User - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:52:44Z http://mathoverflow.net/feeds/user/2035 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100338/eigen-bundles-of-a-trivial-vector-bundle/100341#100341 Answer by a-fortiori for eigen-bundles of a trivial vector bundle a-fortiori 2012-06-22T10:18:34Z 2012-06-22T10:49:23Z <p>No. Let $C$ be an open affine part of an elliptic curve over the complex numbers and $L$ a non-trivial line bundle on $C$. Now, $L\oplus L^{-1}$ is trivial, so let $G=\mathbf Z/2\mathbf Z$ operate by $1$ on $L$ and by $-1$ on $L^{-1}$.</p> <p>For complete varieties (for simplicity, say having a rational point), all endomorphisms of a free bundle are given by constant matrices, so all direct summands are free again.</p> http://mathoverflow.net/questions/100264/c-equivariant-modules-on-a-vector-bundle-vs-graded-modules-on-the-pushforward/100334#100334 Answer by a-fortiori for C^*-equivariant modules on a vector bundle vs graded modules on the pushforward. a-fortiori 2012-06-22T09:14:54Z 2012-06-22T09:14:54Z <p>I think the picture is clearer if you do everything on $X$. With $\mathcal B=\pi_\ast\mathcal O_E$, the operation of $\mathbf G_m$ becomes a homomorphism $\mathcal B\to\mathcal B\otimes\mathbf C[t^{\pm1}]$ which is a $\mathcal O_X\otimes\mathbf C[t^{\pm1}]$-comodule structure; and $\mathbf G_m$-equivariant quasi-coherent sheaves on $E$ correspond to quasi-coherent $\mathcal B$-modules $\mathcal N$ together with a homomorphism $\mathcal N\to\mathcal N\otimes\mathbf C[t^{\pm1}]$ which is a $\mathcal O_X\otimes\mathbf C[t^{\pm1}]$-comodule structure such that the multiplication $\mathcal B\otimes\mathcal N\to\mathcal N$ is a comodule homomorphism. (For this part, $\mathbf G_m$ may be replaced by any affine group.)</p> <p>Now the comodule structures translate into gradings, and the last condition says that the grading on $\mathcal N$ is compatible with the grading on $\mathcal B$.</p> http://mathoverflow.net/questions/92471/is-the-pushforward-via-a-proper-map-of-a-finite-presentation-module-of-finite-pre/92474#92474 Answer by a-fortiori for Is the pushforward via a proper map of a finite presentation module of finite presentation? a-fortiori 2012-03-28T17:46:36Z 2012-03-28T17:46:36Z <p>Consider a ring $A$ and an ideal $I\subseteq A$, then $A/I$ is finitely presented as an $A/I$-module, but only finitely presented as an $A$-module if $I$ is finitely generated.</p> http://mathoverflow.net/questions/89398/are-injective-quasi-coherent-modules-acyclic/89477#89477 Answer by a-fortiori for Are injective quasi-coherent modules acyclic? a-fortiori 2012-02-25T10:45:58Z 2012-02-25T10:45:58Z <p>The case of quasi-compact semi-separated schemes is treated in the references given in the comments above.</p> http://mathoverflow.net/questions/89473/is-hix-f-finitely-generated-over-gammao-x-if-f-is-coherent/89476#89476 Answer by a-fortiori for Is $H^i(X,F)$ finitely generated over $\Gamma(O_X)$ if $F$ is coherent? a-fortiori 2012-02-25T10:28:34Z 2012-02-25T10:33:36Z <p>Counterexample: For $X=\mathbb A^2\setminus\{0\}$, one has $\Gamma(X,O_X)=\mathbb C[u,v]$, but $H^1(X,O_X)\cong\mathbb C[u^{\pm1},v^{\pm1}]/(\mathbb C[u,v^{\pm1}]+\mathbb C[u^{\pm1},v])$ is not finitely generated over $\mathbb C[u,v]$.</p> http://mathoverflow.net/questions/89030/direct-image-sheaf-and-tensor-product-is-the-projection-formula-an-isomorphism/89164#89164 Answer by a-fortiori for Direct image sheaf and tensor product (is the projection formula an isomorphism?) a-fortiori 2012-02-22T07:58:45Z 2012-02-22T07:58:45Z <p>Most of this becomes obvious when translated using affine charts. For the first question, use a presentation of $N$ and the fact that $(f_A)_*$ is exact to reduce to the case $N=A$. The projection formula is stated in EGA I (new edition), Corollaire 9.3.9.