# Curve minus point isomorphic to Curve?

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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I think for curves we are done - I would also advocate the claim that no smooth variety $X$, of any dimension, is isomorphic to $X - x$. –  Xandi Tuni May 20 '10 at 8:58
Note that in the analytic case, we have $\mathbb C$ minus the naturals vs. $\mathbb C$ minus the positive integers. –  Allen Knutson May 23 '10 at 1:18

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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YES! That's the right thing to do! –  Xandi Tuni May 20 '10 at 9:02

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.

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This rules out EVERY constructible subset Z since it's Hodge polynomial is always nontrivial. –  VA. May 20 '10 at 18:39

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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Not quite: in characteristic $0$, $\mathbb{P}^1$ and $\mathbb{A}^1$ are both simply connected. –  Pete L. Clark May 20 '10 at 8:37
@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. –  H. Hasson May 20 '10 at 8:43
@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. –  Pete L. Clark May 20 '10 at 8:48
...hmm, true. Fortunately that's the only example where the fundamental group fails. –  Xandi Tuni May 20 '10 at 8:50

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.

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