Homological algebra studies homology in a general algebraic setting. The purpose is extraction of information about structures involved in terms of tangible objects like rings groups and modules.

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Definition of contractible chain complex

A relatively simple question. A book I'm reading states "a complex homotopic to the zero complex is called contractible"... but I don't understand the statement. I know what it means for chain maps ...
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1answer
36 views

Faithfully flat descent of projectivity and freeness

I am reading this paper. It is proven there that if $f:A\rightarrow B$ is a faithfully flat morphism of rings and $M$ an $A$-module such that the $B$-module $M\otimes_A B$ is projective, then $M$ ...
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28 views

Interaction of a functor with internal hom

An additive functor between abelian categories $F: \mathscr{C} \to \mathscr{D}$ induces a functor on categories of chain complexes $F: \mathscr{C}^\bullet \to \mathscr{D}^\bullet$. The internal hom ...
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24 views

Minus signs in internal hom

Consider the internal hom of chain complexes $A^\bullet$ and $B^\bullet$ $$\operatorname{Hom}^\bullet(A^\bullet, B^\bullet):=\{\text{degree $n$ maps}\} \qquad df:= d^B\circ f - (-1)^{|f|}f \circ d^A$$...
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1answer
33 views

On selfinjectivity of Hopf algebras

Any group algebra $kG$ of a finite group is selfinjective. More generally Gentile proves that for a group ring $RG$ with $R$ commutative and torsion free as a $\Bbb Z$-module, $RG$ is selfinjective ...
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24 views

Choosing an injective resolution of a short exact sequence of complexes

Lemma: Given a short exact sequence of cochain complexes in an abelian category $\mathcal{C}$ with enough injectives, $$0 \to P^\bullet \xrightarrow{f} Q^\bullet \xrightarrow{g} R^\bullet \to 0,$$ ...
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1answer
25 views

Finite product exists implies finite coproduce exist.

Let $C$ be a category such that the law composition of morphisms is bilinear, and there exists a zero object $0$, and the products exists for arbitrary finite sets of objects of $C$. Then the ...
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67 views

Chain Homotopy in abelian category

When dealing with complexes of modules or groups, the following lemma is pretty easy: If $f,g :E\rightarrow E'$ are homotopic, i.e. $f-g=d'h+hd$ for some h, then $f,g$ induce the same homomorphism ...
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103 views

Finitely generated projective modules over polynomial rings with integral coefficients

There is famous Quillen-Suslin theorem which states that every finitely generated projective module over a ring of polynomials $k[x_1,...,x_n]$, where $k$ is a field, is free. I have never carefully ...
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25 views

What properties are preserved by direct limits? [closed]

We know that direct limit of a directed family of flat $R$-modules is also flat ($R$ is a commutative ring with $1$ and all modules are unital). I am looking for other properties of modules which ...
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1answer
52 views

Meaning of functorial

It's known that for a short exact sequence of complexes, $0\rightarrow E'\rightarrow E\rightarrow E''\rightarrow 0$, it associates a homology sequences $...\rightarrow H(E')\rightarrow H(E)\rightarrow ...
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1answer
19 views

Seifert matrix, linking numbers, generators

I have been asked to compute the seifert form of a knot, the twist knot. I know how to compute the seifert surface, and then the seifert matrix seems to be defined accordingly (according to all the ...
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1answer
36 views

Why solving linear equations is taking a quotient by some subspace?

Linear equation can be represented by a linear form, and its solution space is the same thing as kernel of this form. The same is true for system of linear equations. But this lecture notes suggest ...
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31 views

The relation between Weyl character formula and Frobenius characteristic map

Let $\mathfrak{gl}(n)$ be the general linear Lie algebra of rank $n$, and $\mathfrak{S}_d$ be the symmetric group of rank $d$. It is well-known that the Schur-Weyl duality provide a equivalence ...
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28 views

someone desbribe what Homology theory is in 1hour lecture? [closed]

Am self studying abstract algebra.I feel the concept of ideals,cosets is what exactily is what is called homology group a fundamental principles of homology theory?so,I need someone out there to ...
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2answers
35 views

If $\text{Ext}_R^1(A,I) = 0$ for all $A\in ob(_R\text{Mod})$, then $I\in ob(_R\text{Mod})$ is injective. [duplicate]

Let $\text{Ext}^1(A,I)=0$ for all $A\in ob(_R\text{Mod})$, then $I\in ob(_R\text{Mod})$ is injective. I got stuck by this problem. Any ideas?
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1answer
45 views

A left exact functor preserves quasi-isomorphisms between acyclic complexes

A homological algebra theorem states Theorem: Let $T: \mathscr{A} \to \mathscr{B}$ be a left exact functor between abelian categories, and let $X^\bullet \xrightarrow{f} Y^\bullet$ be a quasi-...
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1answer
50 views

How do you find the free resolution of the module $M$ and of $F/M$ where $F=(K[x,y])^3$?

