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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584 views

Projective resolution of tensor product

Let $M,N$ are $R$ modules and $P^\bullet, Q^\bullet$ are their projective resolutions. Can we obtain projective resolution $M\otimes N$ using $P^\bullet, Q^\bullet$. If i understand correctly homology ...
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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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0answers
14 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
20 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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0answers
7 views

On selfinjectivity of Hopf algebras

Any group algebra $kG$ 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 if and only if $G$ ...
3
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3answers
92 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 ...
2
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3answers
263 views

What is the product and coproduct of Morphism category (Arrow category)?

Given a category C, its morphism category D means a category that has 1) "morphisms of C" as its objects 2) "pairs (f,g) s.t. the diagram (square) commutes" as its morphisms. The above definition ...
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1answer
22 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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0answers
52 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 ...
2
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0answers
55 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
51 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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0answers
23 views

What properties are preserved by direct limits? [on hold]

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 ...
0
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1answer
16 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 ...
1
vote
1answer
33 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 ...
1
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1answer
42 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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0answers
30 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? [on hold]

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
47 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 ...
5
votes
1answer
70 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 ...
2
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0answers
31 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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1answer
152 views

If two chain maps over a PID induce the same homomorphism, then they are homotopic

If two chain maps $f,g:\mathcal{X} \rightarrow \mathcal{Y}$, where $\mathcal{X},\mathcal{Y}$ are chain complexes with free modules $X_p$ and $Y_p$ over a PID, $R$, induce the same homomorphism in the ...
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0answers
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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1answer
224 views

Functor preserves kernels iff it's left exact

I'm trying to understand the proof to a statement in Rotman's 'Introduction to Homological Algebra': Proposition 5.25, p. 240: Let $F :_R\text{Mod} \to \text{Ab}$ be a covariant functor. Then $F$ ...
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0answers
16 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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0answers
35 views

projective resolution for an $I$-torsion $R$-module

Let $R$ be a commutative Noetherian ring with non-zero identity, $I$ be an ideal of $R$ and $M$ be an $I$-torsion $R$-module. We know that there exists an injective resolution of $M$ in which each ...
14
votes
2answers
570 views

Vanishing of a certain Tor

I am reading about the construction of the Affine Grassmannian in Dennis Gaitsgory's seminar notes and there are some commutative algebra facts that I am not able to figure out by myself apparently, ...
3
votes
1answer
90 views

Splitness of quotient sequence

Let $A, B, C$ be holomorphic vector bundles over some complex manifold $X$. Let $A', B', C'$ be sub bundles, respectively. Suppose that we have short exact sequences: $$0 \rightarrow A \rightarrow B \...
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0answers
27 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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0answers
100 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 $K_+(\mathcal A)$ ...
0
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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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0answers
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 $R$, ...
5
votes
1answer
109 views

The relationship of $\hom(M\otimes_RN,M'\otimes_RN')$ and $\hom_R(M,M')\otimes\hom_R(N,N')$.

Let $R$ be a ring with identity, $M$ and $M'$ two right $R$-module, $N$ and $N'$ two left $R$-module. There is a natural way to define a homomorphism $$f:\hom_R(M,M') \otimes \hom_R(N,N')\to \...
2
votes
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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42 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 ...
3
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0answers
35 views

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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2answers
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
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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1answer
55 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\\...
3
votes
1answer
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-...
4
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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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0answers
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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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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30 views

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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0answers
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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0answers
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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16 views

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 ...
2
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1answer
41 views

A short exact sequence that cannot be made into an exact triangle. (Weibel 10.1.2)

The following exercise is in Weibel Chapter 10. Regard the groups $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ as cochain complexes in degree 0. Show that the short exact sequence $$ 0 \...
3
votes
1answer
24 views

A spectrum $I$ is $E$-injective iff the map $i:I\rightarrow I\wedge E$ is an inclusion of a retract.

I was reading some notes on stable homotopy theory and I came across the statement in the title of this question. "Suppose $E$ is a ring spectrum, then $I$ is $E$-injective if and only if the ...