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1d
comment When does a homogeneous morphism have only finite fibers?
Thanks a lot, dear Theo.
1d
comment When does a homogeneous morphism have only finite fibers?
Yes, also the twisted cubic is the intersection of three suitably chosen irreducible distinct quadrics (my example was related: slightly simpler but geometrically less interesting). I don't know about the loneliness of the last variable. Could you please give the reference of your article?
1d
comment When does a homogeneous morphism have only finite fibers?
If the hypersurfaces $f'_i=0$ have infinitely many intersection points then their intersection must contain some irreducible variety of positive dimension but that variety will certainly not be a common irreducible component of the hypersurfaces in general. The simplest example would be of three distinct smooth quadrics in $\mathbb P^3$ containing a common line.
2d
comment Does a complex manifold always admit an acyclic cover for the sheaf of holomorphic functions?
No, it isn't, so that a covering by Stein open sets has no reason to be acyclic for $\mathcal O^*$, unfortunately.
Apr
18
answered What is an example of transcendental extension such that a monomorphism cannot be extended?
Apr
18
awarded  Guru
Apr
17
answered Are planes without $n$ points isomorphic as algebraic varieties for different n?
Apr
17
comment Does a complex manifold always admit an acyclic cover for the sheaf of holomorphic functions?
Dear @Michael, I don't see why the intersection of two such generalized polydiscs in a manifold, even if assuymed connected, should be a third polydisc, but I have no counter-example to present...
Apr
17
comment Does a complex manifold always admit an acyclic cover for the sheaf of holomorphic functions?
Dear Michael, "polydisc" in an abstract complex manifold means biholomorphically isomorphic to a standard polydisc in $\mathbb C^n$. What else could it mean in Ted's first comment (or in your own question) , which is about coverings of a manifold by open subsets ? Convexity (in the affine sense) doesn't make sense in a manifold. Apologies if that terminology caused some misunderstanding.
Apr
17
comment Complex hypersurface globally defined
You are welcome, Mister Godfrey. And, yes your example is correct: it is due to Serre (1953) and the hypersurface without equation is any connected component of the intersection of the domain with the complex plane $z_2=iz_1$ .
Apr
17
revised Complex hypersurface globally defined
added 278 characters in body
Apr
17
comment What is the class group of the complement of three lines in the projective plane?
No, no: you have to remove the projective line $z=0$ first and then, in the resulting affine plane, remove the union $xy=0$ of the two affine intersecting lines $x=0$ and $y=0$ .
Apr
17
comment Complex hypersurface globally defined
You are welcome, Mister Godfrey.
Apr
17
comment What is the class group of the complement of three lines in the projective plane?
Not so basic: there is a big temptation to claim that the answer is $\mathbb Z/3$ due to the exact sequence you display, forgetting the irreducibility condition on the closed subvariety.
Apr
17
comment Complex hypersurface globally defined
No, because in a polydisk $H^2(M,\mathbb Z)=0$ .
Apr
17
revised What is the class group of the complement of three lines in the projective plane?
edited body
Apr
17
comment What is the class group of the complement of three lines in the projective plane?
Here $Y$ is not irreducible and your exact sequence does not apply ( irreducibility of $U$ is irrelevant) .
Apr
17
answered What is the class group of the complement of three lines in the projective plane?
Apr
17
comment Does a complex manifold always admit an acyclic cover for the sheaf of holomorphic functions?
Dear Michael, I'm not sure this is true but it doesn't matter: the intersection of two polydiscs is Stein and Theorem B holds for all Stein manifolds.
Apr
17
comment Does a complex manifold always admit an acyclic cover for the sheaf of holomorphic functions?
Dear Michael, the intersection of two polydiscs is most certainly not isomorphic to a polydisc: it needn't even be connected! You can check that already in $\mathbb C$, where any simply connected open subset distinct from $\mathbb C$ is isomorphic to a (poly)disc, according to Riemann's uniformization theorem. However the intersection of two polydiscs is Stein: see my answer.