Bourbaki

Exercise 4, § 6

Tag 062Q, permanent. Cite it as [Bourbaki, Tag 062Q].

Algebra, III, 632

Bourbaki marks this one as harder than the rest.

Let AA be a commutative ring and aa an ideal of AA. The Rees algebra of the ideal aa is the subalgebra R(a)R(a) of the polynomial algebra A[T]A[T] in one indeterminate, consisting of the polynomials c0+c1T+⋯+cnTnc_0 + c_1 T + \cdots + c_n T^n, where c0∈Ac_0 \in A and ck∈akc_k \in a^k for every integer k≥1k \geq 1.

(a) Show that there exists one and only one surjective AA-algebra homomorphism r:S(a)→R(a)r : S(a) \to R(a) such that the restriction of rr to aa is the AA-linear mapping x↦xTx \mapsto xT.

(b) Suppose that the ring AA is an integral domain. Let a1,…,ana_1, \ldots, a_n be elements of AA with an≠0a_n \neq 0; consider the AA-algebra homomorphism

u:A[X1,…,Xn]→A[T] u : A[X_1, \ldots, X_n] \to A[T]

such that u(Xi)=aiTu(X_i) = a_i T. The kernel r=Ker(u)r = \mathrm{Ker}(u) is a graded ideal whose homogeneous component rmr_m of degree mm consists of the homogeneous polynomials ff of degree mm such that f(a1,…,an)=0f(a_1, \ldots, a_n) = 0. If r′⊂rr' \subset r is the ideal of A[X1,…,Xn]A[X_1, \ldots, X_n] generated by the homogeneous component r1r_1 of rr, show that the AA-module r/r′r/r' is a torsion AA-module. (To verify that, for all f∈rmf \in r_m, there exists c≠0c \neq 0 in AA such that cf∈r′cf \in r', argue by induction on mm, by writing

f(X1,…,Xn)=X1f1(X1,…,Xn)+X2f2(X2,…,Xn)+⋯+Xnfn(Xn) f(X_1, \ldots, X_n) = X_1 f_1(X_1, \ldots, X_n) + X_2 f_2(X_2, \ldots, X_n) + \cdots + X_n f_n(X_n)
where the fjf_j are homogeneous of degree m−1m - 1; consider the polynomial
g(X1,…,Xn)=X1f1(a1,…,an)+X2f2(a2,…,an)+⋯+Xnfn(an), g(X_1, \ldots, X_n) = X_1 f_1(a_1, \ldots, a_n) + X_2 f_2(a_2, \ldots, a_n) + \cdots + X_n f_n(a_n),
observe that g∈r1⊂r′g \in r_1 \subset r' and form anm−1f−Xnm−1g.a_n^{m-1} f - X_n^{m-1} g.)

(c) Deduce from (b) that when AA is an integral domain, the kernel of the homomorphism rr of (a) is the torsion module of S(a)S(a). Deduce that, for S(a)S(a) to be an integral domain, it is necessary and sufficient that S(a)S(a) be a torsion-free AA-module. In particular, if the ideal aa is a projective AA-module, S(a)S(a) is an integral domain and isomorphic to R(a)R(a).

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