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An algebraic surface is an algebraic variety of dimension two. In the case of geometry over the field of complex numbers, an algebraic surface has complex dimension two (as a complex manifold, when it is non-singular) and so of dimension four as a smooth manifold.

13 votes

Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$

It may be a bit unfair to compare $X=\mathbb P^1 \times \mathbb P^1$ to $\mathbb P^1$. EDIT: I removed a too optimistic statement about restricting vector bundles on $\mathbb P^3$ to a smooth hypersur …
The Amplitwist's user avatar
5 votes

Q-factorial and rational singularities on surfaces

About the converse: one does need all the assumptions Karl mentioned in his answer. There are $2$-dim. complete local rings which are UFD but does not have rational singularity. One such example (due …
Hailong Dao's user avatar
  • 30.6k
7 votes
Accepted

Algebraic equivalence VS Numerical Equivalence - An Example.

Numerical and algebraic equivalence coincide up to torsion for codimension $1$ cycles in non-singular projective varieties over $\mathbb C$. Also, the group $Alg_{\tau}^1(X)/Alg^1(X)$ can be identifi …
Hailong Dao's user avatar
  • 30.6k
4 votes
1 answer
975 views

Torsion line bundles with non-vanishing cohomology on smooth ACM surfaces

I am looking for an example of a smooth surface $X$ with a fixed very ample $\mathcal O_X(1)$ such that $H^1(\mathcal O(k))=0$ for all $k$ (such thing is called an ACM surface, I think) and a globally …