Timeline for linear independent families in a tensor product
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 21, 2023 at 17:46 | vote | accept | Sebastian Goette | ||
| Aug 21, 2023 at 7:15 | answer | added | Jeremy Rickard | timeline score: 6 | |
| Aug 20, 2023 at 18:19 | answer | added | Libli | timeline score: 0 | |
| Jul 15, 2017 at 10:52 | comment | added | Sebastian Goette | @RobertBryant I wonder wether $R=\mathbb Z/6$ is weird enough to produce a counterexample. $R$ is the sum of two fields, and it seems to me that one can check separately what happens over $F_2$ and over $F_3$. That is, if $(e_i)$ is linearly independent in $M$ over $\mathbb Z_6$, then it has linearly independent images in the localisations of $M$ at $2$ and $3$, and we are in the field situation again. Probably, one needs rings like $\mathbb Z[X]/X^2$ or even worse. | |
| Jul 14, 2017 at 18:35 | comment | added | Robert Bryant | @JeremyRickard: Oops! You are right; I had in mind a bad definition of 'linearly indepdent' for rings that are not integral domains when I suggested $\mathbb{Z}_6$ as a counter-example. I won't delete it, though, since it would make your comment useless. | |
| Jul 14, 2017 at 8:47 | comment | added | Jeremy Rickard | @RobertBryant But $M$ and $N$ don't contain any non-empty linearly independent families of elements. | |
| Jul 14, 2017 at 8:37 | comment | added | Robert Bryant | What about $R = \mathbb{Z}_6 = \mathbb{Z}_2\oplus\mathbb{Z}_3$, with $M=\mathbb{Z}_2$ and $N=\mathbb{Z}_3$? Then $M\otimes_RN = (0)$, even though each of $M$ and $N$ are principal $R$-modules (i.e., generated by a single element). | |
| Jul 14, 2017 at 8:15 | history | asked | Sebastian Goette | CC BY-SA 3.0 |