With the above hypotheses and notation, consider 2n indeterminates T 1 , … , T n , T 1 ′ , … , T n ′ T_1, \ldots, T_n, T'_1, \ldots, T'_n T 1 , … , T n , T 1 ′ , … , T n ′ F ∈ K [ X 1 , … , X n ] F \in K[X_1, \ldots, X_n] F ∈ K [ X 1 , … , X n ] F ( T ) = F ( T 1 , … , T n ) , F ( T ′ ) = F ( T 1 ′ , … , T n ′ ) F(T) = F(T_1, \ldots, T_n), F(T') = F(T'_1, \ldots, T'_n) F ( T ) = F ( T 1 , … , T n ) , F ( T ′ ) = F ( T 1 ′ , … , T n ′ )
F ( T + T ′ ) = F ( T 1 + T 1 ′ , … , T n + T n ′ ) .
F(T + T') = F(T_1 + T'_1, \ldots, T_n + T'_n).
F ( T + T ′ ) = F ( T 1 + T 1 ′ , … , T n + T n ′ ) . Suppose that P ( T + T ′ ) = ∑ i Q i ( T ) R i ( T ′ )
P(T + T') = \sum_i Q_i(T) R_i(T')
P ( T + T ′ ) = i ∑ Q i ( T ) R i ( T ′ ) where the Q i Q_i Q i R i R_i R i K [ X 1 , … , X n ] K[X_1, \ldots, X_n] K [ X 1 , … , X n ] x ∈ A x \in A x ∈ A y ∈ A ′ y \in A' y ∈ A ′ P ( D ) ( x y ) = ∑ i ( Q i ( D ) x ) ( R i ( D ) y ) .
P(D)(xy) = \sum_i (Q_i(D)x)(R_i(D)y).
P ( D ) ( x y ) = i ∑ ( Q i ( D ) x ) ( R i ( D ) y ) . We introduce n n n T 1 ′ ′ , … , T n ′ ′ T_1'', \ldots, T_n'' T 1 ′′ , … , T n ′′ K [ T 1 , … , T n , T 1 ′ , … , T n ′ , T 1 ′ ′ , … , T n ′ ′ ] = B K[T_1, \ldots, T_n, T_1', \ldots, T_n', T_1'', \ldots, T_n''] = B K [ T 1 , … , T n , T 1 ′ , … , T n ′ , T 1 ′′ , … , T n ′′ ] = B K K K M M M A × A ′ A \times A' A × A ′ A ′ ′ A'' A ′′ B B B M M M K K K f ∈ M f \in M f ∈ M 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
{ ( T i f ) ( a , a ′ ) = f ( d i a , a ′ ) ( T i ′ f ) ( a , a ′ ) = f ( a , d i a ′ ) ( T i ′ ′ f ) ( a , a ′ ) = d i ( f ( a , a ′ ) )
\begin{cases}
(T_i f)(a, a') = f(d_i a, a') \\
(T_i' f)(a, a') = f(a, d_i a') \\
(T_i'' f)(a, a') = d_i(f(a, a'))
\end{cases}
⎩ ⎨ ⎧ ( T i f ) ( a , a ′ ) = f ( d i a , a ′ ) ( T i ′ f ) ( a , a ′ ) = f ( a , d i a ′ ) ( T i ′′ f ) ( a , a ′ ) = d i ( f ( a , a ′ )) Since the d i d_i d i F ∈ K [ X 1 , … , X n ] F \in K[X_1, \ldots, X_n] F ∈ K [ X 1 , … , X n ] ( F ( T ) f ) ( a , a ′ ) = f ( F ( D ) a , a ′ ) (F(T)f)(a, a') = f(F(D)a, a') ( F ( T ) f ) ( a , a ′ ) = f ( F ( D ) a , a ′ )
( F ( T ′ ) f ) ( a , a ′ ) = f ( a , F ( D ) a ′ )
(F(T')f)(a, a') = f(a, F(D)a')
( F ( T ′ ) f ) ( a , a ′ ) = f ( a , F ( D ) a ′ ) and ( F ( T ′ ′ ) f ) = F ( D ) ( f ( a , a ′ ) ) (F(T'')f) = F(D)(f(a, a')) ( F ( T ′′ ) f ) = F ( D ) ( f ( a , a ′ ))
( P ( T ′ ′ ) − ∑ i Q i ( T ) R i ( T ′ ) ) . μ = 0
(P(T'') - \sum_i Q_i(T)R_i(T')).\mu = 0
( P ( T ′′ ) − i ∑ Q i ( T ) R i ( T ′ )) . μ = 0 or also ( P ( T ′ ′ ) − P ( T + T ′ ) ) . μ = 0 (P(T'') - P(T + T')).\mu = 0 ( P ( T ′′ ) − P ( T + T ′ )) . μ = 0 B B B M M M d i d_i d i 1 ≤ i ≤ n 1 \leq i \leq n 1 ≤ i ≤ n
( T i ′ ′ − T i − T i ′ ) . μ = 0
(T_i'' - T_i - T_i').\mu = 0
( T i ′′ − T i − T i ′ ) . μ = 0 in the B B B M M M
P ( T 1 ′ ′ , T 2 ′ ′ , … , T n ′ ′ ) − P ( T 1 + T 1 ′ , T 2 ′ ′ , … , T n ′ ′ ) P(T_1'', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2'', \ldots, T_n'') P ( T 1 ′′ , T 2 ′′ , … , T n ′′ ) − P ( T 1 + T 1 ′ , T 2 ′′ , … , T n ′′ ) P ( T 1 + T 1 ′ , T 2 ′ ′ , … , T n ′ ′ ) − P ( T 1 + T 1 ′ , T 2 + T 2 ′ , … , T n ′ ′ ) P(T_1 + T_1', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2 + T_2', \ldots, T_n'') P ( T 1 + T 1 ′ , T 2 ′′ , … , T n ′′ ) − P ( T 1 + T 1 ′ , T 2 + T 2 ′ , … , T n ′′ ) … \ldots … P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n ′ ′ ) − P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n + T n ′ ) P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n'') - P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n + T_n') P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n ′′ ) − P ( T 1 + T 1 ′ , … , T n − 1 + T n − 1 ′ , T n + T n ′ ) it is seen that the difference P ( T ′ ′ ) − P ( T + T ′ ) P(T'') - P(T + T') P ( T ′′ ) − P ( T + T ′ )
∑ i = 1 n ( T i ′ ′ − T i − T i ′ ) G i ( T , T ′ , T ′ ′ )
\sum_{i=1}^n (T_i'' - T_i - T_i') G_i(T, T', T'')
i = 1 ∑ n ( T i ′′ − T i − T i ′ ) G i ( T , T ′ , T ′′ ) where the G i G_i G i B B B
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.