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location  Rennes, France  
age  62  
visits  member for  4 years, 1 month 
seen  4 hours ago  
stats  profile views  4,197 
1d

comment 
Counterexamples to Elkik's theorem in the nonNoetherian case
$B$ is only smooth over $A[a^{1}]$. 
1d

comment 
irreducible etale cover of a blowup
What does the index $i$ mean? 
Jul 16 
awarded  Yearling 
Jul 7 
comment 
R[[X]] flat as a R[X]module?
If $R[[X]]$ is flat over $R[X]$ then it is flat over $R$. So $R^\mathbb{N}$ must be a flat $R$module; in other words, $R$ is a coherent ring. 
Jul 2 
awarded  Curious 
Jun 25 
comment 
Completion of Bezout Domain a Bezout Domain?
To start with, it is not necessarily a domain: if you take $R=\mathbb{Z}$ and $I=6\mathbb{Z}$, the completion is $\mathbb{Z}_2\times\mathbb{Z}_3$. 
May 27 
comment 
Variant of Hilbert 90 for Galois extensions
@Daniel Loughran: If $g=1$, $\mathrm{Aut}(K)$ is an extension of the finite group of (pointed) automorphisms of an elliptic curve by the group of translations of the same, which is also finite since the ground field is. So, it seems that $\mathrm{Aut}(K)$ is finite in all cases. 
May 22 
comment 
Is every element of $\mathrm{SL}(n,R)$ of finite order diagonalizable?
This argument only decomposes the representation as a direct sum of projective submodules on which $G$ acts via a character. These submodules need not be free, so the question is: what does "diagonalizable" mean? For instance, assume there is a nontrivial invertible $R$module $L$ such that $E:=L\oplus L\cong R^2$. Then the endomorphism of $E$ which is $1$ (resp. $1$) on the first (resp. second) factor $L$ is not diagonalizable in the naive sense. 
May 6 
comment 
A strengthened version of Noether's normalisation lemma?
Clearly not if $\mathrm{Frac}(R)$ is not separable over $k$. Otherwise, I think it's true. I suggest you have a look at Bourbaki. 
May 5 
answered  Constructing a ring whose spectrum is given by order ideals of Z with generic point 
May 5 
comment 
Does every reductive group scheme admit a maximal torus?
@Marty: I don't get it . This only works if $T$ is split. 
Apr 29 
awarded  Enlightened 
Apr 29 
awarded  Nice Answer 
Apr 6 
awarded  Nice Answer 
Mar 28 
comment 
Relation between dimension of Proj(S) and dimension of S
In the definition of Proj, don't you want $P$ to be graded? 
Mar 27 
comment 
Is $G_{\operatorname{red}}$ normal in $G$?
@user76758: in fact, you get exactly my example "in nature" with your construction by taking for $H$ the group of affine automorphisms of the line (except this $H$ is not semisimple). 
Mar 27 
answered  Is $G_{\operatorname{red}}$ normal in $G$? 
Mar 20 
comment 
Is the category of schemes wellpowered? regularly wellpowered?
Unfortunately, a morphism of affine schemes which is a regular mono of schemes (i.e. an immersion) is not necessarily a regular mono in $\mathbf{Aff}$ (i.e. a closed immersion). 
Mar 20 
awarded  Nice Answer 
Mar 20 
comment 
Is the category of schemes wellpowered? regularly wellpowered?
You can take $I=\vert Y\vert$ (the underlying space of $Y$, which is smaller than $\vert X\vert$). For each $y\in I$, choose $V_y$ and $U_y$ such that $y\in V_y$. (We don't need the $U$'s to cover $X$). 