Proposition 7

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

Commutative Algebra, chapter VIII, § 3. Dimension des anneaux noethériens

Let $\rho : A \to B$ be a local homomorphism of noetherian local rings, $M$ a finitely generated $A$ -module and $N$ a finitely generated $B$ -module. Put $\overline{B} = B \otimes_A \kappa_A = B/\rho(m_A).B$ and $\overline{N} = N \otimes_B \overline{B} = N/\rho(m_A).N$ . One has

\dim_B(M \otimes_A N) \leq \dim_A(M) + \dim_B(\overline{N}),

and there is equality if $N$ is flat over $A$ .

One may suppose $M$ and $N$ non-zero. By the corollary of th. 1 (no. 2), there exists a subset $S$ (resp. $T$ ) of $m_A$ (resp. $m_B$ ) such that $\mathrm{Card}(S) = \dim_A(M)$ (resp. $\mathrm{Card}(T) = \dim_{\overline{B}}(\overline{N})$ ) and that $M/SM$ is an $A$ -module of finite length (resp. $\overline{N}/T\overline{N}$ is a $B$ -module of finite length). One has $\rho(S) \subset m_B$ . Let $E$ be the $B$ -module $M \otimes_A N$ ; the $B$ -module $E/(\rho(S).E + T.E)$ is isomorphic to $M/SM \otimes_A N/TN$ , hence is of finite length: by props. 18 and 19 of II, § 4, no. 4, one has

\begin{align*} \mathrm{Supp}(M/SM \otimes_A N/TN) &= \mathrm{Supp}(N/TN) \cap {}^a\rho^{-1}(\mathrm{Supp}(M/SM)) \\ &= \mathrm{Supp}(N/TN) \cap {}^a\rho^{-1}(\mathrm{Supp}(\kappa_A)) \\ &= \mathrm{Supp}(N/TN \otimes_A \kappa_A) = \mathrm{Supp}(\overline{N}/T\overline{N}) = \{m_B\}. \end{align*}

It follows, by th. 1, that one has the inequality

\dim_B(E) \leq \mathrm{Card}(S) + \mathrm{Card}(T) = \dim_A(M) + \dim_{\overline{B}}(\overline{N}).

Now suppose $N$ flat over $A$ , and let us show that there is equality in formula (12). Let $a$ (resp. $b$ ) be the annihilator of $M$ (resp. $N$ ). By props. 18 and 19 of II, § 4, no. 4, one has

\begin{align*} \mathrm{Supp}(E) &= \mathrm{Supp}(N) \cap {}^a\rho^{-1}(\mathrm{Supp}(M)) \\ &= V(b) \cap {}^a\rho^{-1}(V(a)) = V(b + \rho(a).B). \end{align*}

Hence

\dim_B(M \otimes_A N) = \dim(B/(b + \rho(a).B))

and also

\dim_A(M) + \dim_{\overline{B}}(\overline{N}) = \dim(A/a) + \dim(B/(b + \rho(m_A).B)).

\dim(B') \geq \dim(A') + \dim(B'/\rho'(\mathfrak{m}_{A'}).B')

and since the ring $B'/\rho'(\mathfrak{m}_{A'}).B'$ is isomorphic to $B/(b + \rho(\mathfrak{m}_A).B)$ , our assertion follows from formulas (13), (14) and (15).

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

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