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談話会・数理科学講演会
過去の記録 ～04/30｜次回の予定｜今後の予定 05/01～
| 担当者 | 足助太郎，寺田至，長谷川立，宮本安人（委員長） |
|---|---|
| セミナーURL | https://www.ms.u-tokyo.ac.jp/seminar/colloquium/index.html |
2023年06月30日(金)
15:30-16:30 数理科学研究科棟(駒場) 大講義室(auditorium)号室
数理科学研究科所属以外の方は、https://forms.gle/Pw6AHaJjqAwaHB8s9から参加登録をお願いいたします。
Guy Henniart 氏 (Université Paris-Saclay)
Did you say $p$-adic? (English)
数理科学研究科所属以外の方は、https://forms.gle/Pw6AHaJjqAwaHB8s9から参加登録をお願いいたします。
Guy Henniart 氏 (Université Paris-Saclay)
Did you say $p$-adic? (English)
[ 講演概要 ]
I am a Number Theorist and $p$ is a prime number. The $p$-adic numbers are obtained by pushing to the limit a simple idea. Suppose that you want to know which integers are sums of two squares. If an integer $x$ is odd, its square has the form $8k+1$; if $x$ is even, its square is a multiple of $4$. So the sum of two squares has the form $4k$, $4k+1$ or $4k+2$, never $4k+3$ ! More generally if a polynomial equation with integer coefficients has no integer solution if you work «modulo $N$» that is you neglect all multiples of an integer $N$, then a fortiori it has no integer solution. By the Chinese Remainder Theorem, working modulo $N$ is the same as working modulo $p^r$ where $p$ runs through prime divisors of $N$ and $p^r$ is the highest power of $p$ dividing $N$. Now work modulo $p$, modulo $p^2$, modulo $p^3$, etc. You have invented the $p$-adic integers, which are, I claim, as real as the real numbers and (nearly) as useful!
I am a Number Theorist and $p$ is a prime number. The $p$-adic numbers are obtained by pushing to the limit a simple idea. Suppose that you want to know which integers are sums of two squares. If an integer $x$ is odd, its square has the form $8k+1$; if $x$ is even, its square is a multiple of $4$. So the sum of two squares has the form $4k$, $4k+1$ or $4k+2$, never $4k+3$ ! More generally if a polynomial equation with integer coefficients has no integer solution if you work «modulo $N$» that is you neglect all multiples of an integer $N$, then a fortiori it has no integer solution. By the Chinese Remainder Theorem, working modulo $N$ is the same as working modulo $p^r$ where $p$ runs through prime divisors of $N$ and $p^r$ is the highest power of $p$ dividing $N$. Now work modulo $p$, modulo $p^2$, modulo $p^3$, etc. You have invented the $p$-adic integers, which are, I claim, as real as the real numbers and (nearly) as useful!
