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location Rennes, France
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visits member for 4 years, 6 months
seen 3 hours ago

2h
comment Extending a model to a given compactification of its generic fiber
@OlivierBenoist: right, but you just have to check that this $R$-scheme is separated, e.g. by the valuative criterion.
1d
comment Elliptic curves and connected components
In Joe's argument, all you need to know is that $E(\mathbb{R})$ is a topological group; the connected component of the identity is then a subgroup.
Jan
21
answered Flatness of a morphism of complex analytic spaces
Jan
20
comment A weak analytic version of the valuative criterion of properness
Are you sure condition (c) is what you mean? The inclusion of an open subset of $Y$ satisfies (a), (b) and (c).
Jan
20
answered What if the base change of an algebraic space is representable
Jan
17
comment on the local structure of schemes
In char. $p>0$, you can have $d>\dim X$, e.g. $X=\mathrm{Spec}\,(k[t]/(t^p))$ (here $\Omega$ is free of rank 1).
Jan
13
comment Isotriviality: two definitions
If $X_0$ is stable and singular, the constant family $X_0\times \mathbb{P}^1$ satisfies the second condition but not the first one.
Jan
13
comment The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$
I don't understand the question, but the upvotes suggest I may be dumb.
Jan
12
comment Universal property of categorical quotients
If an orbit space quotient is a categorical quotient by definition, then the answer is trivially yes, by uniqueness of the categorical quotient.
Jan
10
comment Need information about particular kind of quotients of semisimple algebraic groups by free abelian discrete subgroups
Let us continue this discussion in chat.
Jan
10
comment Need information about particular kind of quotients of semisimple algebraic groups by free abelian discrete subgroups
@მამუკაჯიბლაძე: And sorry, I just realized I had written $T=\mathbb{C}^g$ but meant $T=(\mathbb{C}^\times)^g$.
Jan
10
comment Need information about particular kind of quotients of semisimple algebraic groups by free abelian discrete subgroups
@მამუკაჯიბლაძე: If $\Gamma\cong\mathbb{Z}^{2g}$ is an (additive) lattice in $\mathbb{C}^{g}$, any $\mathbb{Z}$-basis of $\Gamma$ contains a $\mathbb{C}$-basis of $\mathbb{C}^{g}$. This implies that $\mathbb{C}^{g}/\Gamma$ is isomorphic (as complex Lie group) to $(\mathbb{C}^{g}/\mathbb{Z}^{g})/L$ where $L\cong\mathbb{Z}^{g}$ is the image of $\Gamma$. Since $\mathbb{C}/\mathbb{Z}\cong\mathbb{C}^\times$, "your" $T/L$'s are exactly the same as their additive analogues.
Jan
10
comment Need information about particular kind of quotients of semisimple algebraic groups by free abelian discrete subgroups
Also, one should assume $\vert q\vert\neq1$: if $q$ is a root of unity, we get the trivial case where $q^\mathbb{Z}$ is finite, and if $\vert q\vert=1$ but $q$ has infinite order, then $q^\mathbb{Z}$ is not a closed subgroup.
Jan
10
comment Need information about particular kind of quotients of semisimple algebraic groups by free abelian discrete subgroups
Indeed it is something of a miracle that $\mathbb{C}^\times/q^\mathbb{Z}$ "is" algebraic, and in fact the higher-dimensional analogue is false, i.e. if $T=\mathbb{C}^g$ and $L\subset T$ is a lattice, then $T/L$ can be any complex torus (not in general algebraic if $g\geq2$).
Jan
2
comment Smoothness and smoothness over formal neighborhood
@prochet: sorry, I had misread $\mathbb{A}^n$ for $\mathbb{A^N}$. And no, I don't know an example.
Dec
22
awarded  Critic
Dec
19
comment Real algebraic solution
@Mostafa: If you know that your variety is absolutely irreducible and has a real point in the smooth locus, then it even has a totally real algebraic point. This follows from the main result of my paper avaliable here.
Dec
19
comment Real algebraic solution
@Alex and Jason: one "special property of the reals" is this: since the field $K$ of real algebraic numbers is real closed, the embedding $K\subset \mathbb{R}$ is "elementary" in the language of ordered fields. This immediately implies the result.
Dec
15
comment Torsors and twists of algebraic groups
@Mostafa: on second thought, we only have $\underline{\mathrm{Aut}}(\mathbb{G}_{a,\mathbb{Q}})=\mathbb{G}_{m,\mathbb{Q}}$. If $S$ has characteristic $p>0$ and $t\in\mathscr{O}_S$, then $x\mapsto x+tx^p$ is an automorphism of $\mathbb{G}_{a,S}$ iff $t$ is locally nilpotent. Thus we have a compatible family of automorphisms over $\mathbb{F}_p[t]/(t^{n+1})$ that does not extend to $\mathbb{F}_p[[t]]$.
Dec
14
comment Is there a scheme parametrizing the closed subgroups of an algebraic group?
And this raises a natural question: is $F$ a disjoint sum of quasicompact subspaces?