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Answer by a-fortiori for eigen-bundles of a trivial vector bundle

2012-06-22T10:18:34Z

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 $L^{-1}$.

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.

Answer by a-fortiori for C^*-equivariant modules on a vector bundle vs graded modules on the pushforward.

2012-06-22T09:14:54Z

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.)

Now the comodule structures translate into gradings, and the last condition says that the grading on $\mathcal N$ is compatible with the grading on $\mathcal B$.

Answer by a-fortiori for Is the pushforward via a proper map of a finite presentation module of finite presentation?

2012-03-28T17:46:36Z

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.

Answer by a-fortiori for Are injective quasi-coherent modules acyclic?

2012-02-25T10:45:58Z

The case of quasi-compact semi-separated schemes is treated in the references given in the comments above.

Answer by a-fortiori for Is $H^i(X,F)$ finitely generated over $\Gamma(O_X)$ if $F$ is coherent?

2012-02-25T10:28:34Z

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 $\mathbb C[u,v]$.

Answer by a-fortiori for Direct image sheaf and tensor product (is the projection formula an isomorphism?)

2012-02-22T07:58:45Z

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.

Answer by a-fortiori for flatness condition for local noetherian ring without nilpotent elements

2012-02-19T09:54:54Z

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).

Answer by a-fortiori for Non-finitely-generated module is union of countable chain?

2012-02-16T08:15:12Z

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 $M$.

Answer by a-fortiori for Infinite products of representations of the additive group

2012-02-16T06:20:54Z

Question A: Demazure-Gabriel, Groupes Algébriques, II, §2, 2.6

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 $\bigcup_n \prod_i \ker f_i^n$.

Question C: An example where the filtration $\ker f^n$ does not split is $R=k[[T]]$, $M=k((T))/k[[T]]$, $f=T$.

Answer by a-fortiori for Is a reduced, torsion-free module of finite rank over an Henselian ring free?

2012-02-14T08:31:15Z

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.

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.

Answer by a-fortiori for Why is the identity element of a group denoted by $e$?

2012-02-07T14:15:12Z

Heinrich Weber uses Einheit and e in his Lehrbuch der Algebra (1896).

Answer by a-fortiori for Criterions for Reflexiveness of sheaves and a special case

2012-01-30T14:21:57Z

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.

Answer by a-fortiori for Ultraproduct $\prod_p \mathbb{F}_p/\sim$ and $\mathbb{Q}^*$

2012-01-28T10:47:20Z

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.

Answer by a-fortiori for References to SGA 8 and descent theory

2012-01-08T12:11:35Z

For question 1, see the comment above.

Collecting the answers to question 2:

Grothendieck's original FGA, starting with TDTE I
Vistoli's chapter in FGA explained, for the connection with stacks
http://mathoverflow.net/questions/22032/what-is-descent-theory for a very short overview
Bosch-Lütkebohmert-Raynaud, Néron Models (recommended by BCnrd in the above-mentioned thread)
Waterhouse, Introduction to Affine Group Schemes, containing a 20-page introduction primarily concerned with the affine case

"Community wiki" post, feel free to modify.

Answer by a-fortiori for degree 0 vector bundles

2012-01-07T09:22:40Z

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 $-1$.

Answer by a-fortiori for not locally of finite type implies not universally closed?

2012-01-06T22:08:04Z

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.

added in edit: In particular, there is no purely topological condition which implies locally finite type.

Answer by a-fortiori for Proper morphism sending coherent to coherent

2011-12-06T17:16:02Z

Gerd Faltings, Finiteness of coherent cohomology for proper fppf stacks, J. Algebraic Geometry 12 (2003) 357–366

Answer by a-fortiori for affine morphism

2011-11-11T10:29:52Z

$g$ is also affine (EGA II, 1.6.1 (v)), hence finite (EGA III, 4.4.2).

Answer by a-fortiori for some question about base extension

2011-11-11T10:19:05Z

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).

Answer by a-fortiori for Affine scheme on spec(A) of a ring A as the sheafification of a pre-sheave on spec(A)?

2011-11-10T06:17:44Z

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.

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 $\mathscr O_{\mathrm{Spec}(A)}$.)

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 $s\in S_U$.

