>>928 追加

MIT Sutherland先生の講義下記
これ分かり易い
”2 = (2, 2, 2, 2, 2, . . .)
2002 = (0, 42, 287, 2002, 2002, . . .)
?2 = (5, 47, 341, 2399, 16805, . . .)
2^?1 = (4, 25, 172, 1201, 8404, . . .)
√2 = ((3, 10, 108, 2166, 4567 . . .)”
か。なるほどね。

Zpは、?2、2^?1、√2が表現できる、というか、含むわけか
Z_7には、”no cube roots”つまり、2^(1/3)は含まない
なお、√2はZ_7内に二通りの展開を持つとあるね

要するに、実数R中の多くの数が、Zp つまりp進”整数”ってわけね
で、2 = (2, 2, 2, 2, 2, . . .)みたく、ずっと上の桁まで詰まっている

なお、”You can easily recreate ・・ in Sage”とある。Sageは数式システムか
これは、雪江本百回音読しても、自分には分からんだろうなｗ

(参考)
https://math.mit.edu/classes/18.782/lectures.html
LECTURES MIT 18.782 - Arithmetic Geometry
https://math.mit.edu/classes/18.782/LectureNotes4.pdf
18.782 Introduction to Arithmetic Geometry Fall 2013
Lecture #4 09/17/2013
Andrew V. Sutherland
4.1 Inverse limits

4.2 The ring of p-adic integers
Definition 4.3. For a prime p, the ring of p-adic integers Zp is the inverse limit
Zp = lim ←? Z/pnZ
of the inverse system of rings (Z/pnZ) with morphisms (fn) given by reduction modulo p^n
(for a residue class x ∈ Z/pn+1Z, pick an integer x ∈ x and take its residue class in Z/pnZ).
The multiplicative identity in Zp is 1 = (￣1, ￣1, ￣1, . . .), where the nth ￣1 denotes the residue
class of 1 in Z/pnZ. The map that sends each integer x ∈ Z to the sequence (￣x, ￣x, ￣x, . . . ￣ ) is
a ring homomorphism, and its kernel is clearly trivial, since 0 is the only integer congruent
to 0 mod p^n for all n. Thus the ring Zp has characteristic 0 and contains Z as a subring.
But Zp is a much bigger ring than Z.

つづく