# Tag Info

## Hot answers tagged homological-algebra

5

The identification is as follows. Let $M$ be a $G$-module. Then $M$ becomes a $\mathbb{Z}[G]$-module by $$\left( \sum_gn_g g\right)m=\sum_g n_g(gm).$$ Conversely if $M$ is a $\mathbb{Z}[G]$-module, then $M$ becomes a $G$-module by $$gm=(1g)m.$$ Since $R=\mathbb{Z}[G]$, viewed as a left $R$-module, is a projective object , $\mathbb{Z}[G]$-mod, and hence ...

5

Since you are considering not necessarily commutative ring and thus is forced to taking hom-set and tensor product of abelian group, it's not reasonable to expect that $\hom_R(M,M')\otimes\hom_R(N,N')$ and $\hom(M\otimes_RN,M'\otimes_RN')$ are comparable. For example, taking $M:=R_R,N:=_RR$, then the two become $R\otimes_{\mathbb Z}\hom_R(M',N')$ and ...

4

Trivial action of ring $A$ on Abelian group $M$: $am=0$ for all $a\in A$ and $m\in M$. If $A$ has an identity element (and the axiom $1x=x$ is posed), then this forces $m=1m=0$ for all $m$, hence $M=\{0\}$.

4

$\mathbb{Z}/6\mathbb{Z}$ is not a PID, because it's not a domain.

3

Right exactness of a functor only means that short right exact sequences are preserved. Long right exact sequences include short exact sequences, which are only preserved by exact functors. For the functor $- \otimes C$ this means that $\mathrm{Tor}_0(-,C) = - \otimes C$, but not necessarily $\mathrm{Tor}_1(-,C) = 0$. Convince yourself that ...

3

Ok, I believe that the book try to do the following thing (although avoiding a lot of details). As a general case for every category $\mathbf C$ you can build a category $\text{Fam}(\mathbf C)$ whose: objects are families of objects in $\mathbf C$, that is stuff like $(c_i)_{i \in I}$ where each $c_i$ is in $\mathbf C$ morphisms from $(c_i)_{i \in I}$ to ...

3

All maps are morphisms (unless otherwise stated) - you're in an arbitrary additive category, where set maps don't even make sense. By universal, it means that any other $j$ that satisfies that property (i.e., any $j: A' \to B$ with $fj = 0$) factors uniquely through $i$ (i.e., there is a unique map $g: A' \to A$ with $j = ig$.) This is what one means by ...

3

The intended interpretation is absolutely that a $G$-module $M$ becomes a module over the ring $\mathbb{Z} G$. As a group, $\mathbb{Z} G$ is simply a free abelian group of the same rank as the order of $G$, so that structure remembers nothing about how $G$ acted on $M$. As to the trivial $G$ module $\mathbb{Z}$, the action of $\mathbb{Z} G$ is $(a_1 ... 2 (Edit after incorrect first answer) Let$f$be a cocycle (central extension with abelian quotient): writing additively, the cocycle condition is$f(a,b)+f(a+b,c)=f(a,b+c)+f(b,c)$. Then$g:(a,b)\mapsto f(a,b)-f(0,0)$is also a cocycle. When$c=0$the cocycle relation yields$g(a,0)=0$for all$a$, and similarly$g(0,a)=0$for all$a$. When$c=a$the ... 2 I think you mean$\text{Hom}_\Gamma(\text{Hom}_{\Lambda}(N,M),M)$? Since$\text{Hom}_{\Lambda}(N,M)$doesn't have a natural$\Lambda$-module structure. Also, the fact that$\Lambda\cong \text{End}_{\Gamma}(_\Gamma M)$isn't necessary. Try proving that the evaluation map is an isomorphism for$N=M$, and then that the class of modules$N$for which it is an ... 2 You seem to be just confusing different notions of module.$G$-module would be an abelian group$M$with the action of a group$G$compatible with addition. (This can well be the trivial action.)$R$-module, or just module, where$M$is an abelian group and one has scalar multiplication with the elements from the ring$R$similarly to case of ... 2 This follows from the "long" 5 lemma, by extending the diagram to have another column of$0$s on the left. Alternatively, it is easy to see directly: if you identify$B$with$B'$and$C$with$C'$via the isomorphisms, commutativity of the right square says that$B\to C$and$B'\to C'$become the same map, and so you get that$A$and$A'$are kernels of ... 2 It's rarely true. Take$A=\mathbb{Z}$. Then for any group$G$,$H_1(G,\mathbb{Z})$is the abelianization of$G$: i.e.,$G/[G,G]$. So if you take any group$G$whose abelianization is trivial (for example, any non-abelian simple group), and any subgroup$C$with non-trivial abelianization (for example, any non-trivial cyclic subgroup), then ... 2 You should look at Barr, M. Acyclic models, CRM Monograph Series, Volume 17. American Mathematical Society, Providence, RI (2002), and see if that satisfies you for completeness. A theorem involving crossed complexes rather than chain complexes is in Section 10.4 of the book partially titled Nonabelian Algebraic Topology (2011). This version gives in ... 2 Why not? Assuming you write$p$for the row index and$q$for the column index,$d^h(C_{p,q}) \subseteq C_{p+1,q}$and$d^v(C_{p,q}) \subseteq C_{p,q+1}$. Hence$d(C_{p,q}) \subseteq C_{p+1,q} \oplus C_{p,q+1} \subseteq \text{Tot}_{n+1}^{\oplus}(C)$if$p+q = n$. This should appear clear if you draw a plane with coordinates$p,q$and the modules$C_{p,q}$... 2 Let$\phi\colon H \to G$be a group homomorphism,$M$a$\mathbb ZH$-module,$N$a$\mathbb ZG$-module, and$\psi\colon M \to \operatorname{Res}_\phi N$an$H$-equivariant map. We want to understand the induced map $$H_*(H;M) \longrightarrow H_*(G;N).$$ Let$E_*H$and$E_*G$be free$\mathbb ZH$- and$\mathbb ZG$-resolutions of the trivial representation ... 2 The short exact sequences$0\to\mathbb{Z}\to G\to\mathbb{Q}\to0$are classified by the group$\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$. Two exact sequences define the same element in$\operatorname{Ext}(\mathbb{Q},\mathbb{Z})$if and only if there exists a homomorphism$G\to G'$making the diagram$$\require{AMScd} \begin{CD} 0 @>>> \mathbb{Z} ... 2 Note that$\text{End}_\Gamma(M)$is precisely the center of$\Gamma = \text{End}_{\Lambda}(M)$. Let$\varphi \in \text{End}_{\Gamma}(M)$. Consider$D \varphi: DM \rightarrow DM$. I claim that$D \varphi$is a$\Gamma^\text{op}$-module homomorphism. Note that given$\psi \in \Gamma$, the action of$\Gamma^{\text{op}}$on$DM$is given by$\psi \cdot f := f ...

