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Definition by Mumford's red book:

An affine variety is defined as an irreducible algebraic subset of $\mathbb{A}^n$ with a sheaf of rings of $k$-valued functions.

A projective variety is defined as an irreducible algebraic subset of $\mathbb{P}^n$ with a sheaf of rings of $k$-valued functions (page 28).

A variety is defined as a separated prevariety (page 37) and a prevariety is a locally ringed space which is locally isomorphic to an affine variety as locally ringed space.

Question: If we have a non-projective irreducible variety, then is it possible that it is topologically homeomorphic to a closed subset of $\mathbb{P}^n_k$ in the Zariski topology?

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  • $\begingroup$ Why do you care about a topological homeomorphism? This is a very weak and typically inappropriate condition when studying algebraic geometry. $\endgroup$
    – KReiser
    Commented Dec 28, 2020 at 20:15
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    $\begingroup$ It’s also clear that $\mathbb{A}^1_k$ is homeomorphic to $\mathbb{P}^1_k$. $\endgroup$ Commented Dec 28, 2020 at 20:17
  • $\begingroup$ @KReiser Since I want to check some variety is projective, and homeomorphic to a closed subset is easier to prove than closed embedding. $\endgroup$
    – user832207
    Commented Dec 28, 2020 at 20:45
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    $\begingroup$ @Sate No problem. Hopefully that this shows you, as KReiser is suggesting, that homeomorphism isn’t too interesting of a notion in algebraic geometry. In fact, general topology is incredibly useful in algebraic geometry but in a way very different than you might be used to. Usually general topology (discussions of whether spaces $T_i$, simply connected, etc.) are very useful for telling spaces apart. In algebraic geometry this is, for the most part, completely false (except connectedness, which is important). Most algebro-geometric objects don’t have intrinsic topological properties $\endgroup$ Commented Dec 28, 2020 at 20:52
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    $\begingroup$ which distinguish them. Instead, topology is often times more useful purely as a convenient way to phrase certain algebraic properties. This is not to say that general topology is not useful, in fact the opposite is true, and general topology is a secret workhorse behind many deep results in algebraic geometry. But, it is just that, a workhorse—not a marquee feature. Good luck. $\endgroup$ Commented Dec 28, 2020 at 20:53

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Alex's answer in the comments probably suffices, but I'll write one here. As mentioned above, $\Bbb{A}^1_k\cong \Bbb{P}^1_k$ as topological spaces with their Zariski topologies. Actually, the situation is even worse. Any pair of (quasi-compact) curves is homeomorphic because their point sets have the same cardinality and the Zariski topology coincides with the finite complement topology. In particular, given any curve in $\Bbb{P}^n_k$, it is homeomorphic to $\Bbb{A}^1_k$ and we see that your question has an affirmative answer.

As a sidenote: even if we consider curves over $\Bbb{C}$ in the classical topology, smooth curves of genus $1$ (elliptic curves) are all diffeomorphic to a torus, but this does not mean that we regard them as the same in algebraic geometry.

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  • $\begingroup$ Thanks, could you explain why all curves has the same cardinality or could you give me a reference for this fact? $\endgroup$
    – user832207
    Commented Dec 29, 2020 at 0:19
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    $\begingroup$ @Sate I was about to comment on this. Presumably Alekos means quasi-compact curves. For affine curves use Noether normalization to get a finite surjective map to $\mathbb{A}^1_k$. Since $\#(\mathbb{A}^1_k)=\max\{\#(k),\aleph_0\}$, and in particular infinite, it’s easy to see that this implies that the affine curve has the same cardinality. For a quasi-compact curve one can cover by finitely many affine curves to get the result. For non quasi-compact curves you’ll get into trouble if you don’t bound the number of connected components. $\endgroup$ Commented Dec 29, 2020 at 0:29
  • $\begingroup$ @AlexYoucis Thanks for your explanation. $\endgroup$
    – user832207
    Commented Dec 29, 2020 at 0:46
  • $\begingroup$ Yeah, I implicitly assumed quasi-compactness. I'll add it now. Thanks @AlexYoucis $\endgroup$ Commented Dec 29, 2020 at 0:59

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