Applying the usual associated sheaf construction to $\mathscr O'$ seems to be what Hartshorne does when he defines $\mathscr O_{\mathrm{Spec}(A)}$.

Answer by a-fortiori for Does isomorphisms of sheaf of holomorphic sections implies isomorphisms of two holomorphic vector bundles over the same complex space ?

2011-11-01T05:42:22Z

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.

Answer by a-fortiori for Question about affine open covering

2011-10-31T06:32:46Z

No, this is not true, but it is not what Hartshorne uses. You need $Y$ separated, then the cartesian square $$\begin{matrix} U_i\cap f^{-1}(V) & \to & U_i\times V \\ \downarrow && \downarrow \\ Y & \to & Y\times Y \end{matrix}$$ 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.

Answer by a-fortiori for Modules of finite support

2011-10-30T07:45:53Z

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 $\mathrm{supp}(M)=\mathrm{Spec}(A/\mathrm{ann}(M))$.

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.

Answer by a-fortiori for Does a regular pair of elements in a noetherian domain remain regular if their order is switched?

2011-10-27T16:07:51Z

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 $A/(b)$.

Answer by a-fortiori for Do the solutions to the unit equation lie dense in the complex numbers

2011-10-23T09:45:20Z

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.

Answer by a-fortiori for Closed immersion into (relative) projective bundle.

2011-10-10T18:38:21Z

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 $p^*L\to O(1)$.

Answer by a-fortiori for For any $n$, does there exist a number field with at least $n$ solutions to the unit equation

2011-10-09T15:40:48Z

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 $\mathbf Z[\zeta_m]$.

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.

Answer by a-fortiori for Cohomology of fixed point subspaces

2011-10-06T16:58:51Z

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.)

Answer by a-fortiori for Quotient of flat module is flat - a property in Mumford's Red book

2011-10-05T16:02:21Z

This is false without finiteness conditions: let $k$ be a field, $A=k[X,Y]$, $B=A_{(X)}$, $M=B$, $f=X$.

Answer by a-fortiori for Filtrant (not necessarily totally ordered) projective system commuting with direct sums

2011-10-04T12:04:47Z

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.

For how to construct such examples, see G. Bergman, Some Empty Inverse Limits or G. Higman, A.H. Stone, On inverse systems with trivial limits. J. Lond. Math. Soc. 29, 233-236 (1954)

There is also Bourbaki, Topologie generale, III §7 Ex. 2, but the argument given there seems to be incomplete.

Comment by

2012-07-06T11:52:30Z

@Piotr: could you please explain the "boils down" a bit further? The implication "isomorphic duals" $\implies$ "all hom spaces isomorphic" seems to require some implication of the sort $2^\kappa=2^\lambda\implies \alpha^\kappa=\alpha^\lambda$ for all cardinals $\alpha$. Is this true?

Comment by

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 subjective and argumentative.

Comment by

2012-07-04T10:28:10Z

It is also possible that $\mathrm{pd}_B(N)=0\ne\infty=\mathrm{pd}_B(M)$.

Comment by

2012-06-27T12:14:53Z

As suggested in the paper, $A$ can be a finitely generated algebra over $\mathbf C$.

Comment by

2012-06-25T15:01:28Z

Computer says that the ring is reversible at least for $s\leq 4$.

Comment by

2012-06-25T12:36:06Z

The zero set of $f$ is just $p$, not one of the local branches.

Comment by

2012-06-25T07:15:34Z

I guess the first example in S. Kleiman, Misconceptions about $K_X$ (L'Enseignement Mathématique, 25(1979), 203-206) still works.

Comment by

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.

Comment by

2012-06-22T15:19:20Z

Right. .

Comment by

2012-06-22T09:30:36Z

Your first "Added" 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 ( mathoverflow.net/questions/88529/… ).

Comment by

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?

Comment by

2012-06-21T13:52:10Z

Google gives zimmer.csufresno.edu/~mnogin/talks/… which may not be formally published, but is definitely short.

Comment by

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 étale torsors are already Zariski torsors, corresponding to the fact that $H^1_\mathrm{et}(X,\mathcal O_X)=H^1(X,\mathcal O_X)$.

Comment by

2012-06-21T13:19:23Z

Comment by

2012-06-21T13:15:12Z

It is obvious for Cech cohomology.