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bio website math.washington.edu/~kovacs
location Seattle, WA
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visits member for 4 years, 4 months
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I am an algebraic geometer.

Jan
21
revised Is being reduced a generic property of schemes?
typos
Jan
20
revised Is being reduced a generic property of schemes?
added 625 characters in body
Jan
20
revised Is being reduced a generic property of schemes?
typo
Jan
20
answered there exists a hypersurface H ⊂ X such that X \ H is Stein and L is trivial over X \ H
Jan
20
revised Is being reduced a generic property of schemes?
deleted 504 characters in body
Jan
20
comment Is being reduced a generic property of schemes?
You're right. My original answer was different, then I forgot half of the assumptions and made an edit (actually several) and ended up with this. I have a new example which looks much better now. Cheers!
Jan
20
revised Is being reduced a generic property of schemes?
deleted 504 characters in body
Jan
20
revised Is being reduced a generic property of schemes?
deleted 504 characters in body
Jan
19
revised Is being reduced a generic property of schemes?
added 164 characters in body
Jan
19
revised Is being reduced a generic property of schemes?
added 164 characters in body
Jan
19
answered Is being reduced a generic property of schemes?
Jan
17
revised What properties define open loci in excellent schemes?
updated TeX code from the pioneer days of mathoverflow
Jan
11
revised Moving a divisor on a (reducible, non-reduced) curve
corrected typo
Jan
11
comment Moving a divisor on a (reducible, non-reduced) curve
Yes, indeed, that's what I meant. Thanks for spotting it. Cheers!
Jan
11
answered Moving a divisor on a (reducible, non-reduced) curve
Jan
10
comment Moving a divisor on a (reducible, non-reduced) curve
I'll try to write an answer tomorrow to explain what I mean.
Jan
9
comment Moving a divisor on a (reducible, non-reduced) curve
As long as you are not requiring $D$ to be effective (which you can't require for a general $\mathscr L$) you can move the support by adding a principal divisor with the right amount of zeros or poles at the questionable points. Or you could just start with choosing representatives for your divisor that are non-zero and invertible in a neighbourhood of any intersection points. You can always do this for any finite set of points.
Jan
9
comment Leray's theorem up to some degree
@dadexix86: yes. I edited the answer to have it in one place.
Jan
9
revised Leray's theorem up to some degree
added 1540 characters in body
Jan
9
answered Leray's theorem up to some degree