26

You can think about tensor products as a kind of colimit; you're asking the hom functor $\text{Hom}_A(L, -)$ to commute with this colimit in the second variable, but usually the hom functor only commutes with limits in the second variable. Dually, you can think about homs as a kind of limit (in the second variable); you're asking the tensor product functor $(...

answered Dec 17 '15 at 16:49

Qiaochu Yuan

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22

Take $A$ to be Brian Conrad's universal counter example, i.e., the infinite countable product of copies of $\mathbf{F}_2$. Then $A$ is absolutely flat and every finitely presented module is finite locally free. On the other hand, every element of $A$ is an idempotent. Hence if $B$ is a Noetherian ring whose spectrum is connected, then any ring map $A \to B$ ...

22

The example $A=M=\mathbb{Z}$, $L=N=\mathbb{Q}$ shows that the answer is negative: we have $\text{Hom}(L,M)=0$ so $\text{Hom}_A(L,M)\otimes_AN=0$, but $\text{Hom}_A(L,M\otimes_AN)=\text{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q})=\mathbb{Q}\neq 0$

answered Dec 17 '15 at 16:09

Neil Strickland

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19

I think some of the commenters have forgotten the time when they found vector space linear algebra understandable, but triangulated categories confusing.
For someone in such a state, a useful tool to help understand statements about triangulated categories is passage to the Grothendieck group. Recall that this is done by taking the abelian group ...

16

### is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?

Here are a few comments that might be useful. I don't think there is a chance that this can work unless the scheme in question has the resolution property (meaning every coherent sheaf is a quotient of a locally free sheaf of finite rank). Otherwise the category of locally free sheaves does not even form a generator of the category of all quasi-coherent ...

16

This will be a short overview on techniques I am familiar with. For simplicity, I will talk about bounded t-structures, which are determined by their heart $\mathcal{A} = D^{\le 0} \cap D^{\ge 0}$; and on the bounded derived category of coherent sheaves $D^b(X)$ on a variety/stack.
Tilting is in principle extremely powerful: $\mathcal{A}_1$ is obtained by ...

15

For real-analytic manifolds, coherence of the structure sheaf always holds. The 1-sentence reason is that one can pass to real and imaginary parts on Oka's coherence theorem in several complex variables.
Before discussing a proper proof of that 1-sentence executive summary, I should address that in the real-analytic setting however an analogue of one of ...

dg.differential-geometry coherent-sheaves analytic-functions analytic-geometry real-analytic-structures

13

Taking global sections is the same thing as computing the hom from $O_X$. In other words, there is an isomorphism of functors $\Gamma(X,-)\cong\hom(O_X,-)$, so both functors have the same derived functors.
As for using the derived category to define cohomology: yes, simply because that is what the derived category is for!

answered Jun 1 '17 at 2:21

Mariano Suárez-Álvarez

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11

### is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?

Here's a negative answer: there can be no self-dual way to pass from the category of finitely generated projective modules to the category of all finitely generated modules (over, say, a Noetherian ring). To see this, note that the category of finitely generated projective modules is self-dual via the functor $Hom(-,R)$, but the inclusion of projective ...

10

### is the category of coherent sheaves some kind of abelian envelope of the category of vector bundles?

(obviously, in the affine case this question translates into: can the category of (finitely generated) modules be defined via the category of projective modules (of finite rank)?)
Yes (you're assuming Noetherian here, right?). We will need to combine two observations. Let $A$ be an $\text{Ab}$-enriched category and let $\text{Mod}(A)$ be the category of ...

answered Nov 28 '13 at 2:44

Qiaochu Yuan

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10

Dear Rami,
You could see Kollar-Mori, Birational geometry of algebraic varieties. (Page 73)
or
Lazarsfeld, Positivity in Algebraic Geometry I and II. (Page 257)

9

In a smooth case, the reference is Proposition 2.3 in Atiyah and Hirzebruch's Analytic cycles on complex manifolds.
For a non-smooth case, I don't know the general reference, but Theoreme 3 in Henri Cartan's paper Variétés analytiques réelles et variétés analytiques complexes
establishes theorems A and B for coherent (supports of coherent analytic sheaves) ...

ag.algebraic-geometry cv.complex-variables singularity-theory real-algebraic-geometry coherent-sheaves

8

Welcome new contributor. It is best to think about these kinds of examples on one's own. Thus, please try this for yourself before reading the following.
Let $(Y,Z)$ be any pair of a Noetherian scheme $Y$ and a nonempty closed subset $Z$ that is nowhere dense. For instance, let $Y$ be the following Spec of a DVR, $Y=\text{Spec}\ k[t]_{\langle t\rangle}$...

7

Prove that $\chi (\mathcal F \otimes L_1^{n} \otimes L_2^m)$ is a polynomial function of $n$ and $m$ of degree $d$ when $L_1$ and $L_2$ are very ample. You can do this by exactly the same induction argument. (If you have a function of two variables whose successive differences in both variables are degree $d-1$ polynomials, then it is a degree $d$ polynomial....

