Hot answers tagged homological-algebra
6
$\require{AMScd}$The factorization you ask about is an instance of the slogan "distinguished triangles are just like exact sequences".
To wit, the axiom on existence of morphisms yields that in a distinguished triangle
\begin{CD}
X @>u>> Y @>v>> Z @>w>> X[1]
\end{CD}
$v$ is a weak cokernel of $u$. For if $f \colon Y \to W$ is a ...
4
What you are asked to show, namely that $[l] = [l_1 + l_2 + l_3]$ in $H_1(X)$, means precisely that $$l_1+l_2+l_3-l=\partial(\text{something})$$ where $\text{something}\in C_2(X)$, the free abelian group on the set of singular $2$-simplices. That is, you will need to define certain maps $\Delta^2\to X$ and a linear combination of them ($\in C_2(X)$) whose ...
4
What you mentioned is about the Grothendieck's proof of Samuel's Conjecture:
If $R$ is a local domain which is a complete intersection and such that $R_{\mathfrak p}$ is UFD for every prime ideal of height $\le 3$, then $R$ is UFD.
See SGA 2, Corollaire 3.14, page 132. (Maybe the keyword here is parafactorial). But probably the best algebraic ...
3
1) As my understanding, the orientation of g.e depends on both k-cell e and the element g. It could be defined as below:
Suppose that B the chosen basis for the k-dimensional space containing k-cell e and B' the chosen basis for the k-dimensional space containing f. The orientation is defined by the sign of the determinant of the matrix which exactly is the ...
3
HINT The snake lemma tells us that there is an exact sequence $$0 \to L \to M^\prime.$$
HINT 2 Kernels of $R$-modules have a universal property.
Added: The universal property of a kernel is this. A kernel of a morphism $f:M \to N$ is a morphism $i:K \to M$ such that $f \circ i = 0$ and such that is universal with respect to this property, meaning that if ...
3
Here is an abstract way to think about this (which might not be what you're looking for):
The condition that $f\circ v=g\circ v\Rightarrow f=g$ for all $f,g\in\operatorname{Hom}(M'',N)$ says that $v$ is right-cancellative, that is, $v$ is an epimorphism in the category of $A$-modules. Epimorphisms in the category of $A$-modules are surjective on underlying ...
3
I think I have found the solution using Zach L's hint.
Let $N=\operatorname{coker}(v)=M''/\operatorname{Im}(v)$, and let $p\in\operatorname{Hom}(M'', N)$ be the canonical map $p: M''\to M''/\operatorname{Im(v)}=N$. We observe for every $x\in M$, we have
$$p(v(x))=v(x)+\operatorname{Im}(v)=0+\operatorname{Im}(v)=0_{M''/\operatorname{Im(v)}}$$
So $p\circ ...
3
Is there any finite simply-connected CW complex on which $\mathbb{Z}/2$ acts freely? No, since then the classifying space $B(\mathbb{Z}/2)$ would have finite homological dimension, which would imply that the group $\mathbb{Z}/2$ has finite homological dimension. But one can compute $H_p(\mathbb{Z}/2,\mathbb{Z})=\mathbb{Z}/2$ for odd $p$.
3
A more or less tautological example that shows that group cohomology is "needed" is the fact that taking invariants under finite groups in short exact sequences does not preserve exactness.
It is easy to find examples of surjections $f:M\to N$ of $G$-modules such that the induced map $M^G\to N^G$ on the invariant subspaces is not surjective.
For example, ...
3
I think the touchstone for understanding direct limits is understanding directed unions.
A collection $C$ of sets is directed if for every $X,Y\in C$, there exists $Z\in C$ containing both $X$ and $Y$. This becomes a direct system using inclusion mappings.
Now just by using the directness of this collection, you can compare any two sets (and inductively, ...
3
Let $M$ be an $R$-module, $f \in R$ and let $N$ be the colimit of $M \xrightarrow{f} M \xrightarrow{f} \dotsc$. Directed colimits are easy to construct: Elements come from elements of the individual modules, and are identified if they get sent to the same element by some transition map. So in our case, if $i_n : M \to N$ denotes the $n$th colimit inclusion, ...
2
If you ask the question
if $M$ is an $R$-module of projective dimension $n$, is $\operatorname{Ext}^n(M,R)\neq0$?
then the answer is surpring: it depends. When $R=\mathbb Z$, this is essentially known as Whitehead's problem and Shelah proved that its answer depends on the specific set theory that you choose. Indeed, he showed that depending on the ...
2
Another application is called "Galois descent". Roughly, for a Galois extension $K/k$, if two structures are isomorphic over $K$, we may ask whether they are isomorphic over $k$. Galois descent provides an answer in terms of Galois cohomology.
As an example, let $k$ be a field and let $M$ be a matrix with entries in $k$ and let $K/k$ be a Galois extension. ...
2
It might be visually helpful to track the module side in the notation. To do that, I'll write $A_R$ to indicate a right $R$ module $A$, or $_RA$ to indicate a left $R$ module.
It will also help to consider bimodules over two different rings, since it will help keep us from confusing the module operations.
Suppose $_SB_R$ is a bimodule.
$H=Hom(B_R,A_R)$ ...
2
Well, if the group homomorphisms $\gamma_n^*$ are one-to-one, so you're essentially forming a directed union, then this will be infinitely generated, because any finitely many generators would be in one of the groups of your direct system and therefore can't generate the additional elements in the next group of the system.
Presumably, you want sufficient ...
