(i) ⇒ (ii): this follows from Lemma 1. (ii) ⇒ (iii): this is trivial. (iii) ⇒ (iv): in the situation of (iv), for every $A$-module $N$ there is an isomorphism of $\mathrm{Ext}_A^1(K, N)$ onto $\mathrm{Ext}_A^{n+1}(M, N)$ (X, p. 128, Corollary 4); if (iii) holds, then

for every $N$ and $K$ is projective (X, p. 93, Proposition 10). (iv) ⇒ (i): consider the exact sequence (X, p. 50) If (iv) holds, then $Z_{n-1}(M)$ is projective and $M$ has a projective resolution of length $\leq n$.

In the following statement, we agree that $\pm \infty + 1 = \pm \infty - 1 = \pm \infty$.

a) We have

Equality occurs as soon as $\mathrm{dp}_A(M'') \neq \mathrm{dp}_A(M') + 1$.

b) We have

Equality occurs as soon as $\mathrm{dp}_A(M) \neq \mathrm{dp}_A(M')$.

c) We have

Equality occurs as soon as $\mathrm{dp}_A(M) \neq \mathrm{dp}_A(M'')$.

Let us prove for example a), the proofs of b) and c) being analogous. If N is an arbitrary a-module and n an integer $\geq 0$, we have an exact sequence

If $\mathrm{dp}_A(M'), \mathrm{dp}_A(M'') \leq n$, then $\mathrm{Ext}_A^{n+1}(M', N) = 0$ and $\mathrm{Ext}_A^{n+1}(M'', N) = 0$ (prop. 1), hence $\mathrm{Ext}_A^{n+1}(M, N) = 0$ and $\mathrm{dp}_A(M) \leq n$ (prop. 1), so that

If $\mathrm{dp}_A(M)<\sup\bigl(\mathrm{dp}_A(M'),\mathrm{dp}_A(M'')\bigr)$, then necessarily $\mathrm{dp}_A(M)<+\infty$. For every $n>\mathrm{dp}_A(M)$ and every a-module N, we have

by the preceding exact sequence; by prop. 1, this immediately implies $\mathrm{dp}_A(M'')=\mathrm{dp}_A(M')+1$, since one of the quantities $\mathrm{dp}_A(M')$, $\mathrm{dp}_A(M'')$ is $>\mathrm{dp}_A(M)$.

Example. — Let $a$ be an element of A which is neither invertible nor a right zero divisor. Then $\mathrm{dp}_A(A/Aa)=1$.

Indeed, by the exact sequence $0\longrightarrow A_s\xrightarrow{\varphi} A_s\longrightarrow A/Aa\longrightarrow0$, where $\varphi(x)=xa$, we have $\mathrm{dp}_A(A/Aa)\leq1$. If $\mathrm{dp}_A(A/Aa)<1$, then $A/Aa$ is projective, and there exists an a-linear mapping $\psi:A_s\longrightarrow A_s$ such that $\psi\circ\varphi=\mathrm{Id}$; this implies

and $a$ is invertible.

(i) $\mathrm{dp}_A(M)\leq n$.

(i bis) M possesses a projective resolution P of length $\leq n$ such that $P_i$ is a finitely generated a-module for each $i$.

(ii) $\operatorname{Ext}_A^r(M,N)=0$ for every finitely generated a-module N and every integer $r>n$.

(iii) $\operatorname{Ext}_A^{n+1}(M,N)=0$ for every finitely generated a-module N.

(iv) $\operatorname{Tor}^A_r(P,M)=0$ for every right a-module P and every $r>n$.

(v) $\operatorname{Tor}^A_{n+1}(A/a,M)=0$ for every finitely generated right ideal $a$ of A.

(i bis) $\Rightarrow$ (i) : this is trivial.

(i) $\Rightarrow$ (ii) : this follows from Lemma 1.

(ii) $\Rightarrow$ (iii) : this is trivial.

(iii) $\Rightarrow$ (i) : by (iii) and X, p. 107, prop. 5, we have $\operatorname{Ext}_A^{n+1}(M,N)=0$ for every a-module N, whence (i) by prop. 1.

(i) $\Rightarrow$ (iv) : this follows from Lemma 1.

(iv) $\Rightarrow$ (v) : this is trivial.

