The Stacks project

Lemma 51.11.4. Let $A$ be a universally catenary Noetherian local ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Then

\[ s_{A, I}(M) \geq s_{A^\wedge , I^\wedge }(M^\wedge ) \]

If the formal fibres of $A$ are $(S_ n)$, then $\min (n + 1, s_{A, I}(M)) \leq s_{A^\wedge , I^\wedge }(M^\wedge )$.

Proof. Write $X = \mathop{\mathrm{Spec}}(A)$, $X^\wedge = \mathop{\mathrm{Spec}}(A^\wedge )$, $Z = V(I) \subset X$, and $Z^\wedge = V(I^\wedge )$. Let $\mathfrak p' \subset \mathfrak q' \subset A^\wedge $ be primes with $\mathfrak p' \not\in Z^\wedge $ and $\mathfrak q' \in Z^\wedge $. Let $\mathfrak p \subset \mathfrak q$ be the corresponding primes of $A$. Then $\mathfrak p \not\in Z$ and $\mathfrak q \in Z$. Picture

\[ \xymatrix{ \mathfrak p' \ar[r] & \mathfrak q' \ar[r] & A^\wedge \\ \mathfrak p \ar[r] \ar@{-}[u] & \mathfrak q \ar[r] \ar@{-}[u] & A \ar[u] } \]

Let us write

\begin{align*} a & = \dim (A/\mathfrak p) = \dim (A^\wedge /\mathfrak pA^\wedge ),\\ b & = \dim (A/\mathfrak q) = \dim (A^\wedge /\mathfrak qA^\wedge ),\\ a' & = \dim (A^\wedge /\mathfrak p'),\\ b' & = \dim (A^\wedge /\mathfrak q') \end{align*}

Equalities by More on Algebra, Lemma 15.43.1. We also write

\begin{align*} p & = \dim (A^\wedge _{\mathfrak p'}/\mathfrak p A^\wedge _{\mathfrak p'}) = \dim ((A^\wedge /\mathfrak p A^\wedge )_{\mathfrak p'}) \\ q & = \dim (A^\wedge _{\mathfrak q'}/\mathfrak p A^\wedge _{\mathfrak q'}) = \dim ((A^\wedge /\mathfrak q A^\wedge )_{\mathfrak q'}) \end{align*}

Since $A$ is universally catenary we see that $A^\wedge /\mathfrak pA^\wedge = (A/\mathfrak p)^\wedge $ is equidimensional of dimension $a$ (More on Algebra, Proposition 15.109.5). Hence $a = a' + p$. Similarly $b = b' + q$. By Algebra, Lemma 10.163.1 applied to the flat local ring map $A_\mathfrak p \to A^\wedge _{\mathfrak p'}$ we have

\[ \text{depth}(M^\wedge _{\mathfrak p'}) = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'}) \]

The quantity we are minimizing for $s_{A, I}(M)$ is

\[ s(\mathfrak p, \mathfrak q) = \text{depth}(M_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) = \text{depth}(M_\mathfrak p) + a - b \]

(last equality as $A$ is catenary). The quantity we are minimizing for $s_{A^\wedge , I^\wedge }(M^\wedge )$ is

\[ s(\mathfrak p', \mathfrak q') = \text{depth}(M^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) = \text{depth}(M^\wedge _{\mathfrak p'}) + a' - b' \]

(last equality as $A^\wedge $ is catenary). Now we have enough notation in place to start the proof.

Let $\mathfrak p \subset \mathfrak q \subset A$ be primes with $\mathfrak p \not\in Z$ and $\mathfrak q \in Z$ such that $s_{A, I}(M) = s(\mathfrak p, \mathfrak q)$. Then we can pick $\mathfrak q'$ minimal over $\mathfrak q A^\wedge $ and $\mathfrak p' \subset \mathfrak q'$ minimal over $\mathfrak p A^\wedge $ (using going down for $A \to A^\wedge $). Then we have four primes as above with $p = 0$ and $q = 0$. Moreover, we have $\text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'})=0$ also because $p = 0$. This means that $s(\mathfrak p', \mathfrak q') = s(\mathfrak p, \mathfrak q)$. Thus we get the first inequality.

Assume that the formal fibres of $A$ are $(S_ n)$. Then $\text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'}) \geq \min (n, p)$. Hence

\[ s(\mathfrak p', \mathfrak q') \geq s(\mathfrak p, \mathfrak q) + q + \min (n, p) - p \geq s_{A, I}(M) + q + \min (n, p) - p \]

Thus the only way we can get in trouble is if $p > n$. If this happens then

\begin{align*} s(\mathfrak p', \mathfrak q') & = \text{depth}(M^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) \\ & = \text{depth}(M_\mathfrak p) + \text{depth}(A^\wedge _{\mathfrak p'} / \mathfrak p A^\wedge _{\mathfrak p'}) + \dim ((A^\wedge /\mathfrak p')_{\mathfrak q'}) \\ & \geq 0 + n + 1 \end{align*}

because $(A^\wedge /\mathfrak p')_{\mathfrak q'}$ has at least two primes. This proves the second inequality. $\square$

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