2
I shall use the flat base change for Tor:
Let $X$ be an $R$-module, $Y$ be an $(R,S)$-bimodule and $Z$ an $S$-module. If ${}_R Y_S$ is flat on both sides then $$\text{Tor}^S_n(X \otimes_R Y,Z) \simeq \text{Tor}^R_n(X,Y \otimes_S Z).$$
If particular, if $R\to S$ is a flat ring homomorphism, we get $$\text{Tor}^S_n(X \otimes_R S,Z) \simeq ...
2
We had better suppose that $A$ is abelian, and so a $G$-module; otherwise I'm not sure the question makes sense (because the $H^2$ with coeffs. in $A$ would not be defined).
Group cohom. is functorial in the coefficients, so $C \to A$ induces
$f_*: H^2(G,C) \to H^2(G,A)$. This latter map has an interpretation in terms of
extensions: it sends the extension ...
2
Your proof is OK. It will become straight forward with a little bit more experience ;). You have done the only possible thing: Somehow extend the left exact sequence to a short exact sequence, using the image in order to make the last map surjective. By the way, I have given the same proof in SE/292037.
2
I think you got to the point beginning with the sentence "Consider the exact sequence..." You can probably omit the work before that, although there is no mistake there. After that sentence, you get that the sequence is exact at $F(A)$ for free.
The mathematical leap was that you took the given and recognized that there is a relevant short exact sequence ...
1
This is true in general (assuming, as I'm sure was meant, that $B$ is flat "on either side" means on both sides), and I don't think you need the rings to be commutative. If $P_* \to A$ is a free resolution, then $P_* \otimes_R B$ has homology groups $Tor^R(A,B)$, and so it's also exact above degree zero. It's then a resolution too -- not by free or ...
1
I guess we are working with $\Bbb C[x,y]$-modules.
The degree-zero part of the quasi-isomorphism would have to be a morphism $\Bbb C[x,y] \oplus \Bbb C[x,y] \to \Bbb C[x,y]$ inducing an isomorphism between $K$ and $\Bbb C[x,y]$, where $K$ is the kernel of the differential of $C^*$, that is
$$ K = \{ (y R, x R) \, | \, R \in \Bbb C[x,y]\}.$$
But such a ...
1
This new condition is important and without it I don't think the result is true.
Here's an attempt (could be mistakes so be wary!)
We have
$$
M \otimes_{A[[t]]}^{\mathbb{L}} B[[t]] \cong (M \otimes_{A[[t]]}^{\mathbb{L}} A[t]/t^n) \otimes_{A[[t]]}^{\mathbb{L}} B[[t]]
$$
By associativity this is (quasi-isomorphic) to
$$
M \otimes_{A[[t]]}^{\mathbb{L}} ...
1
For any sequence of $R$-modules $M_n$ whatsoever (regardless of whether you obtained them as submodules of $H_n(C)$ for some $C$), the chain complex $D$ defined by
$$\cdots\xrightarrow{\;d_{n+2}\;} M_{n+1}\xrightarrow{\;d_{n+1}\;}M_n\xrightarrow{\;d_{n}\;} M_{n-1}\xrightarrow{\;d_{n-1}\;}\cdots$$
where each $M_n$ is in the $n$th spot, and the maps $d_n$ are ...
1
The answer is no, and counterexamples are easy to construct using the isomorphism from Schreier's theorem, which says that $H^2(G,A)$ is isomorphic as a pointed set $0\in H^2(G,A)$ is the point) to $EXT(G,A)$ which is the set of short exact sequences $1\to A\to\hat G\to G\to 1$ up to equivalence, with the point being the semi-direct product $\hat G=A\rtimes ...
1
I resolved my confusion : it's a straightforward application of the universal property of the free graded-commutative graded algebra on a graded set of generators. My confusion stemmed from the fact that I hadn't really understood that the filtration on $H^*$ was used to filter each factor $H^n$, and that any indeterminacy in choosing representatives of ...
1
As pointed out in the comments that submodules of projective modules are projective does not always hold. If this holds the algebra is called hereditary. An equivalent description of that is that $\operatorname{gldim} A \leq 1$.
But even if it holds, this does not guarantee that submodules of projectives are direct summands. For example take the hereditary ...
1
Find explicit isomorphisms. I assume that $R$ is unitary.
If $f:A\to B$ is a $\Bbb Z$-morphism, then let $g(a):=r\mapsto f(ra)$.
If $g:A\to \hom_{\Bbb Z}(R,B)$ is an $R$-morphism, then $f(a):=g(a)\,(1)$.
Check that these defined mappings $g$ and $f$ are as stated, and that $f\mapsto g$ and $g\mapsto f$ inverses to each other.
1
Your second deduction is untrue. You cannot, in general, conclude from a short sequence
$$
0\longrightarrow A_1 \longrightarrow A_2 \longrightarrow A_3 \longrightarrow 0
$$
an isomorphism
$$
A_2 \cong A_3 \oplus A_1
$$
For instance, this is not the case for the short exact sequence of abelian groups
$$
0 \longrightarrow \mathbb{Z} ...
1
If a group $G$ acts on an abelian group $N$, we can form the semidirect product $N\rtimes G$ and there is a canonical surjection $p:N\rtimes G\to G$.
This surjection is split, and in fact split surjections are of this form with abelian kernel.
Now, a split surjection admits many different splittings. As soon as you try to classify them, you end up with ...
1
The construction of cross product algebras is a very natural problem.
It is very easy to arrive at the $2$-cocycle condition for assocativity and to the condition that such cocycles be cohomologous for the algebras to be isomoorphic.
With sufficient hand waving, this example can be concluded by mentioning the Brauer group of fields and the amazing fact ...
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