Lemma 4

Let $A_0$ be the ring $\mathbf{Z}[T_1,\ldots,T_n]$, and let $u:A_0[X_1,\ldots,X_n]\to A_0[U]$ be the homomorphism of $A_0$-algebras such that $u(X_i)=T_iU$. The kernel of $u$ is the ideal $\mathfrak{d}_0$ of $A_0[X_1,\ldots,X_n]$ generated by the elements $(T_iX_j-T_jX_i)$ for $1\leq i<j\leq n$. If $t$ is the ideal of $A_0$ generated by $(T_1,\ldots,T_n)$, $u$ induces an isomorphism

It is obviously enough to prove the first assertion. For every sequence of natural integers $\alpha=(\alpha_1,\ldots,\alpha_n)$ and every integer $k\in[0,n]$, let $P_{\alpha,k}$ denote the monomial

let $N$ be the sub-$\mathbf{Z}$-module of $A_0[X_1, ..., X_n]$ generated by the $P_{\alpha, k}$ for $\alpha \in \mathbf{N}^n$ and $0 \leq k \leq n$. We shall prove that the restriction of $u$ to $N$ is injective and that $A_0[X_1, ..., X_n] = \mathfrak{d}_0 + N$; since we have $\mathfrak{d}_0 \subset \mathrm{Ker}\,u$, this will entail the lemma.

Observe that $N$ is generated by the set $S$ formed by $P_{0,0} = 1$ and by those of the $P_{\alpha, k}$ for which $\alpha_k \neq 0$. To prove the injectivity of the restriction of $u$ to $N$, it suffices to show that two distinct elements of $S$ have as their images by $u$ distinct monomials in $A_0[U]$. Now we have $u(P_{\alpha, k}) = T^\alpha U^{\sum \alpha_i}_{i \geq k}$, so that the equality $u(P_{\alpha, k}) = u(P_{\alpha', k'})$ entails $\alpha = \alpha'$ and $\sum_{i \geq k} \alpha_i = \sum_{i \geq k'} \alpha_i$. Suppose that $P_{\alpha, k}$ and $P_{\alpha', k'}$ belong to $S$. If $\alpha = 0$, then we have $k = k' = 0$; if $\alpha \neq 0$, we obtain $k = k'$ since $\alpha_k$ and $\alpha_{k'}$ are different from zero, whence the result.

Show that every monomial $T^\alpha X^\beta \in A_0[X_1, ..., X_n]$ is congruent modulo $\mathfrak{d}_0$ to a $P_{\eta, k}$. For every sequence $\lambda \in \mathbf{N}^n$, we shall denote by $i(\lambda)$ (resp. $j(\lambda)$) the smallest (resp. the largest) integer $k \in [1, n]$ such that $\lambda_k \neq 0$. Among the monomials $T^\gamma X^\delta$ which are congruent to $T^\alpha X^\beta$ modulo $\mathfrak{d}_0$, choose one for which the rational integer $j(\gamma) - i(\delta)$ is minimal; let us show that we then have $j(\gamma) - i(\delta) < 0$. Suppose that we have $j(\gamma) \geq i(\delta)$; put $j = j(\gamma), i = i(\delta)$, and let $\varepsilon = \inf (\gamma_j, \delta_i)$. Since $(T_j^\varepsilon X_i^\varepsilon - T_i^\varepsilon X_j^\varepsilon)$ is divisible by $(T_j X_i - T_i X_j)$, and therefore belongs to $\mathfrak{d}_0$, we see that $T^\gamma X^\delta$ is congruent modulo $\mathfrak{d}_0$ to $T^{\gamma'} X^{\delta'}$, where $\gamma'_i = \gamma_i + \varepsilon, \gamma'_j = \gamma_j - \varepsilon, \gamma'_k = \gamma_k$ for $k \neq i, j$, and $\delta'_i = \delta_i - \varepsilon, \delta'_j = \delta_j + \varepsilon, \delta'_k = \delta_k$ for $k \neq i, j$. Since $\gamma'_j$ or $\delta'_i$ is zero, we have $j(\gamma') - i(\delta') < j(\gamma) - i(\delta)$ which contradicts the minimal character of $j(\gamma) - i(\delta)$.

Consequently we have $j(\gamma) < i(\delta)$, whence $T^\gamma X^\delta \in N$, which completes the proof of the lemma.

Let us prove that (iii) implies (iv). Consider the ring $A_0 = \mathbf{Z}[T_1, ..., T_n]$ and the ideal t of $A_0$ generated by $T_1, ..., T_n$. Endow $M$ with the structure of $A_0$-module for which $T_i m = x_i m$ for $m \in M, 1 \leq i \leq n$. According to X, p. 155, $H_i(x, M)$ is identified canonically with $H_i(T, M)$.

Since the sequence $T$ is regular for $A_0$, it follows from the implication (i) $\Rightarrow$ (ii) that the complex $K.(T, A_0)$ defines a free resolution of the $A_0$-module $A_0/t$. Let us remark that the latter is identified with $\mathbf{Z}$, endowed with the structure of $A_0$-module such that $T_i \mathbf{Z} = 0$ for $1 \leq i \leq n$. Thus the condition (iii) is equivalent to $\mathrm{Tor}_1^{A_0}(M, \mathbf{Z}) = 0$.

Let us show that (iii) implies $\mathrm{Tor}_1^{A_0}(M, A_0/t^r) = 0$ for all $r \geq 1$. This results from what precedes for $r = 1$. In the general case, consider the exact sequence

The $A_0$-module $t^r/t^{r+1}$ is isomorphic to a finite product of copies of $\mathbf{Z}$; we therefore have $\mathrm{Tor}_1^{A_0}(M, t^r/t^{r+1}) = 0$. It follows, by induction on $r$, that

The exact sequence (16) therefore provides, by tensor product with $M$, an exact sequence

whence an isomorphism of $(t'/{t'}^{r+1}) \otimes_{A_0} M$ onto $x' M/{x'}^{r+1} M$. By considering the exact sequence $0 \to {t'}^{r+1} \to t' \to t'/{t'}^{r+1} \to 0$, one then sees by induction on $r$ that the mapping $m_r : t' \otimes_{A_0} M \to x' M$, induced by the operation of $A_0$ in $M$, is an isomorphism.

To prove (iv), it remains to observe that the diagram

where $e$ is the canonical isomorphism of extension of scalars (II, p. 85, prop. 6), is commutative, and to apply Lemma 4.