$M$ is a module generated by $$f_1=(xy,y,x), f_2=(x^2+x,y+x^2,y), f_3=(-y,x,y),f_4=(x^2,x,y).$$ We're to use the lex ordering with $x<y$ and $e_1>e_2>e_3$, where terms are given preference ...
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34 views

Ext group of bundles on moduli space of curves

Let $\mathcal{M}_{g}$ be the moduli space of curves of genus $g$. Let's suppose $g \geq 2$. Let $T$ be the tangent bundle of $\mathcal{M}_{g}$. Is the Ext group $\text{Ext}^1(\bigwedge^2T, T)$ trivial?...
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71 views

Homology Groups of Tangent 2-Spheres

I have been trying to compute the Homology Groups $ H_n $ of two tangent 2-Spheres (we will call this space, X). By previous results, I am able to easily determine that $ H_0(X) $ is isomorphic to the ...
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17 views

Mapping cylinder of chain complexes via $-\otimes \Delta$

An instructor gave me a homework set where the mapping cylinder of a chain map $C_\bullet \xrightarrow{f} D_\bullet$ is defined as $(\Delta^1_\bullet \otimes C_\bullet) \oplus_{C_\bullet} D_\bullet$, ...
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61 views

Homotopy Colimit of Truncations

Let $\mathcal{A}$ be an additive category with countable coproducts. I am just starting to learn about homotopy colimits and I am struggling with the following example that I am very interested in ...
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1answer
71 views

Splitting a short exact sequence of complexes of vector spaces

It's well-known that any complex of vector spaces is isomorphic to a direct sum of two types of indecomposable complexes (a one-dimensional space concentrated in one degree, or two one dimensional ...
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28 views

Exercise about Ext functor and 'tri-module'

While trying to understand the Hochschild-Kostant-Rosenberg theorem, I learned that $Ext_{R \otimes R}^1(R, R) = Der_K(R)$, where $R$ is an regular affine (commutative) $\mathbb{C}$-algebra. I am ...
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101 views

The endomorphisms of an $A_\infty$-algebra as a (bi)module over itself

Let $A$ be a unital associative algebra. A well-known exercise states that the ring of $A$-bimodule endomorphisms of $A$ are isomorphic to the center of $A$. That is, $\text{End}_{A-\text{mod}}(A) \...
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26 views

Derived functors in abelian categories and homotopy theory

For two Abelian categories $\mathcal A,\mathcal B$ and a right exact additive function $F\colon\mathcal A\to\mathcal B$, there is a left derived functor $LF$ acts on chain complexes $K_+(\mathcal A)$ ...
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27 views

A embedding of tensor product over semisimple algebras

Let $R$ be a semisimple Artinian algebra over the complex number field $\mathbb{C}$, that is, $R$ is isomorphic a finite direct product of matrix rings over $\mathbb{C}$. Let $S$ be a ideal of $R$, ...
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1answer
57 views

Compute the projective dimension of the given $R$-module

Let $$R=\frac{K[[x,y,z]]}{\left<xz,yz\right>}\text{ and } M=\frac{R}{\left<z+\left<xz,yz\right>\right>}.$$ Compute the projective dimension of $M$ as an $R$-module. My attempt ...
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43 views

Global dimension of power series ring $k[[x_{1}, \cdots, x_{n}]]$

Let $R$ be the power series ring $k[[x_{1}, \cdots, x_{n}]]$ over a field $k$. Notice that $R$ is a noetherian local ring with residue field $k$. Show that $gl. \dim(R)=pd_{R}(k)=n$. By First Change ...
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Stable equivalences preserving injective dimension

Let $A,B$ be two finite dimensional connected algebras and $F$ be a stable equivalence between their stable module categories (module category modulo projectives). Are there some natural conditions ...
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22 views

Are the derived functors of an additive functor additive? [closed]

Are the derived functors of an additive functor additive? Does this follows formally from the definition?
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60 views

Understanding a certain proof about $R$-module homomorphisms and split exact sequences

I'm currently reading "Algebra: Chapter 0" by Paolo Aluffi. Before I state my problem, I would like to give here the definition of split exact sequences from the book: A short exact sequence ...
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1answer
56 views

Splitness of short exact sequences

Consider the following commutative diagram in an abelian category: $$\require{AMScd}\begin{CD} @.0@.0@.0\\ @.@VVV@VVV@VVV\\ 0@>>>A@>>>B@>>>C@>>>0\\ @.@VVV@VVV@VVV\\...
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1answer
210 views

Auslander-Buchsbaum formula - proof

I have a final exam in commutative algebra and Auslander-Buchsbaum formula is one of the theorems that we have to self-study for the exam. Unfortunately, our course did not cover minimal resolutions ...
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2answers
56 views