(v) $\Rightarrow$ (i bis) : let (L,d) be a free resolution of M such that $L_r$ is finitely generated for all $r$ (X, p. 53, prop. 6). Set $K=\mathrm{Z}_{n-1}(L)$; then K is finitely generated as a submodule of $L_{n-1}$ and we have an exact sequence

By (v) and X, p. 131, cor. 3, we have $\operatorname{Tor}^A_1(A/a,K)=0$ for every finitely generated right ideal $a$ of A. By th. 2 of X, p. 74, the a-module K is flat; since it is finitely generated, hence of finite presentation (X, p. 10, prop. 5), it is projective (X, p. 13, cor.), so (1) is a projective resolution of M.

This follows from X, p. 58, th. 1 (note that $\mathscr{C}_0$ and $\mathscr{C}$ are left exact by cor. 2).

We argue by induction on $n$. If $n = 0$, $M$ is projective, the homomorphism (3) reduces to the canonical homomorphism $P \otimes_A M \to \operatorname{Hom}_A(M^*, P)$ of II, p. 77, and the proposition follows from loc. cit., corollary. If $n > 0$, let

(4)

be an exact sequence of $A$-modules where $L$ is free (and of finite type in case b); then $\mathrm{dp}_A(N) \leqslant n - 1$ (X, p. 135, cor. 2, c)) and $N$ is of finite type in case b.

Let $\theta \in \operatorname{Ext}_A^1(M, N)$ be the class associated with the exact sequence (4) (X, p. 117, déf. 1). Denote

the mappings defined by $u_n(\alpha) = \alpha \circ \theta$ and $v_n(\beta) = \theta \circ \beta$. We have

for all $\alpha\in\operatorname{Ext}_A^{n-1}(N,A)$, $\beta\in\operatorname{Tor}_n^A(P,M)$ (X, p. 129, prop. 6), so that the diagram

where the horizontal arrows are the canonical homomorphisms, is commutative.

If $n=1$, we thus have a commutative diagram:

where the columns are exact; this yields the result in this case. If $n\geq 2$, the mappings $u_n$ and $v_n$ are bijective (X, p. 128, cor. 4 and p. 131, cor. 3). By the induction hypothesis, the canonical homomorphism

is injective (resp. bijective); the diagram (5) shows that the same holds for the canonical homomorphism $\operatorname{Tor}_n^A(P,M)\longrightarrow\operatorname{Hom}_A(\operatorname{Ext}_A^n(M,A),P)$, which completes the proof.

(i) $\operatorname{dh}(A)\leq n$,

(ii) for every $A$-module M, we have $\mathrm{dp}_A(M)\leq n$,

(ii′) for every finitely generated $A$-module M, we have $\mathrm{dp}_A(M)\leq n$,

(iii) for every exact sequence

where the $P_i$ are projective, $K$ is projective,

(iv) for every exact sequence

where the $I^i$ are injective, $N$ is injective,

(v) every $A$-module possesses an injective resolution of length $\leq n$.

The equivalence of conditions (i), (ii) and (iii) follows from prop. 1. We have obviously (ii) $\Rightarrow$ (ii’). It is therefore enough to proving (ii’) $\Rightarrow$ (iv) $\Rightarrow$ (v) $\Rightarrow$ (i).

(ii’) $\Rightarrow$ (iv): with the notation of (iv), let $K$ be the kernel of $I^0 \to I^1$. By X, p. 128, cor. 4, we have for every $A$-module $M$ an isomorphism $\mathrm{Ext}_A^1(M, N) \to \mathrm{Ext}_A^{n+1}(M, K)$; it then follows from (ii’) that $\mathrm{Ext}_A^1(M, N)$ is zero for every finitely generated $A$-module $M$. By X, p. 93, prop. 11, this implies that $N$ is injective, whence (iv).

(iv) $\Rightarrow$ (v): let $M$ be an $A$-module. Applying (iv) to the exact sequence

of X, p. 52, we conclude that $K^{n-1}(M)$ is injective, whence (v).

(v) $\Rightarrow$ (i): this follows from X, p. 100, th. 1.

(i) $\mathrm{dh}(A) \leq n$,

(ii) for every pair of a-modules $M$ and $N$ of finite type, we have $\mathrm{Ext}_A^{n+1}(M, N) = 0$,

(iii) for every left a-module $M$ of finite type and every right a-module $P$ of finite type, we have $\mathrm{Tor}_{n+1}^A(P, M) = 0$.

This follows from Props. 2 and 4.

