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Results tagged with homological-algebra
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user 460
(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.
7
votes
Tachikawa conjecture for commutative algebras proven?
On page 2163, the second line, the authors say "therefore ... is exact". I, and some others, believe this is a gap. The authors have been contacted (in 2010) and have not clarified.
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1
vote
Accepted
Decomposition of a quotient module
Edited to add: Well, now I feel embarrassed to have gotten an answer accepted which is absolute garbage, so I think I should offer an actual answer in addition to the indirect proof in a comment I mad …
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5
votes
Accepted
Vanishing of $\mathrm{Ext}^i_R(N, R)$
No.
Let's work with an Artinian local ring $R$. Let $X$ be a module satisfying $\mathrm{Ext}_R^i(X,R)\neq 0$ for all $i$. (These are plentiful; for example, assume $R$ is non-Gorenstein and let $X$ …
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2
votes
Accepted
Disappearing of $Ext^v_A(M,A)$
Yes. One can even restrict to $M=A/I$ for ideals $I$, and if $A$ is Noetherian it is enough to consider $M=A/p$ for prime ideals $p$. This is Lemma 18.1 in Matsumura's book "Commutative ring theory" …
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3
votes
Accepted
Are maximal Cohen-Macaulay modules supported everywhere?
No. $R = k[x,y]/(xy)$, $M = R/(x)$.
The zero module has infinite depth and support of dimension $-\infty$, so should not be considered MCM.
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3
votes
Accepted
A Module with $Ext^i(M,R) = 0$ for all $i > 0$
As regards the second question: consider for example $R = k[[x]]$ where $k$ is a field. By the structure theory for modules over a PID, indecomposable finitely generated $R$-modules are cyclic, of th …
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6
votes
Additivity of projective dimensions, or, help me lower my blood pressure
One problem in the proof is that $\operatorname{pd}_R(R/I)$ might be infinite, so you can't apply Auslander-Buchsbaum the final time.
I can't seem to come up with an example where $M$ has finite proje …
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4
votes
Accepted
Homological dimensions of module
Yes. See Fossum, Foxby, Griffith, and Reiten, "Minimal injective resolutions with applications to dualizing modules and Gorenstein modules" (Theorem 1.1) and also Roberts, "Two applications of dualiz …
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7
votes
Accepted
Selforthogonal modules over Artinian Gorenstein rings
This would be a very strong version of the Auslander-Reiten Conjecture (see here, for example) in the Gorenstein case. The Conjecture is still open, though many partial results are known.
By the way, …
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6
votes
Differences between reflexives and projectives modules
In general, there are many many more reflexive modules than projective modules. As pointed out in another answer, every second syzygy is reflexive, and not necessarily projective (unless your ring ha …
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8
votes
Accepted
Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension
The "Theorem" isn't true with both rings just normal, or just CM, or even normal and CM. Let $A = k[[x,y,z]]/(xz-y^2) \cong k[[a^2,ab,b^2]]$ and let $B = k[[a,b]]$, with $f$ the natural inclusion. T …
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2
votes
Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension
If $A$ is still regular and $B$ is anything at all, then $B$ has finite flat dimension over $A$. So this is strictly weaker on one ring, though not on both.
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11
votes
Some intuition behind the five lemma?
Since long exact sequences come from splicing together short exact sequences, you might as well worry about the case where $A_1=A_5=B_1=B_5=0$ (at least as far as intuition is concerned). This follow …
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