17,490 reputation
13780
bio website people.fas.harvard.edu/…
location Cambridge, MA
age
visits member for 3 years, 9 months
seen 2 days ago

I am a senior studying mathematics at Harvard. I'll start graduate school at Berkeley in the fall.


Mar
9
awarded  Enlightened
Mar
9
awarded  Nice Answer
Jul
30
awarded  Nice Answer
Jul
30
awarded  Enlightened
Jul
30
awarded  Nice Answer
Jul
28
awarded  Nice Answer
Jul
20
awarded  Yearling
Jul
19
awarded  Nice Answer
Jun
17
awarded  Good Answer
Jun
1
awarded  Guru
May
18
awarded  Constituent
May
10
comment Existence of divisors of degree one on a curve over a finite field
Incidentally, (since you also explained this argument to me), you might be interested in Matt E's edit below. There's an extra step to get from "points of $\mathrm{Pic}^0$" to "line bundles on $C$" (which works in the case of a finite field).
May
10
comment Existence of divisors of degree one on a curve over a finite field
Thanks for clarifying! I had forgotten that in defining the Picard scheme, one needs to use either a rational point or a sheafification process. So that's apparently where the $H^2$ of $\mathbb{G}_m$ comes from.
May
9
comment Existence of divisors of degree one on a curve over a finite field
Doesn't the fact that the line bundle is defined over $\mathbb{F}_q$ mean that the (or rather, a) divisor one could associate to it has to be defined over $\mathbb{F}_q$? Perhaps I'm missing something...
May
6
awarded  Caucus
Apr
28
comment Why do complex functions have a finite radius of convergence?
@Tsotsi: There's a singular point at $z = 1$; this is a special case of a theorem of Pringsheim. See books.google.com/…
Apr
25
comment What is the homotopy colimit of the Cech nerve as a bi-simplical set?
@Ronald: The conservativity property follows from the long exact sequence of a fibration. One way to see it, though, is that you can think of fibrations over $X$ as being functors from $X$ (considered as an $\infty$-category) into spaces, and the pull-back of a fibration from $Y$ to $X$ corresponds to restricting the functor from $X$ to $Y$. Since every object in $X$ is isomorphic to something in the image of $Y$, the pull-back on functors is conservative.
Apr
25
answered What is the homotopy colimit of the Cech nerve as a bi-simplical set?
Apr
20
comment Computation of the hom-set of a comodule over a coalgebra: $Ext_{E(x)}(k, E(x)) = P(y)$.
@Elden: $\hom_{E(x)}(K, E(x)) \simeq k$. For a comodule $M$, maps from $k$ into $M$ correspond to primitive elements of $M$: that is, those $m \in M$ that map to $1 \otimes m$ under the structure map (note that primitives are dual to indecomposables). The only primitive elements of $E(x)$ are the constants.
Apr
20
comment Computation of the hom-set of a comodule over a coalgebra: $Ext_{E(x)}(k, E(x)) = P(y)$.
There seems to be a mistake in the statement "$Hom_{E(x)}(k, E(x)) = P(y)$," which should perhaps read "$\mathrm{Ext}^{\bullet}_{E(x)}(k, k) = P(y)$."