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(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.
3
votes
Surjectivity of natural map of rings
Write the right-hand side as $Hom_B(P/P^2,B)$. If the map you are interested in is surjective, then the preimage of the trace ideal of $P/P^2$ in $B$ must be contained in the the trace ideal of $P$ in …
1
vote
On the map $\Phi_M: M\otimes_RM^*\xrightarrow{x\otimes y\mapsto \left\{f\mapsto f(x)y\right\...
It should be noted that the answer is yes if $R$ is normal and $M$ is torsion-free. That is because of the:
Fact: if a map $f:A \to B$ of reflexive modules is locally an isomorphism in codimension on …
5
votes
Accepted
Ideals in Artinian Gorenstein local ring $(R,\mathfrak m)$ with $\mu(\mathfrak m)=2, \mathfr...
We use the fact that in an Artinian Gorenstein ring, any ideal contains the socle. The assumption tells us that the socle of $A$ is $\mathfrak m^2$, which is principal.
Let $I\neq (0)$ be a non-maxima …
7
votes
Accepted
Is Koszul homology of a monomial ideal always generated by the "obvious" things?
This holds for $n\leq 3$ but may fail for $n=4$ and higher. See Proposition 2.6 and Example 2.9 in the paper "On monomial Golod ideals" (but probably known to experts before).
1
vote
Accepted
Indecomposable modules such that the radical is a submodule of the socle
No. Let $(R,\mathfrak m)$ be commutative local Artin ring, then the radical of $M$ is $\mathfrak mM$ and your condition is equivalent to $\mathfrak m^2M=0$. One can not bound the length of such indeco …
2
votes
Projective dimension of a sub-ideal
Interestingly, the equality you seek holds in one important special case. If $I$ is any monomial ideal and $J$ is the radical of $I$, then $pd_S(I)\leq pd_S(J)$. See the proof of Theorem 2.6 in this p …
26
votes
Short exact sequences every mathematician should know
Given a finitely generated module $M$ over a commutative Noetherian ring $R$, there is a short exact sequence $$0\to M_1 \to R^n \to M\to 0$$
where you map $1$ in each $R$ to a generator of $M$ and $M …
6
votes
When are MCM ideals principal?
On question 1, for what rings all MCM ideals are principal, we can say quite a bit more if one knows that $R$ is parafactorial (that is, the Picard group of the punctured spectrum $Spec^o(R):=Spec(R)- …
8
votes
Betti sequence of finite dimensional commutative algebras
I will first show that the general answer is no (in fact one should expect the opposite even in the graded case) by the Pigeonhole principle. Then we shall construct some concrete examples via an old …
11
votes
Accepted
Commutative algebras with modules of small complexity
There are no other examples. This property is equivalent to $A$ being a hypersurface (see Avramov's note "Infinite Free Resolutions"). By Cohen Structure Theorem, an Artinian local hypersurface (which …
8
votes
Accepted
Equivalence of definitions of Cohen-Macaulay type
For the equivalence you need two more assumptions: (a) $M$ to have finite projective dimension and (b) $R$ to be Gorenstein. (a) is implicitly required in the second condition, otherwise you don't hav …
7
votes
Accepted
For every prime ideal $P$ of any Cohen-Macaulay ring $R$, is the sequence $\operatorname{dep...
Yes, for any ideal in a Noetherian local ring. See: this paper.
4
votes
Ext in symmetric algebras and group algebras
Suppose $A$ is commutative local (necessarily artinian) with the only simple $k\neq A$ then $\psi_k=1$, so your statement 1 will say that $Ext^1_A(M,M)=0$ implies $M$ is free. I stated it as a conje …
4
votes
Accepted
Multiplicity of $Ext^{d-t}(M,\omega_R)$, ($d=\dim R, t=\dim M$)
It is equal to the multiplicity of $M$. In fact, you do not need graded or even Cohen-Macaulayness of $M$. Let $N= \textrm{Ext}^{d-t}(M,\omega_R)$. Let $S(M) := \{P \in \textrm{Supp}(M), \dim R/P = t\ …
12
votes
Accepted
example of Local cohomology
Take $M$ to be the second syzygy of $k$ over $S=k[x_1,x_2,x_3]$. Then a graded version of local duality tells us that $H^2_m(M)$ is dual to $Ext^1(M,R)= Ext^3(k,R)$, the last one is $k$ either by dir …