Timeline for What is the preimage of the maximal ideal under certain exponential functions?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 19, 2021 at 19:24 | comment | added | KConrad | Perhaps you don't need exact formulas for the coefficients, but just estimates on their absolute values. For example, if $|x|_p < (1/p)^{1/(p-1)}$ then $|x^{n-1}/n!|_p < 1$ for all $n \geq 1$, so if $f(x) = x + \sum_{n \geq 2} c_nx^n$ where $|c_n|_p = |1/n!|_p$ then $|f(x) - f(y)|_p = |x-y|_p$ when $|x|_p, |y|_p < (1/p)^{1/(p-1)}$ because $|c_n(x^n-y^n)|_p < |x-y|_p$ for $n \geq 2$. | |
| Oct 19, 2021 at 17:01 | comment | added | user474 | I do not know of any counterexamples to this, and I think the multiplicative formal group law is the only formal group law for which I know an explicit formula for the log. In general I know there are other expressions for the logarithm, but I am unaware of these giving explicit formulas. | |
| Oct 17, 2021 at 5:58 | comment | added | KConrad | When $F = \mathbb G_m$ is the formal multiplicative group, $\exp_F(x) = e^x - 1 = \sum_{n \geq 1} x^n/n = x + \cdots$. For each finite extension $L/\mathbf Q_p$, the disc of convergence of $e^x - 1$ in $L$ is $\{x \in L : |x|_p < (1/p)^{1/(p-1)}\}$, on which $e^x - 1$ is an isometry. In particular, ${\rm ord}(e^x - 1) > 0$ for all $x$ in $L$ where $e^x - 1$ converges. Do you know counterexamples to that property for other formal groups? | |
| Oct 14, 2021 at 12:10 | history | asked | user474 | CC BY-SA 4.0 |