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location Rennes, France
age 63
visits member for 5 years, 1 month
seen 14 hours ago

Aug
20
awarded  Popular Question
Jul
28
awarded  Nice Answer
Jul
16
awarded  Yearling
Jul
1
comment Vanishing of higher direct image of a morphism with generic fiber $\Bbb{P}^1$
If $Y$ is a DVR, take $X'=Y\times \mathbb{P}^1$, and $X=X'\smallsetminus\{z\}$ where $z$ is a point in the closed fiber.
Jul
1
answered Vanishing of higher direct image of a morphism with generic fiber $\Bbb{P}^1$
Jul
1
comment The canonical bundle of an infinitesimal deformation
In fact $X$ need not be proper; or even of finite type (EGA II, 4.5.13).
Jun
19
comment If the direct sum of cyclic modules is cyclic, what happens to nontrivial extensions?
Sorry, I had indeed assumed your rings were commutative.
Jun
18
comment If the direct sum of cyclic modules is cyclic, what happens to nontrivial extensions?
The assumption that $M_1\oplus M_2$ is cyclic implies that $\mathrm{Ann}(M_1)+\mathrm{Ann}(M_2)=R$. Since $\mathrm{Ext}^1_R(M_2,M_1)$ is killed by $\mathrm{Ann}(M_1)$ and by $\mathrm{Ann}(M_2)$, it must be zero.
Jun
11
comment Can every commutative ring of characteristic $p\in\mathbb P$ be written as the form $R/(p)$ with $R$ being a ring of characteristic $0$?
For the existence, take $R=S\times\mathbb{Q}$.
Jun
9
awarded  Civic Duty
May
29
comment Excellent rings
I suspect that in general, the universally catenary condition may be a source of trouble.
May
26
comment Varieties acted upon faithfully by an abelian variety
Standard construction: Let $G$ be a finite subgroup of $A$ acting freely on a variety $Y$. Then $G$ (resp. $A$) acts on $A\times Y$ by the diagonal action (resp. translation on $A$), and these actions commute so $A$ acts on $X:=A\times Y/G$, and it is easy to see that this action is free. Moreover, if you choose $Y$ so that every morphism $Y\to A$ is constant (e.g. $Y$ rational) then there is no $A$-equivariant morphism $X\to A$, and in particular $X$ is not a product $A\times Z$.
May
25
comment Automorphisms of a differential field and transcendence degree
Perhaps I am missing something, but from the definition it seems clear that $\mathcal{E}_\Phi$ is just $\Phi(\mathcal{E})$ and is isomorphic to $\mathcal{E}$ as a (differential) field.
May
16
awarded  Citizen Patrol
May
13
answered Degree of sum of integral elements over a UFD
May
7
comment Infinitesimal deformations of the formal group of $\mathbb{G}_m$
Oh, right. I had read the question too fast.
May
7
comment Infinitesimal deformations of the formal group of $\mathbb{G}_m$
"The intervention of formal groups is a red herring": why is that? The question is about formal groups.
May
7
comment Is a normal proper relative curve over a DVR projective?
@Lisa: I suppose you may do that. In a more down-to-earth way, you can pick a closed point $x$ of the closed fiber $X_0$ which is in the regular locus $U$, and then a (Weil) divisor $D$ through $x$, finite over $R$. If $R$ is henselian, $D$ splits as $D'\coprod D"$ where $x$ is the only point of $D'$ in the closed fiber, so $D'\subset U$. Repeat this with one $x$ in each component of $X_0$ and take the sum of the $D'$'s: this is an ample Cartier divisor.
May
6
comment Is a normal proper relative curve over a DVR projective?
I think the answer is also yes if $R$ is henselian, if this helps.
May
1
comment Why are unramified maps not required to be locally of finite presentation?
For one thing, it seems reasonable that all immersions should be unramified.