Suppose that A and B are Noetherian and that B is a flat A-module. Let p be a prime ideal of A, Q⊂E a p-primary submodule and P a prime ideal of B. For Q⊗AB to be a q-primary submodule of E⊗AB, it is necessary and sufficient that pB be a 'p-primary ideal' of B.
Let us apply Theorem 2 to the A-module E/Q and the B-module B; then AssA(E/Q)={p} and (E/Q)⊗AB is isomorphic to (E⊗AB)/(Q⊗AB) and hence AssB((E⊗AB)/(Q⊗AB))=AssB(B/pB). To say that Q⊗AB is P-primary in E⊗AB therefore means that AssB(B/pB) is reduced to P, whence the corollary.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.