All Questions
Tagged with ac.commutative-algebra derived-categories
36 questions
4
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1
answer
227
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Literature Request: The derived category is Krull-Schmidt
I am looking for literature where it is proven that the derived category of bounded complexes over a finite-dimensional algebra is Krull-Schmidt. I found this question
Literature request: $K^b(\text{...
4
votes
2
answers
285
views
Does $\mathbf R\text{Hom}_R(k, -)\otimes_R^{\mathbf L} k$ commute with co-products?
Let $(R, \mathfrak m, k)$ be a commutative Noetherian local ring. Then, is it true that $\mathbf R\text{Hom}_R(k, -)\otimes_R^{\mathbf L} k$ commutes with arbitrary co-products?
11
votes
1
answer
513
views
When does derived tensor product commute with arbitrary products?
Let $R$ be a commutative Noetherian ring. Let $M$ be an $R$-module. It is well-known that $M$ is finitely generated if and only if the functor $M\otimes_R (-)$ preserves arbitrary products (for ...
3
votes
0
answers
120
views
Derived tensor by perfect complex preserves exact triangle in singularity category?
Let $R$ be a commutative Noetherian ring. Let $\operatorname{D}_{sg}(R)$ be the singularity category of $R$, i.e., the Verdier localization of $D_b(\text{mod } R)$ by the thick subcategory of perfect ...
4
votes
1
answer
267
views
A particular morphism being zero in the singularity category
Let $R$ be a commutative Noetherian ring and $D^b(R)$ be the bounded derived category of finitely generated $R$-modules. Let $D_{sg}(R)$ be the singularity category, which is the Verdier localization $...
5
votes
1
answer
248
views
On the bounded derived category of sheaves with coherent cohomology
Let $(X,\mathcal{O}_X)$ be a locally ringed space such that $\mathcal{O}_X$ is locally notherian, and let $\operatorname{Coh}(\mathcal{O}_X)$ be the category of coherent $\mathcal{O}_X$-modules. The ...
2
votes
1
answer
98
views
A perfect complex over a local Cohen--Macaulay ring whose canonical dual is concentrated in a single degree
Let $R$ be a complete local Cohen--Macaulay ring with dualizing module $\omega$. Let $M$ be a perfect complex over $R$. If the homology of $\mathbf R\text{Hom}_R(M,\omega)$ is concentrated in a ...
2
votes
1
answer
131
views
derived completion and flat base change
Let $f:A \to B$ be a flat morphism of commutative $p$-adic completely rings.
We denote by $D_{\text{comp}}(A)$ the derived category of complexes over $A$, which is derived $p$-adic complete.
For a ...
3
votes
0
answers
106
views
Multiplication map by a ring element on an object vs. all its suspensions in singularity category
Let $R$ be a commutative Noetherian ring, consider the bounded derived category of finitely generated $R$-modules $D^b(R)$ and consider the singularity category $D_{sg}(R):=D^b(R)/D^{perf}(R)$. Let $r\...
1
vote
0
answers
111
views
Kunneth formula for hypercohomology
Let $A_{\bullet}$ and $B_{\bullet}$ be two bounded complexes of sheaves over a variety $X$. Is there a Kunneth-like formula for the hypercohomology of the tensor product $A_{\bullet}\otimes B_{\bullet}...
1
vote
1
answer
149
views
Finitely generated module, which is a virtually small complex, embeds into a module of finite projective dimension?
Let $R$ be a commutative Noetherian ring, and let $\text{mod } R$ denote the abelian category of finitely generated $R$-module. Consider the bounded derived category $D^b(\text{mod } R) $ which is a ...
3
votes
0
answers
161
views
A Nakayama type of claim for countably generated modules on complex affine varieties
Let $U\subset \mathbb{A}^n_{\mathbb{C}}$ be any Zariski open affine subvarity. Let $M$ be an $\mathcal{O}(U)$-module. Suppose $M$ satisfies $M\overset{L}{\otimes}\mathbb{C}_{\mathfrak{M}}\cong 0$ for ...
8
votes
1
answer
413
views
Chain complexes split in the derived category over rings of global dimension 1
Let $R$ be a ring of global dimension $1$. Then I have seen the claim (in a paper, and in this MO post When do chain complexes decompose as a direct sum?) that any chain complex over $R$ is equivalent ...
6
votes
0
answers
399
views
Unbounded derived Nakayama lemma
Let $R$ be a (commutative) local ring, which I don't assume to be noetherian. Let $m$ be its maximal ideal, and $k$ its residue field.
Let $X$ be a complex of $R$-modules with finitely generated ...
4
votes
1
answer
558
views
derived tensor product and finite projective dimension
Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M,N$ be non-zero finitely generated $R$-modules.
Is it known that $M\otimes_R^{\mathbf L} N$ has finite projective dimension if and only if $M$ ...
5
votes
0
answers
586
views
When is the cotangent complex perfect?
Let $X\rightarrow S$ be a proper flat morphism of schemes.
When is the cotangent complex $L_{X/S}$ perfect ?
It is well known, that for local complete intersections the cotangent complex is perfect, ...
6
votes
1
answer
175
views
Relative Ext of Avramov-Martsinkovsky as a derived Hom
Avramov-Martsinkovsky (http://mathserver.neu.edu/~martsinkovsky/Relative.pdf) have defined an exotic version of Ext between two modules over (for simplicity) Gorenstein rings. The basic idea of their ...