</p> http://mathoverflow.net/questions/88863/flatness-condition-for-local-noetherian-ring-without-nilpotent-elements/88920#88920 Answer by a-fortiori for flatness condition for local noetherian ring without nilpotent elements a-fortiori 2012-02-19T09:54:54Z 2012-02-19T09:54:54Z <p>For a generalization to arbitrary noetherian reduced rings, see EGA III, 7.6.9. Applied to a projective resolution $P_*\to M$ for a finitely generated $A$-module $M$, it says in particular that $d(x)\colon x\mapsto\dim_{k(x)}(M\otimes k(x))$ is semi-continuous on $\mathrm{Spec}(A)$, and $M$ is flat iff $d$ is locally constant. The statement in SGA is a special case since semi-continuity together with $\dim_K(M\otimes K)=\dim_k(M\otimes k)$ already implies that $d$ is constant. A more elementary statement can be found in Mumford's treatment of semi-continuity in his Abelian Varieties (Lemma 1 in section II.5).</p> http://mathoverflow.net/questions/88608/non-finitely-generated-module-is-union-of-countable-chain/88610#88610 Answer by a-fortiori for Non-finitely-generated module is union of countable chain? a-fortiori 2012-02-16T08:15:12Z 2012-02-16T08:15:12Z <p>Counterexample: let $R$ be a valuation ring with value group $\Gamma=\mathbf Z^{\omega_1}$ with the lexicographic ordering and $M$ the maximal ideal of $R$. Any proper submodule of $M$ is contained in an ideal of the form $\{r\in R\mid v(r)>\gamma\}$ for some $\gamma>0$, but every countable collection of positive elements of $\Gamma$ has a positive lower bound, so no countable union of proper submodules can be $M$.</p> http://mathoverflow.net/questions/88529/infinite-products-of-representations-of-the-additive-group/88602#88602 Answer by a-fortiori for Infinite products of representations of the additive group a-fortiori 2012-02-16T06:20:54Z 2012-02-16T06:20:54Z <p>Question A: Demazure-Gabriel, Groupes Algébriques, II, §2, 2.6</p> <p>Question B: It is easily checked that the maximal submodule of $\prod M_i$ on which $\prod f_i$ acts locally nilpotent is the categorical product. This may also be described as $\bigcup_n \prod_i \ker f_i^n\subseteq\prod_i M_i$. Actually, since the set-valued functor $(M,f)\mapsto\ker f^n$ is represented by $(R[X]/(X^n),X)$, so commutes with products, and since the equality $M=\bigcup_n \ker f^n$ is equivalent to $f$ acting locally nilpotent, it is clear that the underlying set of the categorical product must be $\bigcup_n \prod_i \ker f_i^n$.</p> <p>Question C: An example where the filtration $\ker f^n$ does not split is $R=k[[T]]$, $M=k((T))/k[[T]]$, $f=T$.</p> http://mathoverflow.net/questions/88408/is-a-reduced-torsion-free-module-of-finite-rank-over-an-henselian-ring-free/88411#88411 Answer by a-fortiori for Is a reduced, torsion-free module of finite rank over an Henselian ring free? a-fortiori 2012-02-14T08:31:15Z 2012-02-14T10:23:00Z <p>Theorem 19 in Kaplansky's Infinite Abelian Groups gives an example of a torsion-free, reduced, indecomposable rank 2 module for any incomplete discrete valuation ring.</p> <p>In short, the construction is as follows: choose $\lambda\in\hat R\setminus R$. This induces a homomorphism $\tilde\lambda\colon K\to\hat R/R$. The short exact sequence $0\to R\to\hat R\to\hat R/R\to 0$ induces an injection $\mathrm{Hom}(K,\hat R/R)\to\mathrm{Ext}^1(K,R)$. The image of $\tilde\lambda$ under this injection corresponds to the desired rank 2 module $M$ sitting in a non-split extension $0\to R\to M\to K\to 0$. Any divisible element would induce a splitting, so $M$ is reduced. Since there is no surjection $R^2\to K$, the module $M$ cannot be free.