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Jan
28
comment Complete Local Ring and Fermat's Last Theorem
What does this have to do with FLT?
Jan
10
awarded  Good Answer
Dec
21
comment Is $k(\!(x,y)\!)$ a topological field?
Oh, right. Simple, now you say it ;-). So this answers the second question. Thanks!
Dec
21
comment Does the cohomology comparison part of GAGA hold over the reals?
There are lots of nontrivial $X$'s such that $X(\mathbb{R})$ is empty.
Dec
21
comment Is $k(\!(x,y)\!)$ a topological field?
@YCor: for the topology just defined, $(1/f)k[[x,y]]$ is closed, not open.
Dec
21
asked Is $k(\!(x,y)\!)$ a topological field?
Dec
12
awarded  Popular Question
Nov
29
comment Is the positive existential theory undecidable?
@MaryStar: I am not sure what you mean by "have to". What you have to do is state your problem(s) in precise terms. The point of my (and other people's) comments is that you get different problems by including (or not) some sort of evaluation function in your language. All these problems may be interesting.
Nov
28
comment Is the positive existential theory undecidable?
@MaryStar I was only observing that without an evaluation function, it becomes irrelevant to think of $e^t$ as a function of $t$, and in fact "$t$" makes no sense, so it is better to rename $e^t$ and $d/dt$ with abstract symbols such as $X$ and $D$. For instance, I think there is no way to define the complex number $e$ by a formula of your language.
Nov
27
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 $X^\lambda\mapsto\lambda X^\lambda$.
Nov
27
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 $\exists\mu(\lambda\mu=1)$.
Nov
26
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 0-dimensional 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 0-dimensional locus of bad reduction
@TimoKeller: yes, that's right.
Nov
7
answered curve over higher dimensional basis with 0-dimensional locus of bad reduction
Oct
21
comment When is the flatness locus non-empty
@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 quasi-compact, 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 self-dual?
@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.