Bourbaki

Proposition 4

Tag 0CPW, permanent. Cite it as [Bourbaki, Tag 0CPW].

Commutative Algebra, chapter VIII, § 5. Anneaux locaux réguliers

a) The ring BB is local and noetherian, with maximal ideal

mB=BmA+∑i=1rBti. m_B = Bm_A + \sum_{i=1}^r Bt_i.

One has dim⁡(A)=dim⁡(B)\dim(A) = \dim(B) and [κB:κA]=1[\kappa_B : \kappa_A] = 1. The monomials t1α(1)...trα(r)t_1^{\alpha(1)} ... t_r^{\alpha(r)}, with 0≤α(i)<di0 \leq \alpha(i) < d_i for 1≤i≤r1 \leq i \leq r, form a basis of the AA-module BB.

b) Let λ\lambda be the canonical homomorphism of mA/mA2m_A/m_A^2 into mB/mB2m_B/m_B^2. Then the kernel of λ\lambda is the κA\kappa_A-vector space generated by γ1,...,γr\gamma_1, ..., \gamma_r. The classes of the elements t1,...,trt_1, ..., t_r form a basis over κA\kappa_A of the cokernel of λ\lambda.

c) For BB to be regular, it is necessary and sufficient that AA be regular and that γ1,...,γr\gamma_1, ..., \gamma_r be linearly independent in the κA\kappa_A-vector space mA/mA2m_A/m_A^2.

The AA-algebra BB is isomorphic to the tensor product B1⊗A⋯⊗ABrB_1 \otimes_A \cdots \otimes_A B_r with Bi=A[T]/(Pi)B_i = A[T]/(P_i) for 1≤i≤r1 \leq i \leq r. It follows that the monomials t1α(1)...trα(r)t_1^{\alpha(1)} ... t_r^{\alpha(r)}, with 0≤α(i)<di0 \leq \alpha(i) < d_i for 1≤i≤r1 \leq i \leq r, form a basis of the AA-module BB. In particular, BB is integral over AA, hence AA and BB 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 BB is noetherian, and every maximal ideal of BB contains B.mAB.m_A. Moreover, in view of the assumption made on P1,...,PrP_1, ..., P_r and the relation Pi(ti)=0P_i(t_i) = 0, one has tidi∈B.mAt_i^{d_i} \in B.m_A for 1≤i≤r1 \leq i \leq r. Hence every maximal ideal of BB contains t1,...,trt_1, ..., t_r, and therefore also the ideal q′=B.mA+Bt1+⋯+Btrq' = B.m_A + Bt_1 + \cdots + Bt_r. Now one has mA=A∩q′m_A = A \cap q' and B=A+q′B = A + q', hence B/q′B/q' is isomorphic to A/mAA/m_A and q′q' is a maximal ideal of BB; consequently, BB is local and one has [κB:κA]=1[\kappa_B : \kappa_A] = 1. This proves a).

Put r=mA2+∑i=1rAcir = m_A^2 + \sum_{i=1}^r Ac_i, and denote by φ\varphi the canonical homomorphism of (A/mA2)[T1,...,Tr](A/m_A^2)[T_1, ..., T_r] onto B/mB2B/m_B^2. Since one has mB=B.mA+∑i=1rBtim_B = B.m_A + \sum_{i=1}^r Bt_i, the kernel nn of φ\varphi is the ideal generated by the classes P‾i(Ti)\overline{P}_i(T_i) of the polynomials Pi(Ti)P_i(T_i) modulo mA2.A[T1,...,Tr]m_A^2.A[T_1, ..., T_r] and the monomials TiTjT_iT_j and xTixT_i for 1≤i,j≤r1 \leq i, j \leq r and xx in mA/mA2m_A/m_A^2. According to the assumption made on PiP_i, namely Pi(T)≡Tdimod  mA.A[T]P_i(T) \equiv T^{d_i} \mod m_A.A[T], one can replace P‾i(Ti)\overline{P}_i(T_i) by γi\gamma_i in this description of nn; consequently, the ring B/mB2B/m_B^2 is isomorphic to the quotient of (A/r)[T1,...,Tr](A/r)[T_1, ..., T_r] by the graded ideal generated by the monomials TiTjT_i T_j and xTix T_i for xx in mA/rm_A / r. Let τi\tau_i denote the class of tit_i modulo mB2m_B^2; one therefore has

B/mB2=(A/r)⊕κAτ1⊕⋯⊕κAτr, B / m_B^2 = (A/r) \oplus \kappa_A \tau_1 \oplus \cdots \oplus \kappa_A \tau_r,

whence

mB/mB2=(mA/r)⊕κAτ1⊕⋯⊕κAτr. m_B / m_B^2 = (m_A / r) \oplus \kappa_A \tau_1 \oplus \cdots \oplus \kappa_A \tau_r.

Assertion b) follows at once from this.

According to formula (6) and the relation [κB:κA]=1[ \kappa_B : \kappa_A ] = 1, one has

[mB/mB2:κB]=[mA/m2:κA]+{r−[r/mA2:κA]}. [m_B / m_B^2 : \kappa_B] = [m_A / m^2 : \kappa_A] + \{ r - [r / m_A^2 : \kappa_A] \}.

Now, the κA\kappa_A-vector space r/mA2r / m_A^2 is generated by γ1,...,γr\gamma_1, ..., \gamma_r and one has

dim⁡(B)=dim⁡(A)≤[mA/mA2:κA]. \dim(B) = \dim(A) \leq [m_A / m_A^2 : \kappa_A].

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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