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seen Jul 16 at 20:09

Jul
16
answered Cardinality of variety
Jul
16
comment Cardinality of variety
@Hurkyl The birational map allows me to compare the cardinality of an open set in the given variety with an open set in the projective space. I don't know (yet) what the cardinality of the latter open set is. However, I think I have a better idea: The variety $X$ contains an open affine set $U$. $U$ has a surjective regular function (for example : projection on first coordinate). Hence $|X| \ge |k|$. Does this make sense?
Jul
16
comment Cardinality of variety
@Hurkyl Thanks for your comment. Even ignoring induction, I don't know how birationality with a hypersurface in the projective space helps. Is it somehow possible to show that an open subset of the hypersurface has cardinality $ |k |$? This seems to me the same question I'm trying to prove, and no reduction of dimension happens. I would appreciate a more detailed answer, if possible.
Jul
16
asked Cardinality of variety
Jul
15
comment Unique line through two points in projective space
Thanks. On 1) Is this actually well defined? It looks like I get different points if I pick another representative for $ P$. Also why is this line unique? On 2) I found the book. Trying to locate this sequence now. Thanks again.
Jul
15
asked Unique line through two points in projective space
Jul
15
accepted Empty set in projective space has negative dimension?
Jul
9
asked Empty set in projective space has negative dimension?
Jul
9
accepted Hartshorne exercise 1.3.8
Jul
8
asked Hartshorne exercise 1.3.8
Jul
8
accepted What to study from Eisenbud's Commutative Algebra to prepare for Hartshorne's Algebraic Geometry?
Jul
8
comment What to study from Eisenbud's Commutative Algebra to prepare for Hartshorne's Algebraic Geometry?
This is very helpful. Thank you.
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
29
asked What to study from Eisenbud's Commutative Algebra to prepare for Hartshorne's Algebraic Geometry?
Jun
5
awarded  Popular Question
Jun
5
comment Relative de Rham Cohomology is Homotopy Invariant
@Connor Thank you. So both fit in a long exact sequence with the same maps. The theorems I'm familiar with to establish an isomorphism (like five lemma), require an already existing map between $ \Omega(f) $ and $ \Omega(g) $ to make a diagram commutative. I tried to use the identity but it doesn't take kernel elements to kernel elements. I don't know what to do next...
Jun
4
comment Relative de Rham Cohomology is Homotopy Invariant
@Connor I'm sorry if I'm not making myself clear. I'm just interested in proving the existence of an algebra isomorphism using what the book has introduced so far. Constructing an explicit isomorphism is a plus but not necessary. $ f $ and $ g $ are the same on cohomology but not chains. Which boundary map do you mean? Would you mind spelling out the details?
Jun
4
revised Relative de Rham Cohomology is Homotopy Invariant
added 197 characters in body
Jun
4
comment Relative de Rham Cohomology is Homotopy Invariant
@Connor I know that $ f $ and $ g $ induce the same map on cohomology. How can I use this to prove the isomorphism?