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Questions tagged [homological-algebra]

Homological algebra studies homology and cohomology groups in a general algebraic setting, that of chains of vector spaces or modules with composable maps which compose to zero. These groups furnish useful invariants of the original chains.

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Does the pullback always exist in an abelian category?

I have the following definition of pullback: given any two morphisms $f \colon A \rightarrow C$, $g \colon B \rightarrow C$, we say that the triple $(P, p_1, p_2)$ is a pullback of $f$ and $g$ if $f \...
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Flat Module from book of NS Gopala Krishnan

I was reading flat modules from book by NS Gopalakrishnan. In starting of faithfully flat algebra the below is written Let $A$ be an $R$-algebra, $M$ and $N$ are $R$-modules. Then homomorphism $\phi_M ...
Swaraj Koley's user avatar
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Rotman-Homological Algebra Theorem 6.32

I am reading Rotman's "An Introduction to Homological Algebra" and am stuck on the proof of Theorem 6.32 (page 355-356): I do not understand why the equalities $W = \text{Tor}_1(K_{i-1}, V_{...
Jackson Wilson's user avatar
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2 answers
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Help finding a morphism between injective resolutions

I'd like to prove the following result of homological algebra: In an abelian category $\mathcal{C}$, given two objects $A$ and $B$, a morphism $\varphi \colon A \rightarrow B$ and two resolutions $I^{\...
Paz's user avatar
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Homological Algebra for Analysis

While studying differential forms, I encountered some concepts from homological algebra, such as (co)chain complexes, de Rham cohomology, pullbacks, and others. Is it reasonable to study the basics of ...
veirab's user avatar
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Is there a Baer's criterion for testing injectivity of sheaves of $\mathcal{O}_X$-modules?

In the important paper by Spaltenstein on resolving unbounded complexes, they turn their hand to sheaves. Let us fix a ringed space $(X;\mathcal{O}_X)$. In the proof of Lemma $4.3$ it is implicitly ...
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Second cohomology of Klein four group [closed]

Let $V_4=\{a,b|a^2=b^2=1; ab=ba\}$ be the Klein four group. Then the second cohomology group $H^2(V_4,\mathbb{C}^*)$ is isomorphic to $\mathbb{Z}_2$ (using Schur multiplier). But, I want to compute ...
MANI's user avatar
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Stable Koszul complex depends only on the radical

In the following paper: https://arxiv.org/pdf/1601.02473 in Appendix A, the following definitions are given: for an element $\alpha$ of a commutative Noetherian ring $R$, the stable Koszul complex is ...
Igor Sikora's user avatar
1 vote
2 answers
60 views

Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

Let $I$ be a directed set. Let consider the direct limit in the category of abelian groups. Suppose $A\cong \varinjlim_{i\in I} A_i$. Then, is it true that $A[2]\cong \varinjlim_i (A_i[2])$ ? Here, $[...
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Recommend reference books on homology algebra

I have just started learning homological algebra and hope to get some recommendations for beginner-friendly reference books. Currently, I am reading related content on the Stacks Project and Rotman's &...
jhzg's user avatar
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Understanding how group cohomology classifies extensions using the derived functor point of view

I am rereading some material about group extensions, in particular because I needed to recall the formula $$H^2(G;A)\cong \mathcal{E}(G;A).$$ We have that $G$ is some group acting on an abelian group $...
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${\mathrm{Ext}}_{k[\epsilon]}^1(k, k)$ and ${\mathrm{Ext}}_{k[X]}^1(k, k)$.

For a field $k$, I am calculating ${\mathrm{Ext}}_{k[\epsilon]}^1(k, k)$ and ${\mathrm{Ext}}_{k[X]}^1(k, k)$, where $\epsilon^2 = 0$. However, there seems no complete explanation as far as I checked. ...
Pierre MATSUMI's user avatar
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Totalization of the morphism between the two possible canonical truncations induces a quasi-isomorphism (Lemma 12.5.2 of Kashiwara, Schapira)

