Let $X$ be a curve and $x\in X$ be a closed point. Is it possible that $X$ and $X-x$ are (abstractly) isomorphic? I wouldn't be surprised if the answer is yes. But what about the classical case, say $X$ a smooth curve over an algebraically closed field?
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$\begingroup$ I think for curves we are done - I would also advocate the claim that no smooth variety $X$, of any dimension, is isomorphic to $X - x$. $\endgroup$– Xandi TuniCommented May 20, 2010 at 8:58
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$\begingroup$ Note that in the analytic case, we have $\mathbb C$ minus the naturals vs. $\mathbb C$ minus the positive integers. $\endgroup$– Allen KnutsonCommented May 23, 2010 at 1:18
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4 Answers
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I don't think that any algebraic variety can be isomorphic to a proper open subset. Over a finite field this follows immediately from counting points over extensions. In the general case one reduces to this by spreading over a finitely generated ring.
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$\begingroup$ YES! That's the right thing to do! $\endgroup$ Commented May 20, 2010 at 9:02
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Although your question has been answered several times over, the following simple argument also works in quite some generality: since the (compactly supported) Euler characteristic satisfies the scissor relation $\chi(X) = \chi(X \setminus Z) + \chi(Z)$ whenever Z is a closed subscheme of X, removing a point (or any subscheme whose Euler characteristic is nonzero) will produce a variety with different Euler characteristic.
By allowing the Euler characteristic to take more "exotic" values, e.g. in the Grothendieck ring of Galois representations, you can rule out almost all cases where Z has Euler characteristic zero, too.
answered May 20, 2010 at 12:11
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2$\begingroup$ This rules out EVERY constructible subset Z since it's Hodge polynomial is always nontrivial. $\endgroup$– VA.Commented May 20, 2010 at 18:39
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I don't think this is possible, for the following reasons.
In the case you have, say, a smooth proper curve $C$ of genus $g$ over $\mathbb C$ minus finitely many points, then you can read the number of "missing" points from the fundamental group of the Riemann surface $C(\mathbb C)$.
In the case of an arbitrary (wlog. algebraically closed) base field, the etale fundamental group will do the same trick.
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$\begingroup$ Not quite: in characteristic $0$, $\mathbb{P}^1$ and $\mathbb{A}^1$ are both simply connected. $\endgroup$ Commented May 20, 2010 at 8:37
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2$\begingroup$ @Pete: you can take off an extra point (to avoid your example) and then look at the prime-to-p fundamental group, and I imagine the rest is basic profinite group theory. $\endgroup$ Commented May 20, 2010 at 8:43
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$\begingroup$ @HH: Yes, absolutely XT's answer can be made to work, as I mentioned in my answer below. There are lots of ways to go: removing another point is a nice one, I agree. $\endgroup$ Commented May 20, 2010 at 8:48
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$\begingroup$ ...hmm, true. Fortunately that's the only example where the fundamental group fails. $\endgroup$ Commented May 20, 2010 at 8:50
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The answer is no. Let me restrict to regular integral curves for simplicity. This perhaps follows most naturally from Zariski's perspective of the "complete Riemann surface" of a function field $k(C)/k$ as the set of all discrete valuations on $k(C)$ which are trivial on $k$. A projective curve is not isomorphic to an affine curve, and if you have an affine curve, you can use this valuation theory to "see what points you're missing".
Also Xandi Tuni's answer of using the etale fundamental group works in every case except for comparing $\mathbb{P}^1$ to $\mathbb{A}^1$ in characteristic $0$, but these curves are obviously not isomorphic for any number of other reasons.
answered May 20, 2010 at 8:43