Lemma 3

Assume A commutative and let M be an a[X]-module. a) One has $\mathrm{dp}_A(M) \leq \mathrm{dp}_{A[X]}(M) \leq \mathrm{dp}_A(M) + 1$. b) If the homothety $X_M$ is injective, one has $\mathrm{dp}_A(M/XM) \leq \mathrm{dp}_{A[X]}(M)$. c) If $XM = 0$, one has $\mathrm{dp}_A(M) + 1 = \mathrm{dp}_{A[X]}(M)$. a) One has an exact sequence of $A[X]$-modules (III, p. 106 and VII, § 5, No. 1)

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).