Timeline for $\mathbb Z_n[x_1^\pm,\dots,x_D^\pm]$-modules extended from $\mathbb Z_n$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 20 at 23:51 | comment | added | Andrea Marino | Of course, you are right. I am quite convinced that there is an error somewhere: the adjustment I made of my 'typo' is partial and should be revisited more thoroughly. The devil is in the details... | |
| Jan 19 at 21:56 | comment | added | Blazej | A short exact sequence $0 \to A \to B \to C \to 0$ with $C$ free splits, so $B \cong A \oplus C$. I will rethink your counterexample with the typo corrected later. | |
| Jan 19 at 18:58 | comment | added | Andrea Marino | Yes, you are right: I forgot a -1. I was identifying $x^km$ with $m$, which corresponds of course to quotienting by $x^k-1$. Why do you claim that an extension of free modules by a free module is free? It seems like mine is a counterexample, but maybe you have a simple homological proof in mind, and that's why you are puzzled by my calculation. | |
| Jan 19 at 18:53 | history | edited | Andrea Marino | CC BY-SA 4.0 |
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| Jan 19 at 16:48 | comment | added | Blazej | Unless I am screwing something up, there can't be a counter-example for prime $n$. Then quasi-free is equivalent to free, and an extension of a free module by a free module is free. Characterization of free modules in homological terms is given by the (positive resolution of) Serre's "conjecture". | |
| Jan 19 at 16:40 | comment | added | Blazej | I am a bit confused by the part titled "Observation". $x$ is an invertible element of the ring, so $M/(x^k M)=0$. | |
| Jan 1 at 20:24 | history | edited | Andrea Marino | CC BY-SA 4.0 |
added 158 characters in body
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| Jan 1 at 19:47 | history | answered | Andrea Marino | CC BY-SA 4.0 |