Let ρ:A→B be a local homomorphism of noetherian local rings, M a finitely generated A-module and N a finitely generated B-module; suppose that the A-module N is flat.
a) Let (x1,…,xr) be a sequence of elements of mA which is regular for the A-module M, and (y1,…,ys) a sequence of elements of mB which is regular for the B-module κA⊗AN; then the sequence (y1,…,ys,ρ(x1),…,ρ(xr)) of elements of mB is regular for the B-module M⊗AN.
b) One has the equality
profB(M⊗AN)=profA(M)+profB(κA⊗AN)
Let x denote the ideal of A generated by the sequence x and y the ideal of B generated by y. By Prop. 10, the sequence y is M⊗AN-regular and N/yN is flat over A, so that the sequence ρ(x)=(ρ(x1),…,ρ(xr)) is regular for M⊗A(N/yN)=(M⊗AN)/y(M⊗AN). This proves a).
To prove b), we may suppose that M and N are nonzero. By Nakayama's lemma, κA⊗AN is likewise nonzero, so that profA(M) and profB(κA⊗AN) are finite (No. 4, Cor. 2 of Th. 2). Take maximal regular sequences x and y; then one has r=profA(M), s=profB(κA⊗AN), and there exists an injective a-linear mapping u:κA→M/xM and an injective b-linear mapping v:κB→κA⊗A(N/yN) (No. 4, Cor. 1 of Th. 2). Since N/yN is flat over A, the b-linear mapping (u⊗1N/yN)∘v of κB into (M/xM)⊗A(N/yN)=(M⊗AN)/(ρ(x)+y)(M⊗AN) is injective. This implies the equality profB(M⊗AN)=r+s (loc. cit.).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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