Theorem 1

Tag 0E1C, permanent. Cite it as [Bourbaki, Tag 0E1C].

Algebra, chapter X, § 8. Dimension homologique

Suppose $A$ commutative. Then

dh(A[X]) = dh(A) + 1 .

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

\mathrm{dp}_{A[X]}(M) \leq \mathrm{dp}_A(M) + 1 \leq dh(A) + 1

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)

\mathrm{dp}_A(M) = \mathrm{dp}_{A[X]}(\overline{M}) - 1 \leq dh(A[X]) - 1 ,

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

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

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