most recent 30 from http://mathoverflow.net 2013-05-23T19:07:01Z http://mathoverflow.net/feeds/user/12198 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60070/extensions-of-group-schemes unknown 2011-03-30T14:10:22Z 2011-03-30T23:58:19Z

<p>Hi, I have the following question: why $Ext^1(\mathbb{G}_m,\mathbb{Z})=0$?</p>

http://mathoverflow.net/questions/59081/on-connectness-and-normality unknown 2011-03-21T16:58:30Z 2011-03-22T09:43:35Z

<p>Hi,</p> <p>the situation is the following: I have a projective scheme $\tilde{P}\rightarrow S=Spec(A)$ with $A$ excellent and $I$-adically complete for some ideal of $A$. A group $Y$ acting on $\tilde{P}$ freely in Zariski topology and $P$ is the quotient by $Y$ and it is proper over $S$. Moreover I know that the fiber $\tilde{P}_0$ of $\tilde{P}$ over $S_0=Spec(A/I)$ is connected. I have to prove that $P$ is irreducible. I read that up to replace $\tilde{P}$ with is normalization it can be assumed that $P$ is normal (this is the first thing I do not understand). Assuming this I read that it is enough to show that $P$ is connected. But why?does normal+connected implies irreducible? I have in mind this example: if we take k-planes, $k>2$, in a $\mathbb{P}^n$, for big n, intersecting only in the origin, this is normal (regular in codimension 1 implies normal right?) and connected but not irreducible. Last problem: I read that since $P$ proper over $S$ and $P_0$ connected then $P$ is connected too.</p>

http://mathoverflow.net/questions/51955/references-for-abelian-schemes unknown 2011-01-13T13:45:08Z 2011-03-20T22:19:43Z

<p>Hi, I have a very basic question. I am looking for references explaining how to construct explicitily a Jacobian starting from a curve or examples of projective equations for an abelian scheme. I know a little bit the theory in general so I need examples to fix it, at least in the cases which are not too complicate( or when it is possible). Thank you </p>

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T14:03:10Z 2011-03-15T16:28:32Z

<p>Hi,</p> <p>I have the following situation: $R,H$ schemes (can be assumed noetherian and of finite type) over a field $k$ which we can assume to be algebraically closed, with $H$ reduced, $Y\subset R\times \mathbb{P}_k^n$ an open subset, $p:Y\rightarrow R$ the restriction of the projection onto the first factor and $w:Y\rightarrow H$ a surjective formally smooth morphism. How can I show that $R$ is reduced? Thank you</p>

http://mathoverflow.net/questions/56382/complete-ring-as-union-of-finite-type-algebras unknown 2011-02-23T11:21:29Z 2011-02-23T15:03:54Z

<p>Hi,</p> <p>why the completion of a local ring $R$ can be written as an increasing union of $R$-algebras of finite type?</p>

http://mathoverflow.net/questions/56263/formal-differences unknown 2011-02-22T10:26:01Z 2011-02-23T00:40:50Z

<p>Hi, given a local ring $A$ with maximal ideal $m$ which are differences between $Spec(\hat{A})$ ($\hat{A}$ completion of $A$ along $m$) and $Spf(A)$?</p>

http://mathoverflow.net/questions/54216/local-statement unknown 2011-02-03T17:17:18Z 2011-02-06T21:48:49Z

<p>I have a property which is local and stable for faithfully flat base change over a base scheme $S$. So I need to prove it for $O_{S,s}$ with $s\in S$. Why if I can prove it for a local artinian ring then this give to me the statement for $O_{S,s}$?</p>

http://mathoverflow.net/questions/54214/why-a-reduced-ring-can-be-embedded-into-a-sum-of-integral-rings unknown 2011-02-03T16:58:36Z 2011-02-03T16:58:36Z

<p>Hi, the question is exactly "why a reduced ring (commutative with 1) can be embedded into a sum of integral rings?" Is this simply because in the normalization process we can have many irreducible component?</p>

http://mathoverflow.net/questions/54112/non-discrete-valuation-ring unknown 2011-02-02T17:24:34Z 2011-02-02T18:14:51Z

<p>Hi, I am looking for examples of non-discrete valuation rings. Could you help me? Thanks</p>

http://mathoverflow.net/questions/60070/extensions-of-group-schemes/60084#60084 unknown 2011-03-30T17:03:50Z 2011-03-30T17:03:50Z

@Ralph DVR is it fine, I don't understand why this depends on the ground ring. Are u looking at $\mathbb{Z}$ as the trivial module?

