User charles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:15:51Z http://mathoverflow.net/feeds/user/10541 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71869/an-r-algebra-a-is-r-separable-if-and-only-if-all-derivations-are-inner/72079#72079 Answer by Charles for An R-algebra A is R-separable if and only if all derivations are inner. Charles 2011-08-04T10:21:51Z 2011-08-04T11:55:05Z <p>You should take a look at <em>Théorie de la descente et algèbres d'Azumaya</em>, M.-A. Knus and M. Ojanguren, Théorème 1.4 p 73-74.</p> <p>This contains all the criteria you need and the non-commutative/commutative case</p> http://mathoverflow.net/questions/55244/why-nilpotent-elements-must-be-allowed-in-modern-algebraic-geometry/55257#55257 Answer by Charles for Why nilpotent elements must be allowed in modern algebraic geometry? Charles 2011-02-12T23:54:03Z 2011-02-12T23:54:03Z <p>Another example of this necessity.</p> <p>If $G$ be an algebraic group, you would like it center $\mathcal{Z}(G)$ to be an algebraic group. </p> <p>But for example if you take $SL_n$ defined over a field $k$ (which is absolutely reduced), its center is $\mu_n=Spec\left(\frac{k[X]}{(X^n-1)}\right)$, which is not reduced if $char(k)$ divides $n$.</p> http://mathoverflow.net/questions/54115/when-a-tensor-product-of-two-local-rings-is-a-local-ring/54129#54129 Answer by Charles for When a tensor product of two local rings is a local ring? Charles 2011-02-02T20:00:01Z 2011-02-02T20:18:25Z <p>Here is a partial answer related to connected rings (which is more general than local rings) in the case where rings are not necessarily commutative.</p> <p>If $K$ is a field, let $A$ be an artinian and connected $K$-algebra (in this situation it's equivalent to being local and artinian). Assume that $A$ is endowed with a morphism of algebras $A\longrightarrow K$. Then for any connected $K$-algebra $B$, $A\otimes_K B$ is connected.</p> <p>An trivial counterexample : if $K$ is $\mathbb{F}_2$ and $L=\mathbb{F}_4$, $L\otimes_K L$ is not connected (sorry to write it this way but there seems to be a bug with latex).</p> <p>edit : this results is also true if $K$ is only supposed to be a domain and with some assumption on $B$ (flatness).</p> http://mathoverflow.net/questions/16621/what-do-you-lose-when-passing-to-the-motive/51642#51642 Answer by Charles for What do you lose when passing to the motive? Charles 2011-01-10T12:00:37Z 2011-01-10T21:52:31Z <p>Here the word "motive" will stand for Grothendieck pure motives modulo rational equivalence. Your point 1. is also true for Grassmann bundles. More precisely the following result holds :</p> <p>*Let $E\longrightarrow X$ be a vector bundle of rank $n$, $k\leq n$ and $Gr_k(E)\longrightarrow X$ the associated Grassmann bundle. Then $M(Gr_k(E))\simeq \coprod_{\lambda}M(X)[k(n-k)-\lambda]$, where $\lambda$ runs through all partitions $\lambda=(\lambda_1,...,\lambda_k)$ satisfying $n-k\geq \lambda_1\geq...\geq \lambda_k\geq 0$.*</p> <p>You can prove it in the same fashion as for the projective bundle theorem, as an application Yoneda type lemma for Chow groups.</p> <p>We now know many things on the motives of quadrics. For example if a quadratic form $q$ is isotropic, the motive of the associated quadric $Q$ has a decomposition $\mathbb{Z} \oplus M(Q_1) \oplus \mathbb{Z}[\dim(Q)]$, where $Q_1$ is a quadric of dimension $\dim(Q)-2$ associated to a quadratic form $q_1$ Witt equivalent to $q$. Using it inductively you get the motivic decomposition of split quadrics and for example if $\dim(q)$ is odd and $q$ is split the motive of $Q$ is $\mathbb{Z}\oplus \mathbb{Z}[1]\oplus ... \oplus \mathbb{Z}[\dim(Q)]$. Another very important result is the Rost nilpotence theorem, which asserts that the kernel of the change of field functor on Chow groups of quadrics consists of nilpotents. This result is very fruitfull because it implies that the study of the motive of quadrics can be done over a field which splits the quadric, working with rational cycles in stead of cycles over the base field. Even though these motivic results give severe restrictions on the higher Witt indices of quadrics and have very important applications, the motive does not contain "everything" about the associated quadratic forms (even in terms of higher Witt indices).