give a proof here. Shrinking $Z$ if necessary, we can suppose $Z$ is affine and Zariski closed in some $\mathbb C^n$. Let $I, J$ be the respective definining (radical) ideals in $\mathbb C[z_1,\dots, z_n]$ of $Z$ and $T$. By hypothesis, there exists a complex open subset $U$ of $\mathbb C^n$ such that $U\cap Z=U\cap T\ne\emptyset$.

We can suppose wlog that $p:=(0,..,0)\in U\cap T$. In the local ring $\mathcal O_{(\mathbb C^n)^{an},p}$ of germs of holomorphic functions, by analytic Nullstellensatz, we have $I \mathcal O_{(\mathbb C^n)^{an},p}=J\mathcal O_{(\mathbb C^n)^{an},p}$ because $U\cap Z=U\cap T$. Passing to the formal completion, we obtain $$I\mathbb C[[z_1,\dots, z_n]]=J\mathbb C[[z_1,\dots, z_n]].$$ By the faithfull flatness of the formal completion $\mathbb C[z_1,\dots, z_n]_{\mathfrak m} \to \mathbb C[[z_1,\dots, z_n]]$ (where $\mathfrak m$ is the maximal ideal corresponding to $p$), we have $$I\mathbb C[z_1,\dots, z_n]_{\mathfrak m}=J\mathbb C[z_1,\dots, z_n]_{\mathfrak m}.$$ This means that $T$ and $Z$ coincide in a Zariski open neighborhood of $p$. Contradiction with the hypothesis $T$ nowhere dense in $Z$.

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  • $\begingroup$ I just realized that the proof of the lemma can be simplified if we use the dimension: $$\dim_p T^{an}= \dim_p T <\dim_p Z=\dim_p Z^{an}=\dim_p T^{an},$$ the last equality coming from the hypothesis $U\cap Z=U\cap T$. $\endgroup$
    – Cantlog
    Aug 31, 2013 at 21:12
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In case anyone is looking for a reference for this result (as I was), one can consult [$\text{SGA1}_\text{new}$].

More explicitly, recall that for a finite type $\mathbb{C}$-scheme $X$, one has the analytification $X^\mathrm{an}$ which, by definition, is $X(\mathbb{C})$ with 'the analytic topology' (see [$\text{SGA1}_{\text{new}}$, Exposé XII, §1]). By definition there is a continuous map of topological spaces (in fact of ringed spaces) $\varphi\colon X^\mathrm{an}\to X$.

Moreover, recall that a subset $Z$ of $X$ is called locally constructible if for all affine open subsets (equiv. all quasi-compact open subsets) $U\subseteq X$ one has that $Z\cap U$ is constructible (i.e. is a finite union of locally closed subsets)--see (Tag 04ZC and Tag 054B for details).

One then has the following.

Proposition([$\text{SGA1}_\text{new}$, Exposé XII, Proposition 2.2]): For any constructible subset $Z$ of $X$ one has the following equality $$\varphi^{-1}\left(\overline{T}\right)=\overline{\varphi^{-1}(T)}.$$

Note that (again see [$\text{SGA1}_{\text{new}}$, Exposé XII, §1]) the set $\varphi^{-1}\left(\overline{T}\right)$ is equal to $\overline{T}^\mathrm{an}=\overline{T}(\mathbb{C})$. So, in words this says that for locally constructible subsets of $X$ the operations of 'taking closure and then taking $\mathbb{C}$-points' and 'taking $\mathbb{C}$-points and taking closure' commute.

In general [$\text{SGA1}_{\text{new}}$, Exposé XII] is usually the best place to look for foundational results connecting algebraic and analytic geometry.

References:

[$\text{SGA1}_\text{new}$] : https://arxiv.org/pdf/math/0206203.pdf

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