http://mathoverflow.net/questions/60070/extensions-of-group-schemes/60084#60084 unknown 2011-03-30T16:38:58Z 2011-03-30T16:38:58Z

@Ralph it seems to me that the lemma says that $Ext_{\mathbb{G}_m}^1(k,\mathbb{Z})$, it is not clear to me how $\mathbb{Z}$ is a $\mathbb{G}_m$-module and how to relate this group with my question

http://mathoverflow.net/questions/60070/extensions-of-group-schemes unknown 2011-03-30T16:26:59Z 2011-03-30T16:26:59Z

@Bisi Agboola does happens that answer depends on the base?

http://mathoverflow.net/questions/60070/extensions-of-group-schemes/60079#60079 unknown 2011-03-30T16:23:59Z 2011-03-30T16:23:59Z

@Steven Landsburd exponential map is not algebraic

http://mathoverflow.net/questions/59081/on-connectness-and-normality unknown 2011-03-22T14:15:54Z 2011-03-22T14:15:54Z

@karl Schwede those $\tilde{P},P/S$ are a family of varieties which are smooth outside the special fiber over $S_0$, so in your case you are assuming $\tilde{P}_0$ smooth and so irreducible.

http://mathoverflow.net/questions/59081/on-connectness-and-normality unknown 2011-03-22T09:43:08Z 2011-03-22T09:43:08Z

@Karl Schwede sorry, &quot;I have to prove that $P$ is irreducible&quot;

http://mathoverflow.net/questions/59081/on-connectness-and-normality unknown 2011-03-21T18:36:49Z 2011-03-21T18:36:49Z

@Karl Schwede thanks!!If I have understand a prof coul be the following:the normalization is defined as the disjoint union of the normalization of each irreducible component. So $X$ normal+connected implies $X$ is one of these. If I show that locally this is a $Spec(A)$ with $A$ a domain we have integral implies irreducible. By structure thm for integrally closed domain $A=A/p_1\times\dots\times A/p_r$ for $p_i$ minimal primes, but by connectness of $Spec(A)$ there is only one of these and $A$ is integral. So $X$ is integral and that's all. Is it correct? Any idea for the other questions?

http://mathoverflow.net/questions/59081/on-connectness-and-normality unknown 2011-03-21T17:56:48Z 2011-03-21T17:56:48Z

@Francesco Polizzi ok sorry! so why normal+connected implies irreducible?I have in mind a solution even if it seems to me there is someting wrong: if $X=Y\cup Z$ with $Y,Z$ irreducible then consider $Y\coprod Z \rightarrow X$. Is this birational?(I think so) then $X$ normal + Zariski's Main thm implies connected fibers but on the intersection ($Z\cap Y\subset X$) fibers are not connected so $X$ must be irreducible.Is it correct?

http://mathoverflow.net/questions/58746/basic-question-concerning-linear-systems unknown 2011-03-17T18:56:43Z 2011-03-17T18:56:43Z

@Robert Garbary when you write $(x^3,x^2y,x^y^2)=(x^2,xy,x^2)$ do you mean $(x^3,x^2y,x^y^2)=(x^2,xy,y^2)$(brutaly calnceling a $x$)? If it is so you have the Veronese embedding and actually you are looking at the complete $H^0(\mathbb{P}^1,\mathcal{O}(2))$

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T15:48:40Z 2011-03-15T15:48:40Z

@David Holmes, @Martin Bright ok, we have to assume $p$ surjective

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T15:08:05Z 2011-03-15T15:08:05Z

@David Holmes, @Martin Bright ,I found this in the paper of Oda and Seshadri &quot;compactifications of generalized Jacobians&quot; p. 60, there is this affermation after lemma 11.8..but probably I am not able to understand something

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T14:51:06Z 2011-03-15T14:51:06Z

@Martin Bright ok, what about if we assume also $p$ formally smooth or smooth?

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T14:46:02Z 2011-03-15T14:46:02Z

@David Holmes yes, can assume $R$ connected

http://mathoverflow.net/questions/58534/formal-smoothness-versus-reducedness unknown 2011-03-15T14:44:02Z 2011-03-15T14:44:02Z

@David Holmes why $Y$ smooth$/ H$ should imply $R$ reduced?

http://mathoverflow.net/questions/56382/complete-ring-as-union-of-finite-type-algebras/56384#56384 unknown 2011-02-23T12:06:24Z 2011-02-23T12:06:24Z

@Leo Alonso I found this as I wrote as &quot;well known fact&quot; on a paper. The fact that this is an union is crucial in the proof. So I would like to understand this at least in easy cases.