Library / Algebra / Chapter VII / § 2
Lemma 3
Tag 072L, permanent. Cite it as [Bourbaki, Tag 072L].
Algebra, VII, p. 12, § 2. Torsion modules over a principal ideal domain
Let x, y, k be integers with k ≥ 0 k \geq 0 k ≥ 0 p > 2 p > 2 p > 2 x ≡ 1 + p y m o d p 2 x \equiv 1 + py \mod p^2 x ≡ 1 + p y mod p 2 x p k ≡ 1 + p k + 1 y m o d p k + 2 x^{p^k} \equiv 1 + p^{k+1}y \mod p^{k+2} x p k ≡ 1 + p k + 1 y mod p k + 2 If x ≡ 1 + 4 y m o d 8 x \equiv 1 + 4y \mod 8 x ≡ 1 + 4 y mod 8 x 2 k ≡ 1 + 2 k + 2 y m o d 2 k + 3 x^{2^k} \equiv 1 + 2^{k+2}y \mod 2^{k+3} x 2 k ≡ 1 + 2 k + 2 y mod 2 k + 3
To prove the first assertion, it is enough to show that, if k ≥ 1 k \geq 1 k ≥ 1 x ≡ 1 + p k y m o d p k + 1 x \equiv 1 + p^ky \mod p^{k+1} x ≡ 1 + p k y mod p k + 1 x p ≡ 1 + p k + 1 y m o d p k + 2 x^p \equiv 1 + p^{k+1}y \mod p^{k+2} x p ≡ 1 + p k + 1 y mod p k + 2 a ∈ Z a \in \mathbf{Z} a ∈ Z k ≥ 1 k \geq 1 k ≥ 1
( 1 + p k a ) p ≡ 1 + p k + 1 a m o d p k + 2 ,
(1 + p^ka)^p \equiv 1 + p^{k+1}a \mod p^{k+2},
( 1 + p k a ) p ≡ 1 + p k + 1 a mod p k + 2 , hence
( 1 + p k y + p k + 1 z ) p = ( 1 + p k ( y + p z ) ) p ≡ = 1 + p k + 1 ( y + p z ) ≡ 1 + p k + 1 y m o d p k + 2 .
(1 + p^ky + p^{k+1}z)^p = (1 + p^k(y + pz))^p \equiv \\
= 1 + p^{k+1}(y + pz) \equiv 1 + p^{k+1}y \mod p^{k+2}.
( 1 + p k y + p k + 1 z ) p = ( 1 + p k ( y + p z ) ) p ≡ = 1 + p k + 1 ( y + p z ) ≡ 1 + p k + 1 y mod p k + 2 . Similarly, for
k ≥ 1 k \geq 1 k ≥ 1 we have
( 1 + 2 k + 1 a ) 2 ≡ 1 + 2 k + 2 a m o d 2 k + 3 ,
(1 + 2^{k+1}a)^2 \equiv 1 + 2^{k+2}a \mod 2^{k+3},
( 1 + 2 k + 1 a ) 2 ≡ 1 + 2 k + 2 a mod 2 k + 3 , so
( 1 + 2 k + 1 y + 2 k + 2 z ) 2 ≡ 1 + 2 k + 2 y m o d 2 k + 3 ,
(1 + 2^{k+1}y + 2^{k+2}z)^2 \equiv 1 + 2^{k+2}y \mod 2^{k+3},
( 1 + 2 k + 1 y + 2 k + 2 z ) 2 ≡ 1 + 2 k + 2 y mod 2 k + 3 , whence the second assertion by induction on k.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Transcribed from the printed edition, 2003, Springer. Not translated.