</p> http://mathoverflow.net/questions/87794/why-is-the-identity-element-of-a-group-denoted-by-e/87797#87797 Answer by a-fortiori for Why is the identity element of a group denoted by $e$? a-fortiori 2012-02-07T14:15:12Z 2012-02-07T14:15:12Z <p>Heinrich Weber uses Einheit and e in his Lehrbuch der Algebra (1896).</p> http://mathoverflow.net/questions/87024/criterions-for-reflexiveness-of-sheaves-and-a-special-case/87032#87032 Answer by a-fortiori for Criterions for Reflexiveness of sheaves and a special case a-fortiori 2012-01-30T14:21:57Z 2012-01-30T14:38:09Z <p>Let $R$ be a discrete valuation ring and $V$ a free $\mathcal O_X$-module of infinite rank on $X=\mathrm{Spec}(R)$. Then, neither $\mathcal Hom(V,\mathcal O_X)$ nor $\mathcal Hom(\mathcal Hom(V,\mathcal O_X),\mathcal O_X)$ is quasi-coherent. In particular, $V$ is not reflexive.</p> http://mathoverflow.net/questions/86889/ultraproduct-prod-p-mathbbf-p-sim-and-mathbbq/86893#86893 Answer by a-fortiori for Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$ a-fortiori 2012-01-28T10:47:20Z 2012-01-28T10:47:20Z <p>It is easy to see that at least one of $-1,2,-2$ is a square in that field: the set of primes where neither $-1$ nor $2$ is a quadratic residue is contained in the set of primes where $-2$ is a quadratic residue.</p> http://mathoverflow.net/questions/85134/references-to-sga-8-and-descent-theory/85185#85185 Answer by a-fortiori for References to SGA 8 and descent theory a-fortiori 2012-01-08T12:11:35Z 2012-01-08T12:11:35Z <p>For question 1, see the comment above.</p> <p>Collecting the answers to question 2:</p> <ul> <li>Grothendieck's original <em>FGA</em>, starting with <a href="http://www.numdam.org/numdam-bin/fitem?id=SB_1958-1960__5__299_0" rel="nofollow">TDTE I</a></li> <li>Vistoli's chapter in <em>FGA explained</em>, for the connection with stacks</li> <li><a href="http://mathoverflow.net/questions/22032/what-is-descent-theory" rel="nofollow">http://mathoverflow.net/questions/22032/what-is-descent-theory</a> for a very short overview</li> <li>Bosch-Lütkebohmert-Raynaud, <em>Néron Models</em> (recommended by BCnrd in the above-mentioned thread)</li> <li>Waterhouse, <em>Introduction to Affine Group Schemes</em>, containing a 20-page introduction primarily concerned with the affine case</li> </ul> <p>"Community wiki" post, feel free to modify.</p> http://mathoverflow.net/questions/85121/degree-0-vector-bundles/85122#85122 Answer by a-fortiori for degree 0 vector bundles a-fortiori 2012-01-07T09:22:40Z 2012-01-07T09:22:40Z <p>No. Counterexample: On $\mathbf P^1$, take a (two-element) basis of $\Gamma(\mathbf P^1, \mathcal O(1))$, so that the corresponding homomorphism $\mathcal O^2\to\mathcal O(1)$ is surjective. Its kernel is a subbundle of degree $-1$.</p> http://mathoverflow.net/questions/85087/not-locally-of-finite-type-implies-not-universally-closed/85091#85091 Answer by a-fortiori for not locally of finite type implies not universally closed? a-fortiori 2012-01-06T22:08:04Z 2012-01-07T09:18:21Z <p>Let $k$ be a field, $A=k[X_1,X_2,\dots]$ and $I=(X_1,X_2,\dots)$. Then $\mathrm{Spec}(A/I^2)\to\mathrm{Spec}(k)$ is a universal homeomorphism, but not locally of finite type.</p> <p>added in edit: In particular, there is no purely topological condition which implies locally finite type.