Lemma 1

Let $M$ be a finitely generated $A$-module and $d$ an integer. a) In order that $M$ be of type $C_{\leq d}$, it is necessary and sufficient that $z(M) \in Z_{\leq d}$; the projection $z_d(M)$ of $z(M)$ on $Z_d$ parallel to $Z_{\leq d-1}$ is then given by

b) In order that $M$ be of type $C^{>d}$, it is necessary and sufficient that $z(M) \in Z^{>d}$; the projection $z^d(M)$ of $z(M)$ on $Z^d$ parallel to $Z^{>d+1}$ is then given by

In order that $M$ be of type $C_{\leq d}$, that is to say of dimension $\leq d$, it is necessary and sufficient that for every minimal prime ideal $p$ of $\mathrm{Supp}(M)$, one have $\dim(A/p) \leq d$, which means that $z(M) \in Z_{\leq d}$. Suppose that one has $\dim(M) \leq d$, and let $p \in \mathrm{Spec}(A)$ be such that $\dim(A/p) = d$; then either $p \notin \mathrm{Supp}(M)$ and hence $M_p = 0$, or else $p \in \mathrm{Supp}(M)$, and $p$ is a minimal element of $\mathrm{Supp}(M)$; in both cases the coefficient of $[V(p)]$ in $z(M)$ is $\mathrm{long}_{A_p}(M_p)$, whence a). Part b) is proved analogously; it should be noted that a finitely generated module $M$ is of type $C^{>d}$ if and only if one has $M_p = 0$ for every prime ideal $p$ of height $< d$.

According to Prop. 9, c) and Remark 3 of No. 4, the sets $C_{\leq d}$ and $C^{>d}$ are hereditary (A, VIII, § 10, No. 1, Def. 1), and one may consider the corresponding Grothendieck groups $K(C_{\leq d})$ and $K(C^{>d})$ (loc. cit., No. 2); for every $A$-module $M$ of type $C_{\leq d}$ (resp. $C^{>d}$), let $[M]_{\leq d}$ (resp. $[M]^{>d}$) denote the associated element in $K(C_{\leq d})$ (resp. $K(C^{>d})$). By Lemma 1, the mappings $z_d$ and $z^d$ are additive; there therefore exist (loc. cit., Prop. 3) homomorphisms

such that $\zeta_d([M]_{\leq d}) = z_d(M)$ for every $A$-module $M$ of type $C_{\leq d}$ and $\zeta^d([N]^{>d}) = z^d(N)$ for every $A$-module $N$ of type $C^{>d}$. Moreover, since $C_{\leq d-1} \subset C_{\leq d}$ and $C^{>d+1} \subset C^{>d}$, one has canonical homomorphisms With these notations: