Impact
~217k
people reached
 0 posts edited
 1 helpful flag
 316 votes cast
20h

comment 
Is the positive existential theory undecidable?
@MaryStar Without the evaluation function, the structure becomes "purely algebraic": you can rename the language as $(+,\cdot,D,0,1,X)$ where $X$ is a constant and $D$ is a unary function. The structure makes sense over any field $k$: take the group $k$algebra $k[(k,+)]$, with basis denoted by $(X^\lambda)_{\lambda\in k}$, where $X$ is interpreted as $X^1$ and $D$ acts as by $X^\lambda\mapsto\lambda X^\lambda$. 
20h

comment 
Is the positive existential theory undecidable?
@MaryStar You may not need $k'=0$ but you do need $\lambda\neq0$, which you can formulate as $\exists\mu(\lambda\mu=1)$. 
1d

comment 
If $k$ is an algebraically closed field of any characteristic, then the fundamental group of $A$ is abelian
Another reference is SGA 1, XI, Théorème 2.1. 
Nov
18 
comment 
Maximality of connected components of finite flat group schemes
@nfdc23: OK, so doesn't have connected fibers in general. In SGA, the notation $G^0$ for a group scheme $G$ is used for the "scheme of identity components of fibers" whenever that exists. 
Nov
17 
comment 
Maximality of connected components of finite flat group schemes
In general, $\mathscr{V}^0$ doesn't make sense. 
Nov
8 
comment 
curve over higher dimensional basis with 0dimensional locus of bad reduction
@AriyanJavanpeykar: But we don't know (yet) that $X\in\mathscr{M}_g(S)$. 
Nov
7 
comment 
curve over higher dimensional basis with 0dimensional locus of bad reduction
@TimoKeller: yes, that's right. 
Nov
7 
answered  curve over higher dimensional basis with 0dimensional locus of bad reduction 
Oct
21 
comment 
When is the flatness locus nonempty
@AriyanJavanpeykar : you probably need $f$ to be of finite presentation. In fact if $f$ is locally of finite presentation, then the flat locus in $X$ is open (EGA IV, (11.3.1)). If in addition $f$ is quasicompact, your claim follows. 
Sep
30 
comment 
Scheme of irreducible components
You may also have a look at Matthieu Romagny, Manuscripta 136, 1–32 (2011) (in the context of algebraic stacks). 
Sep
17 
comment 
Is the Jacobian of curve selfdual?
@Felipe: The definition of a polarization involves a positivity condition (it should correspond to an ample divisor class). So, a principally polarized abelian variety is isomorphic to its dual, but I doubt that the converse is true. 
Sep
3 
comment 
Does boundeddegree base extension yield Zariskidense MordellWeil group?
By a theorem of Zarhin, if $A'$ is the dual of $A$, then $(A\times A')^4$ admits a principal polarization. So the principally polarized case (in dimension $8n$) implies the general case. 
Aug
20 
awarded  Popular Question 
Jul
28 
awarded  Nice Answer 
Jul
16 
awarded  Yearling 
Jul
1 
comment 
Vanishing of higher direct image of a morphism with generic fiber $\Bbb{P}^1$
If $Y$ is a DVR, take $X'=Y\times \mathbb{P}^1$, and $X=X'\smallsetminus\{z\}$ where $z$ is a point in the closed fiber. 
Jul
1 
answered  Vanishing of higher direct image of a morphism with generic fiber $\Bbb{P}^1$ 
Jul
1 
comment 
The canonical bundle of an infinitesimal deformation
In fact $X$ need not be proper; or even of finite type (EGA II, 4.5.13). 
Jun
19 
comment 
If the direct sum of cyclic modules is cyclic, what happens to nontrivial extensions?
Sorry, I had indeed assumed your rings were commutative. 
Jun
18 
comment 
If the direct sum of cyclic modules is cyclic, what happens to nontrivial extensions?
The assumption that $M_1\oplus M_2$ is cyclic implies that $\mathrm{Ann}(M_1)+\mathrm{Ann}(M_2)=R$. Since $\mathrm{Ext}^1_R(M_2,M_1)$ is killed by $\mathrm{Ann}(M_1)$ and by $\mathrm{Ann}(M_2)$, it must be zero. 