The Stacks project

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 $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05HV. Beware of the difference between the letter 'O' and the digit '0'.