This follows from X, p. 137, Cor.

By Prop. 4, we have (i) ⇔ (ii) ⇔ (iv); by Corollary 1 to Prop. 4, we have (vi) ⇒ (iv). By X, p. 35, Example 4, we have (i) ⇒ (vii); since (ii) ⇒ (iii) and (vii) ⇒ (viii) are trivial and since (viii) ⇒ (i) follows from II, p. 39, Prop. 4, it remains to be shown (iii) ⇒ (v) and (v) ⇒ (vi); the last assertion follows from VIII, § 5, No. 1, Props. 1 and 2; finally, if every ideal of $A$ is injective, it is a direct factor in $A$, hence (iii) ⇒ (v).

(i) ⇔ (ii) ⇔ (iii) : this follows from Prop. 4 of X, p. 138. (ii) ⇒ (iv) : let $C$ be a projective complex. If (ii) is satisfied, the submodule $B(C)$ of $C$ is projective, hence (iv) by X, p. 35, Remark b). (iii) ⇒ (v) : let $C$ be an injective complex. If (iii) is satisfied, the quotient $B_n(C)$ of $C_{n+1}$ is injective for all $n$, hence (v) by X, p. 35, Remark a). (iv) ⇒ (ii) : let $P$ be a projective a-module, $M$ a submodule of $P$, and $i : M \to P$ the canonical injection. Let $p : L \to M$ be a surjective homomorphism from a free module $L$ onto $M$. Consider the projective complex $C$ such that $C_1 = L$, $C_0 = P$, $C_i = 0$ for $i \ne 0, 1$, and $d_1 = i \circ p$. If (iv) is satisfied, let $\varphi : C \to H(C)$ be a homomorphism such that $H(\varphi) = 1_{H(C)}$. Since $H_1(C) = \mathrm{Ker}\ p$, $\varphi_1$ is a projector of $L$ onto $\mathrm{Ker}\ p$, so the exact sequence

is split and $M$ is isomorphic to a direct factor of $L$, hence projective.

$(v)\Rightarrow(iii)$ : let $I$ be an injective module, $M$ a quotient of $I$, and $\pi : I \to M$ the canonical projection. Let $i : M \to J$ be an injective homomorphism of $M$ into an injective module $J$. Consider the injective complex $C$ such that $C^0 = I$, $C^1 = J$, $C^i = 0$ for $i \ne 0, 1$, and $d^0 = i \circ \pi$. If (v) is satisfied, let $\psi : H(C) \to C$ be a homomorphism such that $H(\psi) = 1_{H(C)}$. Since

$\psi^1$ is a section of the canonical projection $J \to \operatorname{Coker}i$, and $M$ is a direct factor in $J$, hence injective.

$(ii)\Rightarrow(ii')$ : this is trivial.

$(ii')\Rightarrow(ii)$ : this follows from VII, § 3, Corollary 1 to Theorem 1.

Since $Z(C)$, $B(C)$, $Z(\widetilde C)$, and $B(\widetilde C)$ are projective and $B(C')$ is injective, this follows from X, p. 78, Cor. 2, p. 96, Cor. 1, and p. 98, Cor. 2.

The first assertion is clear if $\mathrm{dp}_A(M) = \pm \infty$; if $\mathrm{dp}_A(M) = n \in \mathbf{N}$, there exists an exact sequence of $A$-modules

where the $P_i$ are projective ; the sequence of $A'$-modules

is exact, since $A'_d$ is flat, and the $A'$-modules $A' \otimes_A P_i$ are projective (II, p. 89, Corollary) ; therefore $\mathrm{dp}_{A'}(A' \otimes_A M) \leq n = \mathrm{dp}_A(M)$. The second assertion is clear if $\mathrm{dp}_{A'}(M') = \pm \infty$; if $\mathrm{dp}_{A'}(M') = m \in \mathbf{N}$, there exists an exact sequence of $A'$-modules

where the $P'_i$ are projective ; the underlying sequence of $A$-modules is exact. Moreover each $P'_i$ is a sub-$A'$-module direct factor of a module ${A'_s}^{(1)}$, hence is a projective $A$-module ; therefore one has $\mathrm{dp}_A(M') \leq m = \mathrm{dp}_{A'}(M')$.

