Library / Algebra / Chapter III / § 4
Lemma 1
Tag 060V, permanent. Cite it as [Bourbaki, Tag 060V].
Algebra, chapter III, § 4. Tensor products of algebras
Let A A A E E E A A A ( x i ) 1 ≤ i ≤ n (x_i)_{1 \leq i \leq n} ( x i ) 1 ≤ i ≤ n E E E ( λ i ) 1 ≤ i ≤ n (\lambda_i)_{1 \leq i \leq n} ( λ i ) 1 ≤ i ≤ n A A A y y y E E E
x i y = λ i y x i for 1 ≤ i ≤ n .
x_i y = \lambda_i y x_i \quad \text{for } 1 \leq i \leq n.
x i y = λ i y x i for 1 ≤ i ≤ n . Then ( x 1 x 2 … x n ) y = ( λ 1 λ 2 … λ n ) y ( x 1 x 2 … x n ) .
(x_1 x_2 \ldots x_n) y = (\lambda_1 \lambda_2 \ldots \lambda_n) y (x_1 x_2 \ldots x_n).
( x 1 x 2 … x n ) y = ( λ 1 λ 2 … λ n ) y ( x 1 x 2 … x n ) . The lemma being trivial for n = 1 n = 1 n = 1 n ≥ 2 n \geq 2 n ≥ 2
( x 1 x 2 … x n ) y = ( x 1 … x n − 1 ) ( x n y ) = ( x 1 … x n − 1 ) ( λ n y x n ) = λ n ( ( x 1 … x n − 1 ) y ) x n ,
(x_1 x_2 \ldots x_n) y = (x_1 \ldots x_{n-1})(x_n y)
= (x_1 \ldots x_{n-1})(\lambda_n y x_n) = \lambda_n ((x_1 \ldots x_{n-1}) y) x_n,
( x 1 x 2 … x n ) y = ( x 1 … x n − 1 ) ( x n y ) = ( x 1 … x n − 1 ) ( λ n y x n ) = λ n (( x 1 … x n − 1 ) y ) x n , which, by the induction hypothesis, is equal to
λ n ( λ 1 … λ n − 1 ) y ( x 1 … x n − 1 ) x n = ( λ 1 … λ n − 1 λ n ) y ( x 1 … x n − 1 x n ) ,
\lambda_n (\lambda_1 \ldots \lambda_{n-1}) y (x_1 \ldots x_{n-1}) x_n = (\lambda_1 \ldots \lambda_{n-1} \lambda_n) y (x_1 \ldots x_{n-1} x_n),
λ n ( λ 1 … λ n − 1 ) y ( x 1 … x n − 1 ) x n = ( λ 1 … λ n − 1 λ n ) y ( x 1 … x n − 1 x n ) , whence the lemma.
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, 1998, Springer. Not translated.