7

In addition to the Myshkin's answer. An electronic version of the paper can be found here: http://www.inp.nsk.su/~silagadz/Cartan.pdf

7

The question is local, so it is enough to show that if $A$ is a Noetherian local ring with maximal ideal $\mathfrak{m}$ and $M$ is a finitely generated module such that $Ext^i(M,A/\mathfrak{m}) = 0$ for $i > 0$ then $M$ is free. Let $n = \dim(M/\mathfrak{m}M)$ and let $f:A^{\oplus n} \to M$ be a homomorphism that induces an isomorphism modulo $\mathfrak{m}...

7

Rationality of a variety with a full exceptional collection is a well-know folklore conjecture. In some form a similar open question is mentioned in the paper of Brown and Shipman "The McKay Correspondence, Tilting, and Rationality".

7

Does one have the duality in this setting?
Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward functor $\mathbf{R}f_*\colon \mathbf{D}_{qc}(X) \to \mathbf{D}_{qc}(Y)$. It turns out that it exists for any morphism $f\colon X \to Y$ of qcqs schemes. Look at ...

6

No. For example, let $X = P^1$, $U = A^1$ and $F = O_X$. Then $i^*F = O_U$ and the global sections of $i^*F$ is the algebra of polynomials $k[t]$. Therefore $\Gamma(X,i_*i^*F) = \Gamma(U,i^*F) = k[t]$, while $\Gamma(X,F) = k$.

6

There is a vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{A}^1$ whose restriction to $\mathbb{P}^1 \times (\mathbb{A}^1 - \{ 0\})$ is isomorphic to $\mathcal{O}^2$, but whose restriction to $\mathbb{P}^1 \times 0$ is $\mathcal{O}(-1)\oplus \mathcal{O}(1)$.
One can take $E$ to be the cokernel of the map
$$ (x, y, t) : \mathcal{O}(-1) \to \mathcal{O}\oplus ...

6

Let $X$ be a connected reduced noetherian scheme and $\mathscr F$ a coherent sheaf on $X$. Let $\varrho(x)=\dim_{\kappa(x)}\mathscr F_x\otimes \kappa(x)$ where $x\in X$ is a point and $\kappa(x)$ is the residue field at $x$.
Using Nakayama's lemma you can prove the following:
The function $\varrho$ is upper semi-continuous on $X$ in the sense that the set
...

6

I am just writing my comment above as an answer. This follows from Corollaire 7.7.8, EGA III.
Grothendieck, Alexander
Éléments de géométrie algébrique (rédigés avec la collaboration de Jean Dieudonné) : III. Étude cohomologique des faisceaux cohérents, Seconde partie.
Publications Mathématiques de l'IHÉS, 17 (1963), p. 5-91
http://www.numdam.org/numdam-...

5

As long as the complement of your open set is "big" as in "codimension $1$", there is no hope to do anything like this. Here is a concrete example on $\mathbb P^n$: Let $\mathscr L_1$ and $\mathscr L_2$ be two arbitrary line bundles such that there exists no morphism $\phi:\mathscr L_1\to \mathscr L_2$. On $\mathbb P^n$ any line bundle restricts to the ...

5

You have to be a bit careful what "vanish" means in this context. For $k = 2 \bmod 4$, at an elliptic point of order 4, "vanishing as a section of the sheaf" and "vanishing as a function on the upper half-plane" aren't the same thing; it's easy to check that $E_4$ is a local basis of the sections of $\mathcal{G}_k$ in a neighbourhood of the elliptic point. ...

5

No, that is not true, for essentially the same reason that I explained a couple of days ago in a different question. For simplicity, assume that $R$ contains its residue field $k$.
Let $(X,x)$ be a smooth, projective curve over $k$ of genus $g>0$ together with a $k$-point, so that $\text{Pic}^0_{X/k}$ is a smooth, projective variety of positive ...

5

I am not aware of such results in full generality, but I know that working without the properness assumtpion was in part the main motivation for Grothendieck to write SGA 2.
Let me focus on a related question (but not exactly the same). Let $\pi : X \rightarrow Y$ be a flat morphism of finite type with $Y$ a smooth variety over a field. Given a coherent ...

5

Not in general. For example, take $X=\mathbb{P}^1_R$ and let $x\in X$ be a closed point on the special fiber. Then $I_x$, the ideal sheaf of $x$ is torsion free (and thus flat over $R$), but the restriction to the special fiber is not torsion free.

5

Assuming the manifold is Hausdorff, the category of sheaves of modules over the sheaf of smooth functions do the trick.
This category is equivalent to the category of non-degenerate module over the ring of compactly supported smooth functions on the manifold, as it is a commutative ring, it remembers this ring and hence the manifold.
More simply, the ring ...

5

The map is induced by the right adjoint $i^!$ of the pushforward functor and the adjunction morphism $i_*i^! \to \mathrm{id}$, in view of the formula $i^!(F) \cong i^*(F) \otimes \omega_{Z/X}[\dim Z/X]$. This works for any locally complete intersection closed embedding.

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