2

As @Hoot says, the splitting should be a map $s_n : C_n \to C_{n+1}$ (and it is indeed what's written in my version of the book). Note that in the formula you have written, the domain of the RHS is not equal to the domain of the LHS... If you want to remember how the splitting map works, look at the definition of a chain homotopy below (in the book). A ...

2

If $k$ is a field, and your quiver has no oriented cycles, then there are exactly as many indecomposable projective modules (up to isomorphism) as there are vertices in the quiver. Moreover, any finite-dimensional representation admits a projective cover. (This is all true because of more general results on Artinian algebras). In your specific example, to ...

2

Your method is not going to work unless you're careful about the choice of the projective modules $P$ and $Q$ through which $gf-\text{id}_M$ and $fg-\text{id}_N$ factor. Indeed, you could replace $P$ by $P\oplus P'$ for any projective module $P'$, changing the isomorphism type of $M\oplus P$. Instead, use the fact that $gf-\text{id}_M$ factor through a ...

2

Over any commutative ring, a flat module is torsion-free. Indeed, assume that $R$ is a commutative ring, and $M$ is an $R$-module which is not torsion-free, that is, there exists a non-zero element $m\in M$ and a non-zero divisor $x\in R$ such that $xm=0$. Consider the morphism $f:R\to R: r\to xr$. Since $x$ is not a zero divisor, $f$ is injective. ...

1

Notice that the map $x\mapsto xg$ is an endomorphism of the complex of $H$-modules $E_*G$, and it is in fact a lift of the identity of the right $H$-module $\mathbb Z$ to its resolution $E_*G$. It follows that that map of complexes is homotopic to the identity.

1

The application that convinced me I should care about homotopical ideas is an answer to the following question: Where do long exact sequences come from? My answer is at this MO question.

1

In general, algebraic topology uses homological algebra a lot. Historically, I believe homological algebra was first developed to study what happened in algebraic topology, and only split off as an independent area of study later. Here is one example of an application of (very basic) homological algebra. Theorem [Brouwer's fixed point theorem]. Every ...

1

$\def\H{\operatorname{Hom}}$There is an epimorphism $A^n\to N$, which implies a monomorphism of $B$-modules $\H_A(N,M)\to\H_A(A^n,M)$, where $B=\operatorname{End}_A(M)$. Since $\H_A(A^n,M)\cong M^n$ as $B$-modules, it's sufficient to prove that $M$ has finite length as $B$-module. Saying that $A$ is an artin algebra means that $A$ is an algebra over some ...

1

Let $M$ be the matrix of the graph, $t:E\rightarrow V$ be the target map and $s:E\rightarrow V$ be the source map, so that if $e$ is an edge linking $v_i$ to $v_j$ then $t(e)=v_j$, $s(e)=v_i$ and $Me=t(e)-s(e)$. If $e$ is any edge from $v_i$ to $v_j$, interpret $-e$ as the edge $e$ with the reverse orientation. That is the same edge but from $v_j$ to $v_i$. ...

1

I think Weibel has a typo. Using his notation, he has supposed that $u_n: C_n \to D_n$ and $d_{n-1} : D_{n-1} \to D_{n-2}$. Then the composition $d_{n-1} \circ u_n$ is not defined.

1

Another possibility is the following paper: Samuel Eilenberg and Saunders MacLane. “Acyclic models”. In: Amer. J. Math. 75 (1953), pp. 189–199. ISSN: 0002-9327. JSTOR: 2372628. I read it a few months ago so I may not remember perfectly, but if I recall correctly, everything was done in detail and I didn't have to fill in any significant gap. The most ...

1

First, note that there always exists a left-$add(X)$-approximation $\varphi:C\to X^{\widehat{m}}$ for any module $C$ in mod-$A$. This comes from the fact that $Hom_A(C,X)$ is a finitely generated module over the (implicit) base ring $k$; if $f_1, \ldots, f_r$ are generators, then the map given in matrix form by $(f_1, \ldots, f_r)^t$ from $C$ to $X^r$ is an ...

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