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22h
awarded  Famous Question
1d
comment What do we call the ring homomorphism $R \rightarrow \mathrm{End}_{\mathbf{Ab}}(X)$ associated with an $R$-module $X$?
Serious question, not being snarky: why does it matter? What use is this object? Thanks!
1d
comment In $\Bbb C$, are polynomials open maps?
They are open. This follows from the open mapping theorem from complex analysis. I actually, off-hand, don't see an easier way to do this.
1d
revised unramified quadratic extension of number field
added 142 characters in body
2d
comment Short exact sequence of groups schemes and dimensions
If your finite flat connected group scheme over an algebraically closed field, then you're of $p$-power order. So, $G^\circ\subseteq G_t$ so $\dim G=\dim G^\circ\leqslant \dim G_t\leqslant \dim G$. EDIT: I noticed that you didn't say $p$-power torsion, but literal $p$-torsion. You should be able to modify my comment accordingly.
Aug
30
awarded  Enlightened
Aug
30
awarded  Nice Answer
Aug
30
comment Is $\mathbb{G}_{m,k}$ (the multiplicative group) simply connected?
Simply connected in the sense that the étale fundamental group is trivial, or simply connected in the sense that all finite isogenies are isomorphisms? Either way, no.
Aug
30
comment When is the geometric Picard group $Pic(X_{\overline{K}})$ of finite type?
That's not quite true—in positive characteristic one can have non-reduced Picard scheme.
Aug
30
comment When is the geometric Picard group $Pic(X_{\overline{K}})$ of finite type?
In particular, it is 23234's suggestion that you need the abelian variety to be dimension $0$ which needs $H^1(X,\mathcal{O}_X)=0$.
Aug
30
comment When is the geometric Picard group $Pic(X_{\overline{K}})$ of finite type?
This should almost never happen. Think about the fact that this group is like the $\bar{K}$-points of an abelian variety which have torsion of all orders.
Aug
26
awarded  Nice Answer
Aug
24
comment Dimension zero of Cohomology Group
Have you considered the canonical degree $g$ map to $\mathbb{P}^1$?
Aug
23
awarded  field-theory
Aug
21
comment Riemann Surface of Genus 1
group operation on $\mathbb{C}/\Lambda$.
Aug
21
comment Riemann Surface of Genus 1
That's a good question. Probably the best way is to defines this map relatively, i.e. defining the group structure on $\mathrm{Hom}(Z,X)$ for any complex manifold $Z$. This is done by doing a relative version of what you have written. The other way is to note that $X$ is, as a complex manifold, equal $\mathbb{C}/\Lambda$ for some $\Lambda\subseteq\mathbb{C}$ a lattice. This follows since the universal cover of genus $1$ surface is $\mathbb{C}$, and has fundamental group $\mathbb{Z}^2$, and so you can conclude. Then, try to show that the operation you have defined is the same thing as the
Aug
21
comment In algebraic topology, for a function $f$ what does $f _\ast$ mean?
Usually it means the induced map on the fundamental groups.
Aug
21
comment question about the dimension of the global section space of a vector bundle
I think you made a mistake. $h^0(C,L\otimes L')=0$. Maybe $L=\mathcal{O}(-1)$ and $L'=\mathcal{O}(1)$
Aug
21
comment question about the dimension of the global section space of a vector bundle
No. Have you tried some examples?
Aug
21
answered References about moduli space of abelian varieties with level structure