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Let $\mathsf{k}$ be an algebraically closed field and $X$ an abstract variety (an integral separated scheme of finite type over $\mathsf{k}$).

Is there any way to define the usual dimension of $X$ using only the poset structure of its open sets in the Zariski topology?

I have the feeling that the answer is not (but who knows). If it turns out to really have a negative solution, which is, in any possible mathematically precise sense, the smallest amount of information we have to add to the poset of open sets in order to be able to capture the dimension?

As is clear from the accepted answer, I overlooked a lot of elementary stuff before posting here on mathoverflow, and hence I couldn't classify my question as 'research level'. I apologize for being slippery and thank you for your pattience.

I guess the moderators may close it if they want.

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    $\begingroup$ You can even completely recover sober topological spaces from their poset of opens; see for instance the nlab page on sober topological spaces. Of course this reproduces Antoine Labelle's answer along the way, so it is not really an alternative method. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 9 at 3:01

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If I understood the question correctly, the answer is yes and pretty straightforward. The dual of the lattice of open sets is the lattice of closed sets. You can recover the notion of irreducible closed set from the lattice structure: an irreducible closed set is a closed set that cannot be written as a join of two smaller closed sets. Then the dimension is the length of the longest chain of irreducible closed sets.

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  • $\begingroup$ I appreciate very much your answer because it clarifies a lot things for me. So essentially, if I consider not only the poset structure, but the obvious lattice strucutre on open sets, then the answer is yes. $\endgroup$
    – jg1896
    Commented Nov 9 at 3:00
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    $\begingroup$ Being a lattice is not extra structure on a poset, it's a property. $\endgroup$
    – Antoine Labelle
    Commented Nov 9 at 3:01

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