Timeline for Does $p$-adic Baker theorem holds in the given case?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2023 at 12:41 | comment | added | Oleg Eroshkin | @ANG That is the conclusion of Brumer theorem. | |
| Apr 8, 2023 at 3:56 | comment | added | MAS | @OlegEroshkin, thanks, but i don't see how logs of $\alpha_1$ and $\alpha_2$ are L.I. over the algebraic closure $A$ of $\mathbb Q$ in $\Omega_2$. Though it should be, but I don't see how | |
| Apr 8, 2023 at 3:52 | history | edited | MAS | CC BY-SA 4.0 |
edited body
|
| Apr 7, 2023 at 20:21 | comment | added | Oleg Eroshkin | @ANG The fact that $\sqrt{-1}$ is not in $\mathbb{Q}_2$ is irrelevant here. If logs of $\alpha_1$ and $\alpha_2$ are linearly independent over $\mathbb{Q}$ then they are independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\Omega_2$ (which contains $\sqrt{-1}$). | |
| Apr 6, 2023 at 16:15 | comment | added | MAS | @OlegEroshkin, just confirm me. In my example $\log_2(\alpha_1)$ and $\log_2(\alpha_2)$ are linearly independent over $\mathbb Q$. So Brumer theorem satisfies, and so by Brumer theorem (above) $\log_2(\alpha_1)$ and $\log_2(\alpha_2)$ are also linearly independent over $\bar {\mathbb Q} \cap \mathbb Q_2$. And the last conclusion is indeed true because $\sqrt{-1} \notin \mathbb Q_2$. Am I right ? | |
| Apr 1, 2023 at 12:41 | comment | added | Oleg Eroshkin | @ANG Algebraic independence is a much stronger condition than linear independence. If numbers are linearly dependent over $\mathbb{Q}$ they are (trivially) algebraically dependent as well. | |
| Apr 1, 2023 at 4:10 | history | edited | MAS | CC BY-SA 4.0 |
added 13 characters in body
|
| Apr 1, 2023 at 3:52 | comment | added | MAS | @OlegEroshkin, ahh, thanks. I am silly. Yes my example is not satisfying the conditions of Brumer theorem. Or, we can say my example satisfies Brumer theorem for $n=2$. I got it. Can you make any comment on the last sentence of my previous comment ? thanks | |
| Apr 1, 2023 at 3:49 | comment | added | Oleg Eroshkin | @ANG What do you mean: theorem is failing? A conclusion of a theorem is true if the conditions are satisfied. One of the conditions of Brumer's theorem is " logs are independent over $\mathbb{Q}$". That's false in your case. So the theorem is not failing - it's not applicable. | |
| Apr 1, 2023 at 3:34 | comment | added | MAS | @OlegEroshkin, thank you very much. I understand these logarithms in my example are not LI for $n > 2$, but my question was really in different direction. why Brumer theorem is failing in my example, for $n > 2$ ? Because Brumer theorem is supposed to hold since it is a general statement. I think LI holds on $A =\bar{\mathbb{Q}} \cap \mathbb Q_2$ not simply on $\mathbb Q$ in Brumer theorem. Can you please comment on this ? By the way, there is still a chance that the logarithm in my example can be algebraically independent (AI) because LI and AI are separate things. What do you say ? | |
| Apr 1, 2023 at 1:38 | comment | added | Oleg Eroshkin | @ANG Let me explain with a simple example. Numbers $2+\alpha$, $3+2\alpha$, and $5+11\alpha$ are linearly dependent over $\mathbb{Q}$ even if $\alpha$ is irrational. You assume that if $\sqrt{-1}$ is not in your field, the linear combinations with it are linearly independent is false. | |
| Apr 1, 2023 at 0:18 | history | edited | MAS | CC BY-SA 4.0 |
added 219 characters in body
|
| Apr 1, 2023 at 0:16 | comment | added | MAS | @JoeSilverman, very nice. I understand. So we don't know about algebraic independence of the $p$-adic logarithms. As for linear independence, why my example fails to hold Brumer theorem (above) for $n>2$ ? | |
| Mar 31, 2023 at 23:55 | comment | added | Joe Silverman | I think you may be confusing the definitions of linear dependence and algebraic dependence. Baker's theorem and it's generalizations are about linear independence of logarithms, but of course, one can write $\sum a_i\log(b_i)=0$ as $\prod b_i^{a_i}=1$ with appropriate choices of branches of the exponentials, so one can translate Baker's theorem into products. What is open is algebraic indpendence of the logarithms with appropriate hypotheses. | |
| Mar 31, 2023 at 23:40 | comment | added | MAS | @OlegEroshkin, Sorry, I couldn't understand. In your 1st comment said "algebraic independence of $p$-dic logarithms is an open problem over $p$-adic". what does mean by it ? | |
| Mar 31, 2023 at 23:32 | comment | added | Oleg Eroshkin | Sorry, I don't understand your question. Algebraic independence over some field $K$ is the same as over the algebraic closure $\bar{K}$. Regarding your question, if $\log_p(\alpha_i)$ are linear combinations over $\mathbb{Q}$ of 2 numbers and $n>2$, then they are linearly dependent over $\mathbb{Q}$. It's a basic linear algebra. | |
| Mar 31, 2023 at 23:18 | history | edited | MAS |
edited tags
|
|
| Mar 31, 2023 at 23:17 | comment | added | MAS | @OlegEroshkin, thanks. So algebraic independence is not known over $\mathbb Q_p$ or $\bar{\mathbb Q}_p$ in the $p$-adic setting. Anyway, does my example work here, at least for the Brumer theorem ? | |
| Mar 31, 2023 at 18:51 | comment | added | Oleg Eroshkin | That page on wikipedia is saying that the algebraic independence of logarithms is an open problem both over complex numbers and over p-adic. The Baker's theorem over p-adic is very well known and the estimates are on par with estimates in the complex case. | |
| Mar 31, 2023 at 17:16 | history | asked | MAS | CC BY-SA 4.0 |