bio | website | math.washington.edu/~kovacs |
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location | Seattle, WA | |
age | ||
visits | member for | 4 years, 4 months |
seen | 4 hours ago | |
stats | profile views | 6,968 |
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 |