$\def\A{\mathcal{A}}$Given an abelian category $\mathcal{A}$, we denote $\mathrm{C}(\mathcal{A})$ to the category of cochain complexes with terms in $\mathcal{A}$. I am trying to understand the proof ...
Elías Guisado Villalgordo's user avatar
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1 answer
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For a compact 4-manifold, no 2-torsion in $H_1(M;\Bbb Z)$ implies no 2-torsion in $H_n(M;\Bbb Z)$ for all $n$

Let $M$ be a topological compact connected oriented 4-manifold with nonempty boundary, and suppose that each boundary component of $M$ is a rational homology 3-sphere. Is it true that if $H_1(M;\Bbb Z)...
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Splitting lemma for Non-abelian groups (semi-direct product)

I was reading the wikipedia page to learn about how the splitting lemma partially holds for non-abelian groups. Splitting lemma partially true $$ 0\to A \to B \to C \to 0 $$ If a short exact sequence ...
stoneaa's user avatar
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Behaviour of graded Betti number of a module

Let $M$ be a graded finitely generated $R=\mathbb{K}[x_1,\dots,x_n]$-module and, the graded Betti number of $M$ is defined by $\beta_{i,p}^{R}(M)=\mathrm{dim}_{k}(\mathrm{Tor}_i^{R}(M,k)_p)$. Suppose $...
Raman's user avatar
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The spectral sequence associated to an exact couple without chasing elements

$\def\Ker{\operatorname{Ker}} \def\Im{\operatorname{Im}}$I am trying to prove 011T to myself. It is a result involving an exact couple, its derived exact couple, the spectral sequence one obtains via ...
Elías Guisado Villalgordo's user avatar
2 votes
1 answer
19 views

Question regarding linear resolution of an ideal

Let $R=K[x_1,x_2,\ldots,x_n]$. $I(G_{(x)})$ is edge ideal of $G_{(x)}$. In the proof of theorem 2.13 given in paper uses the fact that $L=xI(G_{(x)})$ has linear resolution if and only if $I(G_{(x)})$ ...
Okky's user avatar
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Homology + Chinese Remainder Theorem =?

Let $M_i, i=1..n$ be a finite collection of pair-wise coprime moduli. The Chinese remainder theorem says that $\Bbb{Z}/M \approx \prod_i \Bbb{Z}/M_i$. Without going into Bezout / Euclidean algorithm,...
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For quasi-coherent sheaves $\mathcal{F}, \mathcal{G}$ on a proper scheme, $\operatorname{Ext}^2(\mathcal{F},\mathcal{G}) = 0$

In their book, Görtz and Wedhorn, claim that the derived category of quasi-coherent sheaves on a proper scheme has the following properties: abelian exact coproducts for all $\mathcal{F},\mathcal{G} \...
fish_monster's user avatar
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Definition of homology group as quotient in chain complex

I am working through some theory about abelian categories and complexes from "An Introduction to Homological Algebra" by Rotman. I don't understand one of the sections which I will explain ...
Flynn Fehre's user avatar
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Module over local ring which is free after cutting down by a non-zero-divisor

Let $(R,\mathfrak m)$ be a commutative Noetherian local ring and $M$ be an $R$-module such that $M\neq \mathfrak m M$. Let $x\in \mathfrak m$ be a non-zero-divisor on both $R$ and $M$. If $M/xM$ is a ...
Alex's user avatar
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Quasi isomorphic left-bounded and right-bounded complexes of projectives

Let $(A, \mathfrak{m}, k)$ be a complete Noetherian local commutative ring and suppose $\varphi : P_{\bullet} \to Q_{\bullet}$ is a quasi-isomorphism between complexes of finite free $A$-modules such ...
Brendan Murphy's user avatar
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Computing Auslander Reiten Quivers using Computer Algebra software

I was wondering if there is any algebra software that can be used to compute the Auslander Reiten graph for a given path algebra, and find out related details like the exterior algebra structure etc.
PlayerUnknown1098's user avatar
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Tensor product of two modules that are projective over different rings

This is in regards to the proof of Weibels Corollary 9.1.5: Suppose $A$ is a projective module over a comm. ring $k$. I have already proven, that this implies $A^{\otimes n}$ is also $k$-projective. ...
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Does reverse homology $\ker \delta\subset \operatorname{im} \delta$ have a defining equation? I don't think it does.

