Timeline for Formal group as a limit of its finite subgroups
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2019 at 14:30 | comment | added | Lubin | Ah, @PiotrAchinger, thanks. I searched my leaky brain for the word “cofinal” and it wasn’t there. | |
| Oct 15, 2019 at 14:10 | comment | added | ali | @piotr achinger you are right thanks | |
| Oct 15, 2019 at 9:37 | comment | added | Piotr Achinger | As you observe, $m^{(p^n)} \neq m^{p^n}$ if $m$ is not principal. However, since $m$ is finitely generated, the two sequences of ideals are "cofinal": $$ m^{p^N} \subseteq m^{(p^n)} \subseteq m^{p^n} $$ for some $N$ depending on $n$. Indeed, if $m = (x_1, \ldots, x_r)$, then in any monomial of degree $\geq N=r(p^n-1) +1$ in the $x_i$, one factor will appear with exponent $\geq p^n$. So the inverse limits of quotient rings are the same. | |
| Oct 15, 2019 at 5:57 | comment | added | ali | @Lubin I don't understand why it most contains such an ideal?I know it's true for pid but for general complte ring and ideals it is false | |
| Oct 15, 2019 at 0:52 | comment | added | Lubin | But surely the ideal generated by the $x^{p^n}$ contains an ideal of form $m^N$ and is contained in $m^{p^n}$. Isn’t that enough? | |
| Oct 14, 2019 at 18:58 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag, fixed English
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| Oct 14, 2019 at 18:49 | history | edited | ali | CC BY-SA 4.0 |
deleted 47 characters in body
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| Oct 14, 2019 at 18:41 | history | asked | ali | CC BY-SA 4.0 |