The Stacks project

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15.20 Flattening over a Noetherian complete local ring

The following three lemmas give a completely algebraic proof of the existence of the “local” flattening stratification when the base is a complete local Noetherian ring $R$ and the given module is finite over a finite type $R$-algebra $S$.

Lemma 15.20.1. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume

  1. $(R, \mathfrak m)$ is a complete local Noetherian ring,

  2. $S$ is a Noetherian ring, and

  3. $M$ is finite over $S$.

Then there exists an ideal $I \subset \mathfrak m$ such that

  1. $(M/IM)_{\mathfrak q}$ is flat over $R/I$ for all primes $\mathfrak q$ of $S/IS$ lying over $\mathfrak m$, and

  2. if $J \subset R$ is an ideal such that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for all primes $\mathfrak q$ lying over $\mathfrak m$, then $I \subset J$.

In other words, $I$ is the smallest ideal of $R$ such that (15.18.0.1) holds for $(\overline{R} \to \overline{S}, \overline{\mathfrak m}, \overline{M})$ where $\overline{R} = R/I$, $\overline{S} = S/IS$, $\overline{\mathfrak m} = \mathfrak m/I$ and $\overline{M} = M/IM$.

Proof. Let $J \subset R$ be an ideal. Apply Algebra, Lemma 10.98.11 to the module $M/JM$ over the ring $R/J$. Then we see that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for all primes $\mathfrak q$ of $S/JS$ if and only if $M/(J + \mathfrak m^ n)M$ is flat over $R/(J + \mathfrak m^ n)$ for all $n \geq 1$. We will use this remark below.

For every $n \geq 1$ the local ring $R/\mathfrak m^ n$ is Artinian. Hence, by Lemma 15.17.1 there exists a smallest ideal $I_ n \supset \mathfrak m^ n$ such that $M/I_ nM$ is flat over $R/I_ n$. It is clear that $I_{n + 1} + \mathfrak m^ n$ is contains $I_ n$ and applying Lemma 15.16.1 we see that $I_ n = I_{n + 1} + \mathfrak m^ n$. Since $R = \mathop{\mathrm{lim}}\nolimits _ n\ R/\mathfrak m^ n$ we see that $I = \mathop{\mathrm{lim}}\nolimits _ n\ I_ n/\mathfrak m^ n$ is an ideal in $R$ such that $I_ n = I + \mathfrak m^ n$ for all $n \geq 1$. By the initial remarks of the proof we see that $I$ verifies (1) and (2). Some details omitted. $\square$

Lemma 15.20.2. With notation $R \to S$, $M$, and $I$ and assumptions as in Lemma 15.20.1. Consider a local homomorphism of local rings $\varphi : (R, \mathfrak m) \to (R', \mathfrak m')$ such that $R'$ is Noetherian. Then the following are equivalent

  1. condition (15.18.0.1) holds for $(R' \to S \otimes _ R R', \mathfrak m', M \otimes _ R R')$, and

  2. $\varphi (I) = 0$.

Proof. The implication (2) $\Rightarrow $ (1) follows from Lemma 15.18.1. Let $\varphi : R \to R'$ be as in the lemma satisfying (1). We have to show that $\varphi (I) = 0$. This is equivalent to the condition that $\varphi (I)R' = 0$. By Artin-Rees in the Noetherian local ring $R'$ (see Algebra, Lemma 10.50.4) this is equivalent to the condition that $\varphi (I)R' + (\mathfrak m')^ n = (\mathfrak m')^ n$ for all $n > 0$. Hence this is equivalent to the condition that the composition $\varphi _ n : R \to R' \to R'/(\mathfrak m')^ n$ annihilates $I$ for each $n$. Now assumption (1) for $\varphi $ implies assumption (1) for $\varphi _ n$ by Lemma 15.18.1. This reduces us to the case where $R'$ is Artinian local.

Assume $R'$ Artinian. Let $J = \mathop{\mathrm{Ker}}(\varphi )$. We have to show that $I \subset J$. By the construction of $I$ in Lemma 15.20.1 it suffices to show that $(M/JM)_{\mathfrak q}$ is flat over $R/J$ for every prime $\mathfrak q$ of $S/JS$ lying over $\mathfrak m$. As $R'$ is Artinian, condition (1) signifies that $M \otimes _ R R'$ is flat over $R'$. As $R'$ is Artinian and $R/J \to R'$ is a local injective ring map, it follows that $R/J$ is Artinian too. Hence the flatness of $M \otimes _ R R' = M/JM \otimes _{R/J} R'$ over $R'$ implies that $M/JM$ is flat over $R/J$ by Algebra, Lemma 10.100.7. This concludes the proof. $\square$

Lemma 15.20.3. With notation $R \to S$, $M$, and $I$ and assumptions as in Lemma 15.20.1. In addition assume that $R \to S$ is of finite type. Then for any local homomorphism of local rings $\varphi : (R, \mathfrak m) \to (R', \mathfrak m')$ the following are equivalent

  1. condition (15.18.0.1) holds for $(R' \to S \otimes _ R R', \mathfrak m', M \otimes _ R R')$, and

  2. $\varphi (I) = 0$.

Proof. The implication (2) $\Rightarrow $ (1) follows from Lemma 15.18.1. Let $\varphi : R \to R'$ be as in the lemma satisfying (1). As $R$ is Noetherian we see that $R \to S$ is of finite presentation and $M$ is an $S$-module of finite presentation. Write $R' = \mathop{\mathrm{colim}}\nolimits _\lambda R_\lambda $ as a directed colimit of local $R$-subalgebras $R_\lambda \subset R'$, with maximal ideals $\mathfrak m_\lambda = R_\lambda \cap \mathfrak m'$ such that each $R_\lambda $ is essentially of finite type over $R$. By Lemma 15.18.3 we see that condition (15.18.0.1) holds for $(R_\lambda \to S \otimes _ R R_\lambda , \mathfrak m_\lambda , M \otimes _ R R_\lambda )$ for some $\lambda $. Hence Lemma 15.20.2 applies to the ring map $R \to R_\lambda $ and we see that $I$ maps to zero in $R_\lambda $, a fortiori it maps to zero in $R'$. $\square$


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