Background Material. Is "reverse homology" $\ker g \subset \text{im} f$ possible? Question. Forward (usual / historical) homology always has the defining equation $\delta^2 = 0$ for a ...
SeekingAMathGeekGirlfriend's user avatar
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2 answers
111 views

If $f:X \to Y$ induces isomorphism in cohomology, then it induces isomorphism in homology

Let $f:X \to Y$ and $G$ a group (I'm interested specifically in the case $G=\mathbb{Q}$) such that $f^\ast:H^q(X;G) \to H^q(Y;G)$ is an isomorphism for all $q$. It it true that $f_*:H_q(X;G) \to H_q(...
marc's user avatar
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Seemingly incorrect definition of weak convergence in McCleary "A user's guide to ...".

On p.62 of McCleary's "A user's guide to spectral sequences" the following definition A filtration $F$ of a differential graded module, $(A, d),$ is said to be weakly convergent if, for all ...
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cokernel of acyclic chain complex of projective modules after tensoring

Let $R$ be a commutative Noetherian ring, $(\mathbf F,d)$ be an acyclic chain complex of projective $R$-modules and $M$ be a finitely generated $R$-module. Let $i$ be an integer and put $N:=\text{...
Alex's user avatar
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Question about homology of k[x]

I want to compute $Tor_1^A(k,k)$, where A = k[x],k - algebraically closed field. I know how to compute Tor functor through Koszul resolution, but I want to do it strictly through tensoring bar ...
VadimStacheff's user avatar
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Is $\operatorname{Hom}_k(R,k) \cong E(k)$ for a local ring $R$?

Let $(R,m)$ be a local noetherian ring, which is also a $k$-algebra over a filed $k$. Suppose $R/m \cong k$. We can consider module $M=\operatorname{Hom}_k(R,k)$. Using the standard argument (the same ...
Alex's user avatar
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Establish a short exact sequence $0\to Z_p\to Z_{p^2}\to Z_p\to 0$ does not split, every submodule of the projective module need not be projective. [closed]

Establish a short exact sequence $0\to Z_p\to Z_{p^2}\to Z_p\to 0$ does not split, every submodule of the projective module need not be projective.
Haval Mohammed's user avatar
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1 answer
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Local finite dimensional algebras are $\tau$-tilting finite

Let $A$ be a finite dimensional algebra over an algebraically closed field $\mathbb{K}$. If $A$ is a local algebra (has unique maximal left ideal), then it is well known that $A$ is $\tau$-tilting ...
It'sMe's user avatar
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3 votes
1 answer
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Rotman-Homological algebra Chapter 2 exercise 2.33 [closed]

This problem is from Rotman-Homological algebra Ch-2-exer-2.33. Consider the following commutative diagram: If the above diagram is exact, then the map $C' \to C$ is also injective and hence, the ...
PDevi's user avatar
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1 answer
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Equivalent conditions of $\textrm{Ext}^2(A,B) = 0$ for all $A$ and $B$

I am studying homological algebra and I am having difficulty proving the equivalences of the following: (i) If $0 \to A \xrightarrow{f} B$ is exact and $B$ is projective, then $A$ is projective (ii) ...
Squirrel-Power's user avatar
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A quick question on equivalence between cellular and singular homology

We know that for a CW complex $X$, the cellular and singular homology groups are isomorphic: $H_n (X) \cong H_n ^{CW} (X)$ for each $n \in \mathbb{N}$. My question is the following: suppose $Y$ is ...
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group cohomology question

It's well known that an exact sequence 0 -> A -> B -> H -> 1 of groups (with A abelian) yields an exact sequence 0 -> A -> M -> IH -> 0 of H-modules, where IH is the ...
Lorenzo Traldi's user avatar
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1 answer
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Information about inverse limits in R-mod category

I would like to find a book or some references about inverse limits in R-Mod category but I can’t find it.If you can give me some references I will apareciate it.
Aridane Aridane's user avatar
6 votes
1 answer
107 views

If $F$ is an endofunctor on the category of abelian groups $G \mapsto G_{tors}$, then $R^1F(A) \cong A \otimes_{\mathbb{Z}} \mathbb{Q}/\mathbb{Z}$

