bio | website | |
---|---|---|
location | Rennes, France | |
age | 63 | |
visits | member for | 4 years, 3 months |
seen | 2 hours ago | |
stats | profile views | 4,333 |
Oct 8 |
answered | About the Dimension of a complete local ring |
Oct 7 |
comment |
Reciprocal polynomials with roots off the unit circle
"Irreducible": over what field? |
Sep 29 |
comment |
Classification of rings satisfying $a^4=a$
I think you mean that $T_0$ is the fixed ring of $\alpha$, not $\alpha(R)$ (the latter is $R$ since $\alpha$ is an involution). |
Sep 23 |
answered | Non trivial family of hyperelliptic curves |
Sep 20 |
comment |
Locally Closed Orbits in Real Algebraic Geometry
Take $G=SL_2$ and $Y=SL_2/(\pm1)=PGL_2$, with $x_0=1$. Then $\Omega=PSL_2(\mathbb{R})=SL_2(\mathbb{R})/(\pm1)$, while $Y(\mathbb{R})$ has another component, containing the class of $\begin{pmatrix}0&i\\ i&0\end{pmatrix}$ (the matrix is not real but its image in $Y$ is). I doubt very much that you can find interesting conditions on $G$ alone (well, "$G$ unipotent" would do, I guess). By the above discussion, the key point is the structure of stabilizers. |
Sep 20 |
answered | Locally Closed Orbits in Real Algebraic Geometry |
Sep 18 |
comment |
Locally Closed Orbits in Real Algebraic Geometry
@Daniel: I think that the "$G$-orbits" here are the orbits of $G(\mathbb{R})$ acting on $X(\mathbb{R})$. |
Sep 17 |
comment |
preservation of localness among certain Krull domains
I agree with Mathias, but I can't prove anything yet. I suggest you look at the 2-dimensional case. In this case, if you denote by $U$ the complement of the closed point in $\mathrm{Spec}(R)$, then $S$ is the ring of global functions on $V:=U\smallsetminus\{\mathfrak{p}\}$, and I would rephrase Mathias's intuition by saying that $V$ might be affine (in which case, of course, $S$ is not local since $\mathrm{Spec}(S)=V$). |
Sep 6 |
revised |
Existence of affine hulls
Added some results (in particular Theorem 1). |
Aug 31 |
answered | Existence of affine hulls |
Aug 28 |
comment |
Counterexamples to Elkik's theorem in the non-Noetherian case
$B$ is only smooth over $A[a^{-1}]$. |
Aug 28 |
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 |