</p> http://mathoverflow.net/questions/82792/proper-morphism-sending-coherent-to-coherent/82803#82803 Answer by a-fortiori for Proper morphism sending coherent to coherent a-fortiori 2011-12-06T17:16:02Z 2011-12-06T17:16:02Z <p>Gerd Faltings, Finiteness of coherent cohomology for proper fppf stacks, J. Algebraic Geometry 12 (2003) 357–366</p> http://mathoverflow.net/questions/80669/affine-morphism/80672#80672 Answer by a-fortiori for affine morphism a-fortiori 2011-11-11T10:29:52Z 2011-11-11T10:29:52Z <p>$g$ is also affine (EGA II, 1.6.1 (v)), hence finite (EGA III, 4.4.2).</p> http://mathoverflow.net/questions/80668/some-question-about-base-extension/80670#80670 Answer by a-fortiori for some question about base extension a-fortiori 2011-11-11T10:19:05Z 2011-11-11T10:19:05Z <p>A morphism between finite schemes is finite itself (EGA II, 6.1.5 (v)), so $f'$ is finite as well, hence closed (EGA II, 6.1.10). Now apply EGA IV, 2.3.5 (ii).</p> http://mathoverflow.net/questions/80548/affine-scheme-on-speca-of-a-ring-a-as-the-sheafification-of-a-pre-sheave-on-spe/80560#80560 Answer by a-fortiori for Affine scheme on spec(A) of a ring A as the sheafification of a pre-sheave on spec(A)? a-fortiori 2011-11-10T06:17:44Z 2011-11-10T06:27:50Z <p>For any open subset $U\subseteq\mathrm{Spec}(A)$ let $S_U=A\setminus\bigcup_{\mathfrak p\in U}\mathfrak p$ and $\mathscr O'(U)=A[S_U^{-1}]$. It is obviously a presheaf.</p> <p>Claim: For open subsets of the form $U=\mathrm{Spec}(A_f)$ with $f\in A$ we have $\mathscr O'(U)=A_f$. (This shows that the associated sheaf of $\mathscr O'$ is indeed $\mathscr O_{\mathrm{Spec}(A)}$.)</p> <p>Proof: Assume there is an $s\in S_U$ which does not divide $f^n$ for any $n$. The ideal $(s)$ does not meet the multiplicative set $S_f=\{1,f,f^2,\dots\}$, so it is contained in an ideal $\mathfrak q$ which is maximal with respect to this property, but it is well-known that such an ideal $\mathfrak q$ is prime. By construction, $s\in\mathfrak q\in U$, contradicting $s\in S_U$.</p> <p>Applying the usual associated sheaf construction to $\mathscr O'$ seems to be what Hartshorne does when he defines $\mathscr O_{\mathrm{Spec}(A)}$.</p> http://mathoverflow.net/questions/79658/does-isomorphisms-of-sheaf-of-holomorphic-sections-implies-isomorphisms-of-two-ho/79682#79682 Answer by a-fortiori for Does isomorphisms of sheaf of holomorphic sections implies isomorphisms of two holomorphic vector bundles over the same complex space ? a-fortiori 2011-11-01T05:42:22Z 2011-11-01T05:42:22Z <p>Another way of looking at the problem is to consider the (set-valued) sheaf of isomorphisms $E\to F$ and the sheaf of isomorphisms $\mathcal O(E)\to\mathcal O(F)$. There is clearly a map from the former to the latter, so we only need to show that it is an isomorphism locally, i.e., we may assume that both $E$ and $F$ are trivial. But then, isomorphisms $E\cong F$ as well as isomorphisms $\mathcal O(E)\cong\mathcal O(F)$ both have an explicit description by elements of $GL_n(\Gamma(U,\mathcal O))$, and it is easily checked that they correspond.</p> http://mathoverflow.net/questions/79591/question-about-affine-open-covering/79593#79593 Answer by a-fortiori for Question about affine open covering a-fortiori 2011-10-31T06:32:46Z 2011-10-31T06:32:46Z <p>No, this is not true, but it is not what Hartshorne uses. You need $Y$ separated, then the cartesian square $$\begin{matrix} U_i\cap f^{-1}(V) &amp; \to &amp; U_i\times V \\ \downarrow &amp;&amp; \downarrow \\ Y &amp; \to &amp; Y\times Y \end{matrix}$$ exhibits $U_i\cap f^{-1}(V)$ as a closed subscheme of the affine scheme $U_i\times V$. See also EGA I (Springer edition), 5.3.10.