Timeline for Is there an example of a variety over the complex numbers with no embedding into a smooth variety?
Current License: CC BY-SA 2.5
10 events
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| Dec 28, 2009 at 0:35 | comment | added | VA. | I agree that gluing an affine scheme along a closed subscheme gives an affine scheme via a dual construction for rings. And having an invariant affine cover reduces the problem to the affine case. Very nice explicit example! | |
| Dec 27, 2009 at 23:20 | history | edited | Maharana | CC BY-SA 2.5 |
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| Dec 27, 2009 at 23:19 | comment | added | Maharana | @t3suji: yes you are right. The correct hypothesis is that $Z$ must be such that any finite set of closed points is contained in an affine open set. This is how your example will not work. I will insert this into my answer now. Thanks! | |
| Dec 27, 2009 at 23:04 | comment | added | t3suji | I am almost sure that your Claim is false in general (without assuming existence of an invariant cover, which of course exists in our case). For example, let $Z$ be the affine line with doubled points, $Y=Spec(k[x]/(x^2)\oplus k[y]/(y^2))$, where $x$ and $y$ are the coordinates near the two copies of the doubles point, and $Y'=Spec(k[x,y]/(x^2,xy,y^2))$. | |
| Dec 27, 2009 at 22:51 | history | edited | Maharana | CC BY-SA 2.5 |
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| Dec 27, 2009 at 21:47 | history | edited | Maharana | CC BY-SA 2.5 |
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| Dec 27, 2009 at 21:19 | comment | added | Maharana | @VA: I have edited my answer to include the proof. | |
| Dec 27, 2009 at 21:17 | history | edited | Maharana | CC BY-SA 2.5 |
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| Dec 27, 2009 at 15:46 | comment | added | VA. | "There would surely exist an effective (and nonzero) line bundle on X" for the following reason: take an open affine neighborhood U on the ambient smooth scheme Y which intersects X. Take a divisor $D_U$ on Y intersecting X, and let D be its closure. Then the restriction of $O_Y(D)$ to X is the required effective nonzero line bundle. But why is it obvious that the quotient X is a scheme? It is an algebraic space by general Artin's results but why a scheme? The usual trick of proving that X is projective (and therefore a scheme) does not work. | |
| Dec 27, 2009 at 15:26 | history | answered | Maharana | CC BY-SA 2.5 |