5
votes
1
answer
379
views
Jacobson radical of a derived $I$-complete ring
Let $A$ be a commutative ring and $I \subseteq A$ a finitely generated ideal (I am not assuming that $A$ is Noetherian).
Assume that $A$ is derived $I$-complete, meaning, let's say, that $\mathrm{...
10
votes
0
answers
312
views
Triangle $X'\to X\to X''\to\Sigma X'$ splits if $X\simeq X'\oplus X''$?
Given a commutative ring $R$ and a distinguished triangle $X'\to X\to X''\xrightarrow e\Sigma X'$ in the derived category $D(R)$, where $X',X,X''$ are perfect complexes. If we have an equivalence $X\...
4
votes
1
answer
503
views
Mapping cone and derived tensor product
This question is in some sense a continuation to this question: Derived Nakayama for complete modules
For the setting: Let $A$ be a ring and let $I$ be some finitely generated ideal in $A$. Let $f\...
1
vote
1
answer
124
views
Tensoring with complex of finite flat dimension in derived category
Let $(R,m)$ be a Noetherian local ring, and $X$, $Y$ be complexes of finitely generated $R$ modules. Suppose $X$ is bounded above and $Y$ is bounded below. Let $S$ be an $R$-algebra of finite flat ...
3
votes
1
answer
92
views
Pure complex isomorphic to complex of free modules in derived category
This is a basic question, but I don't know the answer. Suppose $M$ is an $R$-module, considered as a complex concentrated in degree $0$. Let $F^*$ be an complex consisting of free modules. Recall that ...
4
votes
0
answers
113
views
Determining whether a morphism is the induced morphism?
Let $F\colon \mathcal A \to \mathcal B$ be a left exact functor between Grothendieck abelian categories. Given a morphism $f\colon A\to B$ in $\mathcal A$ and a morphism $g\colon RF(A)\to RF(B)$ in ...
10
votes
0
answers
241
views
Has anyone seen this construction of dg algebras?
Let $A$ be an associative algebra, $M$ a right $A$-module. Suppose we are given an $A$-module homomorphism $M \to A$. Then we can make $M$ itself into an associative algebra via the multiplication
$$ ...
11
votes
1
answer
1k
views
Do I need to know what an infinity-Gerstenhaber algebra is, and if so, what is it?
I am in the following situation. I have two (rather explicit and specific) dg commutative algebras $R,S$ over a field of characteristic $0$. In fact, $S$ is an $R$-algebra, in that I have a map $R \...
2
votes
1
answer
361
views
If tensor product has finite length, do higher Tors also have finite length?
Let R be a local noetherian ring and let M, N be two finitely generated modules.
Is it true that if $M \otimes_R N$ has finite length, then $Tor_i^R(M,N)$ also has finite length for all i?
I know ...
3
votes
1
answer
173
views
Characterizing the image of $D(A_f) \rightarrow D(A)$
Let $A$ be a ring and $f \in A$ an element. If $M$ is an $A$-module on which multiplication by $f$ is an isomorphism, then $M$ is in fact an $A_f$-module.
Now suppose that $C \in D(A)$ is a complex ...
16
votes
1
answer
754
views
When is every "solid" perfect complex faithful?
Let $R$ be a noetherian commutative ring. Consider $D^{perf}(R)=K^b(R-proj)$ the category of bounded complexes of finitely generated projective $R$-modules, with maps of complexes up to homotopy.
...
1
vote
1
answer
169
views
perfect modules over polynomial algebra
This may be obvious. My question is short:
$R$ is the polynomial algebra $\mathbb{k}[X_{1},\dots , X_{n}]$. Is the $R$-module $\mathbb{k}$ perfect in the sense that $\mathbb{k}$ is a compact object ...
6
votes
0
answers
607
views
On a theorem of Hopkins-Neeman-Thomason on generators of thick subcategories of perfect complexes
Notations and background. Let $R$ be a commutative noetherian local ring and let $D(R)$ denote the derived category of the category of R-modules. A strictly perfect complex on $R$ is a bounded complex ...
8
votes
0
answers
337
views
flatness and derived completion
Let $A$ be a local ring of maximal ideal $\mathfrak{m}$. Let $\hat{A}$ be its completion.
If $A$ is noetherian , then we know that $A\rightarrow\hat{A}$ is faithfully flat.
If $A$ is not noetherian, ...
3
votes
1
answer
378
views
Reference for comparison of heart cohomology with standard cohomology
I'm looking for a reference for the following fact (which I believe to be true and should be easy for people who understand how spectral sequences arise from filtrations).
Let A,B be two hearts of ...
1
vote
0
answers
359
views
Thickness of the category of perfect complexes with finite length homology
Let $R$ be a commutative Noetherian local ring and let $D(R)$ be the derived category of $R$-modules. Recall that a chain complex $C_\bullet$ of modules over $R$ is called perfect if it is isomorphic ...
12
votes
1
answer
1k
views
Fullness of pullback functor in algebraic geometry
Given $f:X\to Y$ a morphism of schemes (or stacks if it's not harder), I am interested in a geometric reformulation of the condition that the functor $f^*:D^b(Coh(Y))\to D^b(Coh(X))$ is full. I can ...
7
votes
1
answer
533
views
Lifting isomorphisms between derived categories
(Remark: I first asked this question at math.stackexchange. As it received no answer, I'm posting it here).
Suppose $A$ and $B$ are commutative rings. Let $A\to B$ be a surjective ring homomorphism. ...
52
votes
7
answers
5k
views
What does a projective resolution mean geometrically?
For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine ...