Text only print
Full screen print
Colloquium
Seminar information archive ～04/27｜Next seminar｜Future seminars 04/28～
| Organizer(s) | ASUKE Taro, TERADA Itaru, HASEGAWA Ryu, MIYAMOTO Yasuhito (chair) |
|---|---|
| URL | https://www.ms.u-tokyo.ac.jp/seminar/colloquium_e/index_e.html |
2023/06/30
15:30-16:30 Room #大講義室(auditorium) (Graduate School of Math. Sci. Bldg.)
If you do not belong to Graduate School of Mathematical Sciences, the University of Tokyo, please apply from the form at https://forms.gle/z22nKn1NUrT41qiR7
Guy Henniart (Université Paris-Saclay)
Did you say $p$-adic? (English)
If you do not belong to Graduate School of Mathematical Sciences, the University of Tokyo, please apply from the form at https://forms.gle/z22nKn1NUrT41qiR7
Guy Henniart (Université Paris-Saclay)
Did you say $p$-adic? (English)
[ Abstract ]
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!
