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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 …

One can not bound the length of such indecomposable modules in general. … In particular, if you take $R=k[x,y]/(x,y)^2$ then it has $\mathfrak m^2=0$, so any module works, and there are indecomposable modules of arbitrary length over $R$, as it does not have finite representation …

As I proved in the answer to this question, the following holds: Proposition: For an Artinian ring $A$, the following are equivalent: $A$ is Gorenstein. Any finitely generated faithful module $M$ ov …

Here is a big class of negative examples to your question. Let $A$ be a non-Gorenstein Artinian ring and $M=w_A$ the canonical module of $A$. Then $ann(M)=0$, as can be seen because $Hom(M,M)\cong A$. …

In general, I think for a non-Gorenstein domain of dimension one, torsion free modules are rarely reflexive. …

Let $G$ denote the Grothendieck group of all finitely generated $R$-modules. … Thus $f$ induces a group homomorphism $\bar G \to \mathbb Z$, where $\bar G$ is $G/H$ where $H$ is the subgroup generated by torsion modules. …

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\ …

Hi Pete, this sounds like a lot of fun! I wish I could be there (-: Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension. T …