</p> <p>Another interesting class of varieties to motivic computations are the cellular spaces, i.e. schemes $X$ endowed with a filtration by closed subschemes $\emptyset \subset X_0\subset ... \subset X_n= X$ and affine bundles $X_i\setminus X_{i-1}\rightarrow Y_i$. In this situation the motive of $X$ is isomorphic to the direct sum of (shifts) of the motives of the $Y_i$. For example the filtration of $\mathbb{P}^n$ given by $X_i=\mathbb{P}^i$ and those affine bundles are given by the structural morphism of $\mathbb{A}^{i}$ imply the motivic decomposition $M(\mathbb{P}^n)=\mathbb{Z}\oplus ... \oplus \mathbb{Z}[n]$, and as you can see this is the same motive as odd dimensional split quadrics, so you certainely loose information.</p> <p>The situation is much more complicated replacing quadratic forms by projective homogeneous varieties, but still under some assumption you can recover some results such as Rost nilpotence theorem, and we now begin to have a good description of their motive. Under these assumption the motive of projective homogeneous varieties encodes informations about the underlying variety, such as the canonical dimension, with the example of the computation of those of generalized Severi-Brauer varieties. Some works have also been done to link motives in this case with the higher Tits indices of the underlying algebraic groups.</p> <p>Just to cite a few mathematicians from who we owe these great results : V. Chernousov, N. Karpenko, A. Merkurjev, V.Petrov, M. Rost, N.Semenov, A. Vishik, K. Zainoulline and probably many others that i forgot to mention.</p> <p>edit : to add more precision to the nice answers of Mr. Chandan Singh Dalawat and Mr. Evgeny Shinder, motives of (usual) Severi-Brauer varieties of split algebras are indeed the same as projective space and split quadrics (in odd dimension) but it is obvious that on the base field they're are not necessarily isomorphic since the Severi-Brauer variety is totally split as long as there is a rational point, whereas an isotropic quadratic form is not completely split.</p> http://mathoverflow.net/questions/28496/what-should-be-learned-in-a-first-serious-schemes-course/50141#50141 Answer by Charles for What should be learned in a first serious schemes course? Charles 2010-12-22T09:02:06Z 2010-12-22T09:08:07Z <p>As David and Anweshi told before, think it could be very interesting to deal with functor of points, with main example being subfunctors of Grassmannians. I would make some general statements on functor of points (Yoneda lemma, definition of functor of points, vector bundles) and then begin to study as soon as possible classical examples, such as Grassmanians, Severi-Brauer varieties and their tautological vector bundle, varieties of flag of subspaces...</p> <p>Finally it would lead to a glimpse on group schemes and algebraic groups.</p> http://mathoverflow.net/questions/46367/chow-group-of-a-fiber-product-of-grassmann-bundles Chow group of a fiber product of grassmann bundles Charles 2010-11-17T15:21:58Z 2010-11-17T15:49:10Z <p>Let $X$ be a smooth projective variety and $E\longrightarrow X$ a vector bundle of rank $n$. For any $0\leq k\leq n$ the associated Grassmann bundle $G_k(E)\longrightarrow X$ yields and we have the so-called "basis theorem" (see Fulton "Intersection theory", Proposition 14.6.5) which asserts that for any $s\geq 0$, $$CH_s(G_k(E))=\bigoplus_{\lambda}CH_{s-k(n-k)+|\lambda|}(X)$$ where $\lambda$ runs over all partitions $\lambda=(\lambda_1,...,\lambda_k)$ with $n-k\geq \lambda_1\geq...\geq\lambda_k\geq 0$.</p> <p>I would like to know if there is some similar result for fiber product of Grassman bundles, i.e. consider a vector bundle $E\longrightarrow X$ of rank $n$ and $0\leq k_1\leq k_2\leq n$ two integer : does the $s$-dimensional Chow group $CH_s(G_{k_1}(E)\times_{X} G_{k_2}(E))$ is isomorphic to a direct sum of $\bigoplus_{p\in \mathcal{P}}CH_p(X)$ for some set $\mathcal{P}$ ?</p> <p>I tried to look a an answer considering the immersion $G_{k_1}(E)\times_X G_{k_2}(E)\longrightarrow G_{k_1k_2}(E\times_X E)$ without success.</p> <p>This question studying the case where $X=\mathbb{P}^1$ and $E$ the tautological vector bundle.</p> http://mathoverflow.net/questions/72500/one-side-ideals-in-end-x Comment by Charles Charles 2011-08-09T18:56:45Z 2011-08-09T18:56:45Z Not really suited to mathoverflow...