Colloquium

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Seminar information archive ～05/24｜Next seminar｜Future seminars 05/25～

Organizer(s)	AIDA Shigeki, OSHIMA Yoshiki, SHIHO Atsushi (chair), TAKADA Ryo
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)

[ 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!

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