</p> http://mathoverflow.net/questions/79503/modules-of-finite-support/79506#79506 Answer by a-fortiori for Modules of finite support a-fortiori 2011-10-30T07:45:53Z 2011-10-30T08:01:31Z <p>If $M$ is finite-dimensional, the ring $A/\mathrm{ann}(M)$ is as well, so it is Artinian, hence it has only finitely many prime ideals, and we have $\mathrm{supp}(M)=\mathrm{Spec}(A/\mathrm{ann}(M))$.</p> <p>Conversely, suppose the support of $M$ is finite. For any $\mathfrak p\in\mathrm{supp}(M)$, the subset $\mathrm{Spec}(A/\mathfrak p)\subseteq\mathrm{supp}(M)$ is finite, but since $A/\mathfrak p$ is finitely generated over $F$, this implies that $\mathfrak p$ is maximal. $M$ has a finite filtration $M=M^0\supseteq M^1\supseteq\dots$ such that each $M^i/M^{i+1}$ is isomorphic to $A/\mathfrak p$ for some $\mathfrak p\in\mathrm{supp}(A)$, and each $A/\mathfrak p$ is finite-dimensional since $\mathfrak p$ is maximal, so we conclude that $M$ itself is finite-dimensional.</p> http://mathoverflow.net/questions/79276/does-a-regular-pair-of-elements-in-a-noetherian-domain-remain-regular-if-their-or/79280#79280 Answer by a-fortiori for Does a regular pair of elements in a noetherian domain remain regular if their order is switched? a-fortiori 2011-10-27T16:07:51Z 2011-10-27T16:07:51Z <p>The only part to be shown is that $a$ is not a zero-divisor on $A/(b)$. Consider some $s\in A$ such that $as\in(b)$, say $as=bt$. Since $b$ is not a zero-divisor on $A/(a)$, we conclude that $t$ maps to zero in $A/(a)$, i.e., $t=au$. Since $A$ is a domain, it follows that $s=bu$, so $s$ maps to zero in $A/(b)$.</p> http://mathoverflow.net/questions/78876/do-the-solutions-to-the-unit-equation-lie-dense-in-the-complex-numbers/78880#78880 Answer by a-fortiori for Do the solutions to the unit equation lie dense in the complex numbers a-fortiori 2011-10-23T09:45:20Z 2011-10-23T10:30:08Z <p>If $f\in\mathbf Z[X]$ is any monic polynomial, the solutions of $x(1-x)\cdot f(x)=1$ are solutions of the unit equation. Take some $y\in U\setminus\mathbf R$. Since the substitution $z\mapsto1/(1-z)$ leaves $S$ invariant, we may assume $|y|>1$. For $n$ given, choose $u,v\in\mathbf R$ such that $y(1-y)\cdot(y^n+uy+v)=1$. Now if $n$ is sufficiently large, Rouché's theorem shows that the number of solutions in a suitable neighbourhood of $y$ in $U$ does not change if we replace $u$ and $v$ by the nearest integers. Hence, $S\cap U$ is nonempty. Since $U$ was arbitrary, this implies that $S\cap U$ is infinite.</p> http://mathoverflow.net/questions/77713/closed-immersion-into-relative-projective-bundle/77722#77722 Answer by a-fortiori for Closed immersion into (relative) projective bundle. a-fortiori 2011-10-10T18:38:21Z 2011-10-10T18:38:21Z <p>For the fact that $\mathbf P(F)\to\mathbf P(E)$ is a closed embedding, see EGA II, 4.1.2. By 4.2.9, an $X$-morphism $f\colon T\to\mathbf P(E)$ factors over $\mathbf P(F)$ iff $(pf)^*E\to f^*O(1)$ factors over $(pf)^*F$. This is equivalent to $(pf)^*L\to f^*O(1)$ being zero, and this again is equivalent to $f$ factoring over the zero locus of $p^*L\to O(1)$.</p> http://mathoverflow.net/questions/77604/for-any-n-does-there-exist-a-number-field-with-at-least-n-solutions-to-the-u/77608#77608 Answer by a-fortiori for For any $n$, does there exist a number field with at least $n$ solutions to the unit equation a-fortiori 2011-10-09T15:40:48Z 2011-10-09T15:40:48Z <p>A slightly more general form of the above mentioned lemma states: whenever $m$ has at least two distinct prime factors and $\zeta_m$ is a primitive $m$-th root of unity, $1-\zeta_m$ is a unit in $\mathbf Z[\zeta_m]$.