Timeline for Reference request: conditions for the cardinality of the kernel of a linear map from $\mathbb{Z}_m^n \to \mathbb{Z}_m^k$ is a power of $m$
Current License: CC BY-SA 4.0
13 events
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Apr 4, 2019 at 6:08 | history | edited | Jianrong Li |
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Apr 2, 2019 at 15:55 | vote | accept | Jianrong Li | ||
Apr 1, 2019 at 20:52 | answer | added | tj_ | timeline score: 3 | |
Apr 1, 2019 at 20:44 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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Apr 1, 2019 at 20:42 | comment | added | Jianrong Li | @KConrad, thanks. Yes, I care about only the size of the kernel. | |
Apr 1, 2019 at 20:23 | comment | added | KConrad | You changed the question in a significant way (which is okay, since nobody really answered it yet). All you care about is the size of the kernel having a particular type of formula? | |
Apr 1, 2019 at 20:14 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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Apr 1, 2019 at 20:13 | comment | added | Jianrong Li | @tj_, thank you very much for your comments. They are not the same. I will edit the question. | |
Apr 1, 2019 at 19:04 | comment | added | Sam Hopkins | One obvious condition: $m$ is prime. | |
Apr 1, 2019 at 18:58 | history | edited | Jianrong Li | CC BY-SA 4.0 |
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Apr 1, 2019 at 18:55 | comment | added | Jianrong Li | @KConrad, thank you very much for your comments. I edited the question. | |
Apr 1, 2019 at 17:25 | comment | added | KConrad | Strictly speaking, $\ker(f) \subset \mathbf Z_m^4$ while $\mathbf Z_m^2$ is not in a canonical way a subgroup of $\mathbf Z_m^4$, so it is incorrect to say $\ker(f)$ equals $\mathbf Z_m^2$. | |
Apr 1, 2019 at 15:19 | history | asked | Jianrong Li | CC BY-SA 4.0 |