Mathematics Stack Exchange
Mathematics Stack Exchange

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.

A chain complex is a sequence of abelian groups, vector spaces, or modules, with linear maps connecting them which compose to zero.

Homological algebra is the study of chain complexes and their homology groups.

5114 questions
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Exact sequences and derived functors

Let $R$ be a commutative ring and $ r\in R$. Denote $R_r = \{ s\in R \mid s r =0 \}$. If $R_r $ is nonzero $0 \rightarrow R_r \rightarrow R\stackrel{r\cdot} \rightarrow R \rightarrow R/ rR \rightarrow 0$ is exact. Deduce …
Problemsolving
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Are chain homotopies for chain complexes of vector spaces special?

Question: Is it true that, for a chain complex of vector spaces, two chain maps induce the same homomorphisms on homology if and only if they are chain homotopic? More importantly, as a sanity check for myself, isn't it true that only one direction…
Chill2Macht
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Localization of Tor of an algebra

Let $R$ be a commutative ring, let $A$ be an $R$-algebra, let $S\subset A$ be a multiplicative subset, and let $M$ be an $R$-module. Is it true that $\mathrm{Tor}_p^R(S^{-1}A,M) \cong S^{-1}\mathrm{Tor}_p^R(A,M)$ ? The first problem is that I don't…
user46225
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Four Lemma(proof)

I am proving four lemma: I want to show that if the rows are in the commutative diagram are exact and m and p are surjective, and q is injective, then n is surjective. See the following link. http://en.wikipedia.org/wiki/Five_lemma When they are…
Reader
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Differential on total complex

Given a bicomplex $C_{\ast,\ast}$, i.e. objects $C_{p,q}$ in an abelian category with horizontal $$d_{p,q}^h:C_{p,q}\to C_{p-1,q}$$ and vertical $$d_{p,q}^v:C_{p,q}\to C_{p,q-1}$$ differentials that anticommute. Then there is the notion of the total…
ilil
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On changing bases while working with Hom: is $\text{Hom}_R (M, R \otimes_I M) \simeq \text{Hom}_I(M, M)$?

Let's say I have a ring $R$ and a subring $I$. Let $M$ be a $R$-module. Is it true that, as $R$-modules (or as groups): $$\text{Hom}_R (M, R \otimes_I M) \simeq \text{Hom}_I(M, M)$$ If so, why? I am aware of a particular case where: $$\text{Ext}_R…
Catherine Ray
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How do I prove $\operatorname{Hom}_R(A, B)$ torsion-free and divisible under the given conditions, where $A$ and $B$ are abelian groups.

Let $A,B$ be abelian groups. I want to show that (a) the group $\operatorname{Hom}_R(A,B)$ is torsion-free when $A$ is divisible, and (b) the group $\operatorname{Hom}_R(A,B)$ is divisible when $A$ is torsion-free and divisible.
Mark
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Need help understanding the proof that every free module is projective.

We have a free module $F$ with basis $X$. If $f: F \to B$ and $\alpha: A \to B$ are $R$-linear and $\alpha$ is surjective, to find an $R$-linear map $g:F \to A$ we would have to use the universal property of free modules, so to this end we're first…
Mark
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Is this a projective resolution?

Let $k$ be a field, $R=k[x]$. Is this a projective resolution of $k$ over $R$? $$0\to k[x]\to k[x]\to k\to 0$$ where the left map is $x\mapsto x-1$ and the right map is $x\mapsto 1$ ? If not, what is a projective resolution in this case?
DontBeLikeMe
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Is a chain map inducing zero on all homology groups necessarily a boundary?

Given the category $\mathbf{Chain}_A$ of chain complexes over an abelian group $A$. The $n$-th Homology functor is: $$H_n:\mathbf{Chain}_A\to\mathbf{Ab}$$ Clearly it is additive: $$H_n(\varphi+\varphi')=H_n(\varphi)+H_n(\varphi')$$ There are special…
C-star-W-star
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Two question about Ext functor in mod-A

As we know, if $\textrm{Ext}_{A}^{1}(M,N)=0$ with $M,N\in \textrm{mod-}A$, then all short exact sequences with first item as $N$ and last item as $M$ are split. My question is, what is the meaning of $\textrm{Ext}_{A}^{n}(M,N)=0, n>1$? Is there…
jin zhang
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Regarding Yoneda description of Ext group

This question is regarding the Yoneda description of $Ext^{n}$ group of r modules M and N. I want to know that what is the inverse element of an n-extension of M by N. Let $X.= 0 \to N \to X_n \to X_{n-1}\to \ldots \to X_1 \to M \to 0$ be a…
budi
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Kunneth formula for arbitrary tensor products of chain complexes

Over a field, taking homology commutes with the tensor product of two chain complexes by the Künneth formula: $H(C_*\otimes D_*)\cong H(C_*)\otimes H(D_*)$. Does this extend to arbitrary tensor products of chain complexes, i.e. do we have…
iwriteonbananas
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flat chain complexes

Let $X$ be a flat chain complex of R modules where R is a PID. Let $Z_n=\ker(d_n)$ and $B_n=\operatorname{Im}(d_{n+1})$ where $d_n :\to X_n \to X_{n-1}$. Then how can I conclude that the short exact sequence $0 \to Z_n \to X_n \to B_{n-1} \to 0$ is…
budi
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$\mbox{Hom}_R(M,N)$ as a left R-module.

I am reading the book of "Bland - Rings and their Modules (2011)". Note: $R$ is not (necessary) commutative. I want to take you to example 1 of section 1.5 on page 33. $M$ and $N$ are right R-modules. Bland claims that $\mbox{Hom}_R(M,N)$ is a…
Steenis
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