Bourbaki

Proposition 8

Tag 0E1H, permanent. Cite it as [Bourbaki, Tag 0E1H].

Algebra, chapter X, § 8. Dimension homologique

Let MM be a graded AA-module bounded below.

a) The following conditions are equivalent : (i) MM is isomorphic to a graded AA-module of the form A⊗A0NA \otimes_{A_0} N, where NN is a graded projective A0A_0-module (resp. graded free) ; (ii) MM is a projective AA-module (resp. graded free) ; (iii) M/J0MM/J_0 M is a projective A0A_0-module (resp. graded free) and Tor1A(A0,M)=0\mathrm{Tor}_1^A(A_0, M) = 0.

b) Assume furthermore that MM has a generator system consisting of homogeneous elements of bounded degrees. Then the following conditions are equivalent : (i) The graded AA-module MM has a finite composition series whose quotients are isomorphic to graded AA-modules of the form A⊗A0NA \otimes_{A_0} N, where NN is a graded A0A_0-module that is flat ; (ii) MM is a flat AA-module ; (iii) M/J0MM/J_0 M is a flat A0A_0-module and Tor1A(A0,M)=0\mathrm{Tor}_1^A(A_0, M) = 0.

In each case, there is obviously (i) ⇒ (ii) ⇒ (iii). Therefore, it remains to prove (iii) ⇒ (i).

a) The graded AA-module A⊗A0M/J0MA \otimes_{A_0} M/J_0 M is a projective AA-module since M/J0MM/J_0 M is a projective A0A_0-module. The canonical homomorphism of AA-modules

p:M→M/J0M p : M \to M/J_0 M
is surjective ; hence there exists a graded AA-homomorphism of degree zero
s:A⊗A0M/J0M→M s : A \otimes_{A_0} M/J_0 M \to M
such that p∘s(a⊗x)=axp \circ s(a \otimes x) = ax for a∈Aa \in A and x∈M/J0Mx \in M/J_0 M.

By Lemma 5, ss is an isomorphism of AA-modules from A⊗A0M/J0MA\otimes_{A_0}M/J_0M onto MM, whence (i).

b) By the hypothesis on MM, there exist integers a,ba,b with a≤ba\leq b such that MM is generated by ⨁a≤i≤bMi\displaystyle\bigoplus_{a\leq i\leq b}M_i. We argue by induction on the positive integer b−ab-a.

If b−a=0b-a=0, then MM is generated by MaM_a and the canonical A0A_0-homomorphism

Ma⟶M/J0M M_a\longrightarrow M/J_0M
is bijective; we then deduce from the AA-homomorphism A⊗A0Ma⟶MA\otimes_{A_0}M_a\longrightarrow M defined by the AA-module structure of MM a graded AA-homomorphism
s:A⊗A0M/J0M⟶M s:A\otimes_{A_0}M/J_0M\longrightarrow M
satisfying the condition of Lemma 5. Then, by Lemma 5, ss is bijective, whence (i).

In the general case, let M(a)M^{(a)} be the (graded) sub-AA-module of MM generated by MaM_a and M′M' the quotient M/M(a)M/M^{(a)}. We have an exact sequence

0⟶M(a)→fM→gM′⟶0, 0\longrightarrow M^{(a)}\xrightarrow{f}M\xrightarrow{g}M'\longrightarrow0,
whence, since Tor⁡1A(A0,M)=0\operatorname{Tor}_1^A(A_0,M)=0 by hypothesis, exact sequences
Tor⁡2A(A0,M′)⟶Tor⁡1A(A0,M(a))⟶0(6) \tag{6} \operatorname{Tor}_2^A(A_0,M')\longrightarrow\operatorname{Tor}_1^A(A_0,M^{(a)})\longrightarrow0
0⟶Tor⁡1A(A0,M′)⟶M(a)/J0M(a)→1⊗fM/J0M→1⊗gM′/J0M′⟶0.(7) \tag{7} 0\longrightarrow\operatorname{Tor}_1^A(A_0,M')\longrightarrow M^{(a)}/J_0M^{(a)} \xrightarrow{1\otimes f}M/J_0M\xrightarrow{1\otimes g}M'/J_0M'\longrightarrow0.

But the canonical homomorphism Ma⟶M(a)/J0M(a)M_a\longrightarrow M^{(a)}/J_0M^{(a)} is bijective. It follows that the homomorphism 1⊗f:M(a)/J0M(a)⟶M/J0M1\otimes f:M^{(a)}/J_0M^{(a)}\longrightarrow M/J_0M is injective and that its image is a direct factor sub-A0A_0-module of M/J0MM/J_0M. It then follows from the exact sequence (7) that Tor⁡1A(A0,M′)=0\operatorname{Tor}_1^A(A_0,M')=0 and that the A0A_0-module M′/J0M′M'/J_0M' is flat since isomorphic to a direct factor of M/J0MM/J_0M. By the induction hypothesis (which applies to M′M' since it is generated by the Mi′M'_i for a<i≤ba<i\leq b), M′M' satisfies condition (i), hence is flat. We then deduce from the exact sequence (6) that Tor⁡1A(A0,M(a))=0\operatorname{Tor}_1^A(A_0,M^{(a)})=0, but M(a)/J0M(a)M^{(a)}/J_0M^{(a)} is identified with MaM_a which is a flat A0A_0-module (as a direct factor submodule of M/J0MM/J_0M) ; by what has already been proved, the graded AA-module M(a)M^{(a)} is isomorphic to A⊗A0MaA\otimes_{A_0}M_a, hence also satisfies (i), which completes the proof.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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