</p> <p>Choosing $a=1-\zeta_m$ and $b=\zeta_m$ for the various primitive roots of unity, we get $\varphi(m)$ solutions for $K=\mathbf Q(\zeta_m)$. So any such $m$ satisfying $\varphi(m)\geq n$ will do.</p> http://mathoverflow.net/questions/77372/cohomology-of-fixed-point-subspaces/77374#77374 Answer by a-fortiori for Cohomology of fixed point subspaces a-fortiori 2011-10-06T16:58:51Z 2011-10-06T16:58:51Z <p>Example: Let $N\subset\mathbb R^n$ be a compact submanifold. Choose a diffeomorphism $h\colon\mathbb R\to\mathbb R$ satisfying $h(t)\geq t$ for all $t$, and $h(t)=t$ iff $t=0$. Let $M=\mathbb R^n\times\mathbb R$ and $\phi\colon M\to M$, $(x,t)\mapsto(x,h(t)+d(x,N))$. The fixed point set of $\phi$ is $N\times0$, but the homotopy class of $\phi$ does not depend on $N$. (You can easily modify this example so that $M$ is compact.)</p> http://mathoverflow.net/questions/77244/quotient-of-flat-module-is-flat-a-property-in-mumfords-red-book/77250#77250 Answer by a-fortiori for Quotient of flat module is flat - a property in Mumford's Red book a-fortiori 2011-10-05T16:02:21Z 2011-10-05T16:02:21Z <p>This is false without finiteness conditions: let $k$ be a field, $A=k[X,Y]$, $B=A_{(X)}$, $M=B$, $f=X$.</p> http://mathoverflow.net/questions/77035/filtrant-not-necessarily-totally-ordered-projective-system-commuting-with-direc/77124#77124 Answer by a-fortiori for Filtrant (not necessarily totally ordered) projective system commuting with direct sums a-fortiori 2011-10-04T12:04:47Z 2011-10-04T12:04:47Z <p>No. There are nontrivial filtrant projective systems $(M_i)$ of $R$-modules with surjective transition maps such that $\varprojlim M_i=0$. Both $(\varprojlim M_i)^{(X)}$ and $\varprojlim(M_i^{(X)})$ may be identified with submodules of $(\varprojlim M_i)^X=0$, so the canonical map $(\varprojlim M_i)^{(X)}\to\varprojlim(M_i^{(X)})$ is an isomorphism.</p> <p>For how to construct such examples, see <a href="http://math.berkeley.edu/~gbergman/papers/unpub/emptylim.pdf" rel="nofollow">G. Bergman, Some Empty Inverse Limits</a> or G. Higman, A.H. Stone, On inverse systems with trivial limits. J. Lond. Math. Soc. 29, 233-236 (1954)</p> <p>There is also Bourbaki, Topologie generale, III §7 Ex. 2, but the argument given there seems to be incomplete.</p> http://mathoverflow.net/questions/101431/are-there-two-non-isomorphic-modules-such-that-all-the-hom-sets-are-isomorphic Comment by 2012-07-06T11:52:30Z 2012-07-06T11:52:30Z @Piotr: could you please explain the &quot;boils down&quot; a bit further? The implication &quot;isomorphic duals&quot; $\implies$ &quot;all hom spaces isomorphic&quot; seems to require some implication of the sort $2^\kappa=2^\lambda\implies \alpha^\kappa=\alpha^\lambda$ for all cardinals $\alpha$. Is this true? http://mathoverflow.net/questions/101420/music-mathematical-point-of-view-revised Comment by 2012-07-05T19:22:01Z 2012-07-05T19:22:01Z There are certainly attempts at connecting mathematics and music, but the question whether they actually get anywhere near making a connection is <i>subjective and argumentative</i>. http://mathoverflow.net/questions/101294/extension-of-projective-module/101300#101300 Comment by 2012-07-04T10:28:10Z 2012-07-04T10:28:10Z It is also possible that $\mathrm{pd}_B(N)=0\ne\infty=\mathrm{pd}_B(M)$. http://mathoverflow.net/questions/100567/complex-spaces-with-a-global-holomorphic-function-being-a-zero-divisor-in-a-local