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Oct 15 at 19:03 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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Jan 1 at 19:47 answer added Andrea Marino timeline score: 0
Jan 1 at 17:33 comment added Andrea Marino (false proof that every module is quasi-free: consider $M_0 = M \otimes_R \mathbb{Z}_n$. Then $M_0 \otimes_{\mathbb{Z}_n} R \simeq (M \otimes_R \mathbb{Z}_n ) \otimes_{\mathbb{Z}_n} R \simeq M \otimes_R (\mathbb{Z}_n \otimes_{\mathbb{Z}_n} R ) \simeq M \otimes_R R \simeq M$ )
Jan 1 at 16:40 comment added Andrea Marino I think $M$ quasi-free and fgen can be characterized quite explicitly. Note that $M \otimes_R \mathbb{Z}_n \simeq M_0$ must be fgen too. By the classification theorem, $M_0 \simeq \bigoplus_{\lambda \in \Lambda} \mathbb{Z}_{q_{\lambda}}$ with $\Lambda$ finite and $q_{\lambda} \mid n$ prime powers. This implies $M \simeq \bigoplus R/q_{\lambda} R$, that is a finite sum of quotients for primes coming from $R \to \mathbb{Z}_n$.
S Jan 1 at 14:09 history bounty started Blazej
S Jan 1 at 14:09 history notice added Blazej Draw attention
Jun 7, 2022 at 10:27 comment added Blazej In the first version of this question I mentioned that quasi-free modules $M$ satisfy $\mathrm{Ext}^i_R(M,N)=0$ for every $i>0$ and every $N$ free over $\mathbb Z_n$ (equivalently, $N$ of finite projective dimension) and asked if the converse is true. Then I realized that module $M = \mathrm{im} \begin{pmatrix} 2 & 0 \\ 1-x & 2 \end{pmatrix}$ with $n=4$ and $D=1$ provides a counterexample to this converse.
Jun 7, 2022 at 10:23 history edited Blazej CC BY-SA 4.0
I realized that one point in my question is not true.
Jun 7, 2022 at 8:20 history asked Blazej CC BY-SA 4.0