Cohomology of free group acting trivially

I read here the formula for cohomology of a free group with trivial action: $$H^q(G, M) = \begin{cases} M &\text{for } q = 0\\ M^n &\text{for } q = 1\\ 0 &\text{for } q \geq 2\\ \end{cases}...
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Prove that $A \otimes_{\mathbb{Z}[G]} B = (A \otimes B)_G$

$\newcommand{\Z}{\mathbb{Z}}$ This identity occurs in Cassels and Frohlich, page 98. Let me recall the context: If $A, B$ are left $G$--module i.e. $\Z[G]$-module then we turn $A$ into right $G$-...
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129 views

Is the proof of Auslander-Buchsbaum formula in Rotman's Advanced Modern Algebra wrong?

Suppose that $(R,m,k)$ is a commutative, local, Noetherian ring. If $F$ is a free $R$-module then $\operatorname{Ext}^i_R(k,F)=0$ for $i\geq0$. This statement is part of a proof for the Auslander-...
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16 views

Confusion about right $G$-module structure to compute group homology

$\newcommand{\Z}{\mathbb{Z}}$ According to Cassels and Frohlich, Chapter IV, homology $H_q(G, A)$ can be computed as homology of the chain complex obtained by tensoring the standard free resolution $$\...
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1answer
57 views

If $D$ is a triangulated category, and $E_i$ is a set of generators, then $D$ is equivalent to $D(End(\oplus E_i))$?

I am looking for a result along the lines of the following statement: If $D$ is a triangulated category, and $E_i$ is a set of generators (every object can be obtained up to isomorphism by shifts and ...
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23 views

Question on resolution

let $A$ be a dga, that is an algebra that is the direct sum of modules with a differential $d$ such that $d^2=0$. Clearly any gives rise to a chain complex. For any module $X$, suppose there is a ...
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23 views

How are $B_k(E^{pq}_r)$ and $Z_k(E^{pq}_r)$ defined ? (Spectral sequences)

Daniel Murfet in his notes on spectral sequences gives the definition of cohomological version of spectral sequence ((a),(b),(c) on page 2). After that he states the following: "For $k\geqslant r+...
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Functor Derived, homological algebra, projective modules

Anyone help me with this question Question: Show that if M is module projective, then $L_0FM = FM$ and $L_nFM =0$ for n > 0. p.s. exercise 1, page 346 Basic algebra, vol. 2, Jacobson Thanks in ...
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9 views

Iff conditions for acyclic, free, positive chain complexes with augmentation

I have some doubts about the formulation of the following lemma (from Ferrario, Piccinini - Simplicial structures in topology) and its proof. (II.3.8, page 72) Lemma. Let $(C,\partial) $ be a ...
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1answer
39 views

$\mathrm{Frac}(R)/R$ as a direct limit

Let $R$ be an integral domain, let $F=\mathrm{Frac}(R)$ be its field of fractions. Then $R$ is a submodule of $F$, hence we have quotient module $F/R$. It is true that $$F/R \cong \varinjlim R/rR$$ ...
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1answer
32 views

One of characterizations of projective modules over noetherian ring of finite global dimension

Let $A$ be noetherian ring of finite global dimension, $M$ be finitely generated module. Then i want to prove that $\mathop{Ext^i}(M,A) = 0, i>0 \implies M -$ projective. Since in this case ...
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11 views

Proving there exists an isomorphism between the direct sum of the homological groups of path components and the homological group of the whole space

Let $X$ be a space, let $\{X_\alpha\}_{\alpha\in A}$ be the set of path components of $X$, and let $\iota_{\alpha}:X_\alpha\to X$ be inclusion. Then for each $p\geq 0$, the maps $(\iota_\alpha)_*: H_p(...
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1answer
24 views

Why is $\mathbb{Z}/p^2\mathbb{Z}$ indecomposable in the homotopy category of chain complexes

I want to understand the accepted answer to this question. The answer is supposed to work for the homotopy category of chain complexes of abelian groups too. (i.e. it shows that that category is not ...
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151 views

Why does excision imply this?

In exercise $4$, page 230 of Bredon, he asks for a proof of the Mayer-Vietoris sequence using a commutative braid diagram which substitutes some terms by others using excision. I've solved the ...
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25 views

$\mathbf{Ch}(\mathcal{A})$ has enough projectives

This is the Exercise 2.2.2 from Weibel's book. Suppose an Abelian category $\mathcal{A}$ has enough projectives, then so does the category $\mathbf{Ch}(\mathcal{A})$ of chain complexes over $\mathcal{...
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2answers
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Auslander-Buchsbaum formula without minimal/finite resolutions

Does anybody know a proof of Auslander-Buchsbaum's formula that uses only projective/injective/flat resolutions and homological functors Ext and Hom without using minimal/finite resolutions?