Comment by 2012-06-27T12:14:53Z 2012-06-27T12:14:53Z As suggested in the paper, $A$ can be a finitely generated algebra over $\mathbf C$. http://mathoverflow.net/questions/100550/rings-of-quaternions/100564#100564 Comment by 2012-06-25T15:01:28Z 2012-06-25T15:01:28Z Computer says that the ring is reversible at least for $s\leq 4$. http://mathoverflow.net/questions/100567/complex-spaces-with-a-global-holomorphic-function-being-a-zero-divisor-in-a-local/100587#100587 Comment by 2012-06-25T12:36:06Z 2012-06-25T12:36:06Z The zero set of $f$ is just $p$, not one of the local branches. http://mathoverflow.net/questions/100567/complex-spaces-with-a-global-holomorphic-function-being-a-zero-divisor-in-a-local Comment by 2012-06-25T07:15:34Z 2012-06-25T07:15:34Z I guess the first example in S. Kleiman, Misconceptions about $K_X$ (L'Enseignement Math&#233;matique, 25(1979), 203-206) still works. http://mathoverflow.net/questions/100264/c-equivariant-modules-on-a-vector-bundle-vs-graded-modules-on-the-pushforward/100334#100334 Comment by 2012-06-23T10:24:31Z 2012-06-23T10:24:31Z Yes. If you already know the argument for the affine case, you can also just check that it works identically for the relatively affine case. http://mathoverflow.net/questions/100338/eigen-bundles-of-a-trivial-vector-bundle/100341#100341 Comment by 2012-06-22T15:19:20Z 2012-06-22T15:19:20Z Right. . http://mathoverflow.net/questions/100228/torsors-of-the-additive-group Comment by 2012-06-22T09:30:36Z 2012-06-22T09:30:36Z Your first &quot;Added&quot; question is vague. Torsors and other global representatives of cohomology classes (extensions as mentioned by Angelo, line bundels) have certainly studied before. As for Taylor's formula, its appearance here is not directly related to torsors, but rather to operations of $\mathbf G_a$, as you already know ( <a href="http://mathoverflow.net/questions/88529/infinite-products-of-representations-of-the-additive-group" rel="nofollow" title="infinite products of representations of the additive group">mathoverflow.net/questions/88529/&hellip;</a> ). http://mathoverflow.net/questions/100264/c-equivariant-modules-on-a-vector-bundle-vs-graded-modules-on-the-pushforward Comment by 2012-06-21T18:41:56Z 2012-06-21T18:41:56Z Could you please explain what problems arise if you just try to verify that the obvious candidate for the inverse functor actually works? http://mathoverflow.net/questions/100245/no-injective-groups-with-more-than-one-element Comment by 2012-06-21T13:52:10Z 2012-06-21T13:52:10Z Google gives <a href="http://zimmer.csufresno.edu/~mnogin/talks/regAMSapril2004.pdf" rel="nofollow">zimmer.csufresno.edu/~mnogin/talks/&hellip;</a> which may not be formally published, but is definitely short. http://mathoverflow.net/questions/100228/torsors-of-the-additive-group Comment by 2012-06-21T13:32:22Z 2012-06-21T13:32:22Z (Sheaf) torsors correspond to Cech H^1 if you take the same topology for both. Angelo's point is that &#233;tale torsors are already Zariski torsors, corresponding to the fact that $H^1_\mathrm{et}(X,\mathcal O_X)=H^1(X,\mathcal O_X)$. http://mathoverflow.net/questions/100228/torsors-of-the-additive-group Comment by 2012-06-21T13:19:23Z 2012-06-21T13:19:23Z see also <a href="http://mathoverflow.net/questions/19339/torsors-in-algebraic-geometry" rel="nofollow" title="torsors in algebraic geometry">mathoverflow.net/questions/19339/&hellip;</a> http://mathoverflow.net/questions/100228/torsors-of-the-additive-group Comment by 2012-06-21T13:15:12Z 2012-06-21T13:15:12Z It is obvious for Cech cohomology.