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location  Rennes, France  
age  63  
visits  member for  4 years, 6 months 
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2h

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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

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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 
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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 
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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 
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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 
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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 
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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 
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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 
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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 
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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 
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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 
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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 higherdimensional 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 
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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 
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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 
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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 
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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 
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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? 