The assertion a) follows from X, p. 135, Corollary 2 and Lemma 2.

b) If $\mathrm{dp}_{A[X]}(M) = \pm \infty$, the assertion is trivial. If $M$ is projective and nonzero, then the $A$-module $M/XM$ is identified with $A \otimes_{A[X]} M$, hence is projective, and one has $\mathrm{dp}_A(M/XM) \leq 0 = \mathrm{dp}_{A[X]}(M)$. Let us argue by induction on $\mathrm{dp}_{A[X]}(M) = n$, assumed $> 0$. Consider an exact sequence of $A[X]$-modules $0 \to N \to L \to M \to 0$, where $L$ is a free $A[X]$-module ; applying to the diagram

Proposition 2 of X, p. 4, one sees that $X_N$ is injective and that one has the exact sequence

Since $L$ is free over $A[X]$, $L/XL$ is free over $A$ and one has

since $\mathrm{dp}_A(N/XN) \leq n - 1$ by the induction hypothesis, one deduces $\mathrm{dp}_A(M/XM) \leq n$, which was to be proved.

c) The assertion is trivial if $\mathrm{dp}_{A[X]}(M) = \pm \infty$, and also if $\mathrm{dp}_{A[X]}(M) = 0$ (which is impossible since $XM = 0$). We may therefore suppose $\mathrm{dp}_{A[X]}(M) = n > 0$. Considering as above an exact sequence $0 \to N \to L \to M \to 0$, where $L$ is a free $A[X]$-module, we obtain an exact sequence of $A$-modules

By b), we have $\mathrm{dp}_A(N/XN) \leq \mathrm{dp}_{A[X]}(N) = \mathrm{dp}_{A[X]}(M) - 1 = n - 1$; since $\mathrm{dp}_A(L/XL) = 0$, we deduce from the preceding exact sequence, applying $X$ twice, p. 135, cor. 2, that $\mathrm{dp}_A(M) \leq n - 1$. But, by a), we have

whence c).

For every $A[X]$-module $M$, we have (lemma 3)

therefore $dh(A[X]) \leq dh(A) + 1$; conversely, if $M$ is an $A$-module, let $\overline{M}$ be the $A[X]$-module obtained by equipping $M$ with the structure for which $XM = 0$, then (lemma 3)

therefore $dh(A) \leq dh(A[X]) - 1$.

This follows from the theorem by induction on $n$.

Since $A_0 \otimes_A M$ is isomorphic to $M/J_0 M$, this is none other than ll, p. 171, prop. 6.

We have an exact sequence

and the graded $A$-modules $\mathrm{Ker}\ s$ and $\mathrm{Coker}\ s$ are bounded below. From the exact sequence $A_0 \otimes_A (A \otimes_{A_0} M/J_0 M) \xrightarrow{1 \otimes s} A_0 \otimes_A M \to A_0 \otimes_A \mathrm{Coker}\ s \to 0$, we deduce that $A_0 \otimes_A \mathrm{Coker}\ s = 0$, hence $s$ is surjective (Lemma 4). We then have an exact sequence If $\mathrm{Tor}_1^A(A_0, M) = 0$, then $A_0 \otimes_A \mathrm{Ker}\ s = 0$ and $s$ is injective (Lemma 4).

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

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

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

a) The graded $A$-module $A \otimes_{A_0} M/J_0 M$ is a projective $A$-module since $M/J_0 M$ is a projective $A_0$-module. The canonical homomorphism of $A$-modules

is surjective ; hence there exists a graded $A$-homomorphism of degree zero such that $p \circ s(a \otimes x) = ax$ for $a \in A$ and $x \in M/J_0 M$.

By Lemma 5, $s$ is an isomorphism of $A$-modules from $A\otimes_{A_0}M/J_0M$ onto $M$, whence (i).

b) By the hypothesis on $M$, there exist integers $a,b$ with $a\leq b$ such that $M$ is generated by $\displaystyle\bigoplus_{a\leq i\leq b}M_i$. We argue by induction on the positive integer $b-a$.

If $b-a=0$, then $M$ is generated by $M_a$ and the canonical $A_0$-homomorphism

is bijective; we then deduce from the $A$-homomorphism $A\otimes_{A_0}M_a\longrightarrow M$ defined by the $A$-module structure of $M$ a graded $A$-homomorphism satisfying the condition of Lemma 5. Then, by Lemma 5, $s$ is bijective, whence (i).

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

whence, since $\operatorname{Tor}_1^A(A_0,M)=0$ by hypothesis, exact sequences

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