I am trying to find all right derived functors of the functor $F: Ab \to Ab$ defined as an endofunctor on the category of abelian groups $G \mapsto G_{tors}$. I have proven that $R^iF = 0$ for all $i &...
Squirrel-Power's user avatar
1 vote
1 answer
68 views

Prove that for $F$ an additive functor of abelian categories, $R^0F$ is exact iff $R^1F = 0$ iff $R^iF = 0$ for all $i > 0$

I am beginning to study homological algebra. Let $F: \mathcal{A} \to \mathcal{B}$ be an additive functor of abelian categories. Prove that the following are equivalent: The functor $R^0F$ is exact $R^...
Squirrel-Power's user avatar
1 vote
1 answer
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Is there any example a non self-injective , semi-perfect ring withthe projective cover of every simple module is also injective?

Is there any example a non self-injective, semi-perfect ring $R$ such that the projective cover of every simple module is also injective? In the article "When projective covers and injective ...
Cos's user avatar
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3 votes
1 answer
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Examples of uniserial rings which are non self-injective.

In abstract algebra, a uniserial module M is a module over a ring $R$, whose submodules are totally ordered by inclusion. This means simply that for any two submodules $N_{1}$ and $N_{2}$ of $M$, ...
Cos's user avatar
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1 vote
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Can everything in the bounded derived category of $k[x]/(x^2)$ be built in one step?

Consider the Artinian local ring $R=\mathbb C[x]/(x^2)$. Then, does there exist $G\in \text{D}^b(\text{mod } R)$ such that $ \text{D}^b(\text{mod } R)=\langle G \rangle_1$ ? Here, I am using the ...
Alex's user avatar
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1 vote
1 answer
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Proof that quasi-isomorphisms form a localizing class in the homotopy category of complexes

I'm currently reading Gelfand and Manin's book Methods of Homological Algebra. Theorem III.4.4 says that the class of quasi-isomorphism in a homotopy category of complexes is localizing and I am ...
Albert's user avatar
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3 votes
1 answer
33 views

Giving examples of matrix rings with just one or a few maximal ideals and with easy lattices to compute.

I'm looking for matrix rings with only one maximal ideal. Or maybe with a few maximal ideals, the thing is that I've noticed that these type of rings allow me to find simple modules $R/I$ where is a ...
Cos's user avatar
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0 votes
0 answers
25 views

Unbounded Double Complex with Diagonal Boundary

I am new to spectral sequences and am not very familiar with convergence results once we lose boundedness. Suppose we have a differential cochain complex $X^n$ of $R$-modules with differential $D$, ...
VertexVexed's user avatar
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40 views

Relation between minimal free resolution and free resolution

Let $R=k[x_1,\ldots, x_n]$ and $M$ be a $R-$module. Every free resolution of a module contains a minimal free resolution of module as direct summand. My thoughts: A free resolution is said to be ...
Okky's user avatar
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1 vote
1 answer
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The Prüfer group $\mathbb{Z}(p^{\infty})$ as a $\mathbb{Z}$ module is not projective.

The Prüfer group $\mathbb{Z}(p^{\infty})$ as a $\mathbb{Z}$ module is not projective. Suppose that $\mathbb{Z}(p^{\infty})$ is a $\mathbb{Z}$ projective module, then $\mathbb{Z}(p^{\infty})$ is ...
Sok's user avatar
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1 vote
0 answers
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Higher Extensions and Cohomology

Let $\mathfrak{g}$ be a Lie algebra (over a field $k$) and $A$ e $B$ be $\mathfrak{g}-$módules, there is a well-know bijection $$ H^1(\mathfrak{g}, \mathrm{Hom}_k(A,B)) \simeq E^1(A,B) = \{SES \...
espacodual's user avatar
1 vote
0 answers
34 views

How can Lie algebra cohomology be nontrivial for a semisimple algebra?

Let $\mathfrak{g}$ be a semisimple Lie algebra over an (algebraically closed) field $k$ of characteristic zero. I am going by the definitions in Weibel, chapter 7. Here's my logic: A finite-...
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