a) The ring B is local and noetherian, with maximal ideal
mB=BmA+i=1∑rBti.
One has dim(A)=dim(B) and [κB:κA]=1. The monomials t1α(1)...trα(r), with 0≤α(i)<di for 1≤i≤r, form a basis of the A-module B.
b) Let λ be the canonical homomorphism of mA/mA2 into mB/mB2. Then the kernel of λ is the κA-vector space generated by γ1,...,γr. The classes of the elements t1,...,tr form a basis over κA of the cokernel of λ.
c) For B to be regular, it is necessary and sufficient that A be regular and that γ1,...,γr be linearly independent in the κA-vector space mA/mA2.
The A-algebra B is isomorphic to the tensor product B1⊗A⋯⊗ABr with Bi=A[T]/(Pi) for 1≤i≤r. It follows that the monomials t1α(1)...trα(r), with 0≤α(i)<di for 1≤i≤r, form a basis of the A-module B. In particular, B is integral over A, hence A and B have the same dimension by Th. 1 of § 2, No. 3.
According to Cor. 3 of Prop. 9 of IV, § 2, No. 5, the ring B is noetherian, and every maximal ideal of B contains B.mA. Moreover, in view of the assumption made on P1,...,Pr and the relation Pi(ti)=0, one has tidi∈B.mA for 1≤i≤r. Hence every maximal ideal of B contains t1,...,tr, and therefore also the ideal q′=B.mA+Bt1+⋯+Btr. Now one has mA=A∩q′ and B=A+q′, hence B/q′ is isomorphic to A/mA and q′ is a maximal ideal of B; consequently, B is local and one has [κB:κA]=1. This proves a).
Put r=mA2+∑i=1rAci, and denote by φ the canonical homomorphism of (A/mA2)[T1,...,Tr] onto B/mB2. Since one has mB=B.mA+∑i=1rBti, the kernel n of φ is the ideal generated by the classes Pi(Ti) of the polynomials Pi(Ti) modulo mA2.A[T1,...,Tr] and the monomials TiTj and xTi for 1≤i,j≤r and x in mA/mA2. According to the assumption made on Pi, namely Pi(T)≡TdimodmA.A[T], one can replace Pi(Ti) by γi in this description of n; consequently, the ring B/mB2 is isomorphic to the quotient of (A/r)[T1,...,Tr] by the graded ideal generated by the monomials TiTj and xTi for x in mA/r. Let τi denote the class of ti modulo mB2; one therefore has
B/mB2=(A/r)⊕κAτ1⊕⋯⊕κAτr,
whence
mB/mB2=(mA/r)⊕κAτ1⊕⋯⊕κAτr.
Assertion b) follows at once from this.
According to formula (6) and the relation [κB:κA]=1, one has
[mB/mB2:κB]=[mA/m2:κA]+{r−[r/mA2:κA]}.
Now, the κA-vector space r/mA2 is generated by γ1,...,γr and one has
dim(B)=dim(A)≤[mA/mA2:κA].
Assertion c) follows at once from formulas (7) and (8).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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