## 38.6 Translation into algebra

It may be useful to spell out algebraically what it means to have a complete dévissage. We introduce the following notion (which is not that useful so we give it an impossibly long name).

Definition 38.6.1. Let $R \to S$ be a ring map. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak p$ of $R$. A elementary étale localization of the ring map $R \to S$ at $\mathfrak q$ is given by a commutative diagram of rings and accompanying primes

$\xymatrix{ S \ar[r] & S' \\ R \ar[u] \ar[r] & R' \ar[u] } \quad \quad \xymatrix{ \mathfrak q \ar@{-}[r] & \mathfrak q' \\ \mathfrak p \ar@{-}[u] \ar@{-}[r] & \mathfrak p' \ar@{-}[u] }$

such that $R \to R'$ and $S \to S'$ are étale ring maps and $\kappa (\mathfrak p) = \kappa (\mathfrak p')$ and $\kappa (\mathfrak q) = \kappa (\mathfrak q')$.

Definition 38.6.2. Let $R \to S$ be a finite type ring map. Let $\mathfrak r$ be a prime of $R$. Let $N$ be a finite $S$-module. A complete dévissage of $N/S/R$ over $\mathfrak r$ is given by $R$-algebra maps

$\xymatrix{ & A_1 & & A_2 & & ... & & A_ n \\ S \ar[ru] & & B_1 \ar[lu] \ar[ru] & & ... \ar[lu] \ar[ru] & & ... \ar[lu] \ar[ru] & & B_ n \ar[lu] }$

finite $A_ i$-modules $M_ i$ and $B_ i$-module maps $\alpha _ i : B_ i^{\oplus r_ i} \to M_ i$ such that

1. $S \to A_1$ is surjective and of finite presentation,

2. $B_ i \to A_{i + 1}$ is surjective and of finite presentation,

3. $B_ i \to A_ i$ is finite,

4. $R \to B_ i$ is smooth with geometrically irreducible fibres,

5. $N \cong M_1$ as $S$-modules,

6. $\mathop{\mathrm{Coker}}(\alpha _ i) \cong M_{i + 1}$ as $B_ i$-modules,

7. $\alpha _ i : \kappa (\mathfrak p_ i)^{\oplus r_ i} \to M_ i \otimes _{B_ i} \kappa (\mathfrak p_ i)$ is an isomorphism where $\mathfrak p_ i = \mathfrak rB_ i$, and

8. $\mathop{\mathrm{Coker}}(\alpha _ n) = 0$.

In this situation we say that $(A_ i, B_ i, M_ i, \alpha _ i)_{i = 1, \ldots , n}$ is a complete dévissage of $N/S/R$ over $\mathfrak r$.

Remark 38.6.3. Note that the $R$-algebras $B_ i$ for all $i$ and $A_ i$ for $i \geq 2$ are of finite presentation over $R$. If $S$ is of finite presentation over $R$, then it is also the case that $A_1$ is of finite presentation over $R$. In this case all the ring maps in the complete dévissage are of finite presentation. See Algebra, Lemma 10.6.2. Still assuming $S$ of finite presentation over $R$ the following are equivalent

1. $M$ is of finite presentation over $S$,

2. $M_1$ is of finite presentation over $A_1$,

3. $M_1$ is of finite presentation over $B_1$,

4. each $M_ i$ is of finite presentation both as an $A_ i$-module and as a $B_ i$-module.

The equivalences (1) $\Leftrightarrow$ (2) and (2) $\Leftrightarrow$ (3) follow from Algebra, Lemma 10.36.23. If $M_1$ is finitely presented, so is $\mathop{\mathrm{Coker}}(\alpha _1)$ (see Algebra, Lemma 10.5.3) and hence $M_2$, etc.

Definition 38.6.4. Let $R \to S$ be a finite type ring map. Let $\mathfrak q$ be a prime of $S$ lying over the prime $\mathfrak r$ of $R$. Let $N$ be a finite $S$-module. A complete dévissage of $N/S/R$ at $\mathfrak q$ is given by a complete dévissage $(A_ i, B_ i, M_ i, \alpha _ i)_{i = 1, \ldots , n}$ of $N/S/R$ over $\mathfrak r$ and prime ideals $\mathfrak q_ i \subset B_ i$ lying over $\mathfrak r$ such that

1. $\kappa (\mathfrak r) \subset \kappa (\mathfrak q_ i)$ is purely transcendental,

2. there is a unique prime $\mathfrak q'_ i \subset A_ i$ lying over $\mathfrak q_ i \subset B_ i$,

3. $\mathfrak q = \mathfrak q'_1 \cap S$ and $\mathfrak q_ i = \mathfrak q'_{i + 1} \cap A_ i$,

4. $R \to B_ i$ has relative dimension $\dim _{\mathfrak q_ i}(\text{Supp}(M_ i \otimes _ R \kappa (\mathfrak r)))$.

Remark 38.6.5. Let $A \to B$ be a finite type ring map and let $N$ be a finite $B$-module. Let $\mathfrak q$ be a prime of $B$ lying over the prime $\mathfrak r$ of $A$. Set $X = \mathop{\mathrm{Spec}}(B)$, $S = \mathop{\mathrm{Spec}}(A)$ and $\mathcal{F} = \widetilde{N}$ on $X$. Let $x$ be the point corresponding to $\mathfrak q$ and let $s \in S$ be the point corresponding to $\mathfrak p$. Then

1. if there exists a complete dévissage of $\mathcal{F}/X/S$ over $s$ then there exists a complete dévissage of $N/B/A$ over $\mathfrak p$, and

2. there exists a complete dévissage of $\mathcal{F}/X/S$ at $x$ if and only if there exists a complete dévissage of $N/B/A$ at $\mathfrak q$.

There is just a small twist in that we omitted the condition on the relative dimension in the formulation of “a complete dévissage of $N/B/A$ over $\mathfrak p$” which is why the implication in (1) only goes in one direction. The notion of a complete dévissage at $\mathfrak q$ does have this condition built in. In any case we will only use that existence for $\mathcal{F}/X/S$ implies the existence for $N/B/A$.

Lemma 38.6.6. Let $R \to S$ be a finite type ring map. Let $M$ be a finite $S$-module. Let $\mathfrak q$ be a prime ideal of $S$. There exists an elementary étale localization $R' \to S', \mathfrak q', \mathfrak p'$ of the ring map $R \to S$ at $\mathfrak q$ such that there exists a complete dévissage of $(M \otimes _ S S')/S'/R'$ at $\mathfrak q'$.

Proof. This is a reformulation of Proposition 38.5.7 via Remark 38.6.5 $\square$

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