The Stacks project

91.3 Thickenings of ringed spaces

In the following few sections we will use the following notions:

  1. A sheaf of ideals $\mathcal{I} \subset \mathcal{O}_{X'}$ on a ringed space $(X', \mathcal{O}_{X'})$ is locally nilpotent if any local section of $\mathcal{I}$ is locally nilpotent. Compare with Algebra, Item 29.

  2. A thickening of ringed spaces is a morphism $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ of ringed spaces such that

    1. $i$ induces a homeomorphism $X \to X'$,

    2. the map $i^\sharp : \mathcal{O}_{X'} \to i_*\mathcal{O}_ X$ is surjective, and

    3. the kernel of $i^\sharp $ is a locally nilpotent sheaf of ideals.

  3. A first order thickening of ringed spaces is a thickening $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ of ringed spaces such that $\mathop{\mathrm{Ker}}(i^\sharp )$ has square zero.

  4. It is clear how to define morphisms of thickenings, morphisms of thickenings over a base ringed space, etc.

If $i : (X, \mathcal{O}_ X) \to (X', \mathcal{O}_{X'})$ is a thickening of ringed spaces then we identify the underlying topological spaces and think of $\mathcal{O}_ X$, $\mathcal{O}_{X'}$, and $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp )$ as sheaves on $X = X'$. We obtain a short exact sequence

\[ 0 \to \mathcal{I} \to \mathcal{O}_{X'} \to \mathcal{O}_ X \to 0 \]

of $\mathcal{O}_{X'}$-modules. By Modules, Lemma 17.13.4 the category of $\mathcal{O}_ X$-modules is equivalent to the category of $\mathcal{O}_{X'}$-modules annihilated by $\mathcal{I}$. In particular, if $i$ is a first order thickening, then $\mathcal{I}$ is a $\mathcal{O}_ X$-module.

Situation 91.3.1. A morphism of thickenings $(f, f')$ is given by a commutative diagram
\begin{equation} \label{defos-equation-morphism-thickenings} \vcenter { \xymatrix{ (X, \mathcal{O}_ X) \ar[r]_ i \ar[d]_ f & (X', \mathcal{O}_{X'}) \ar[d]^{f'} \\ (S, \mathcal{O}_ S) \ar[r]^ t & (S', \mathcal{O}_{S'}) } } \end{equation}

of ringed spaces whose horizontal arrows are thickenings. In this situation we set $\mathcal{I} = \mathop{\mathrm{Ker}}(i^\sharp ) \subset \mathcal{O}_{X'}$ and $\mathcal{J} = \mathop{\mathrm{Ker}}(t^\sharp ) \subset \mathcal{O}_{S'}$. As $f = f'$ on underlying topological spaces we will identify the (topological) pullback functors $f^{-1}$ and $(f')^{-1}$. Observe that $(f')^\sharp : f^{-1}\mathcal{O}_{S'} \to \mathcal{O}_{X'}$ induces in particular a map $f^{-1}\mathcal{J} \to \mathcal{I}$ and therefore a map of $\mathcal{O}_{X'}$-modules

\[ (f')^*\mathcal{J} \longrightarrow \mathcal{I} \]

If $i$ and $t$ are first order thickenings, then $(f')^*\mathcal{J} = f^*\mathcal{J}$ and the map above becomes a map $f^*\mathcal{J} \to \mathcal{I}$.

Definition 91.3.2. In Situation 91.3.1 we say that $(f, f')$ is a strict morphism of thickenings if the map $(f')^*\mathcal{J} \longrightarrow \mathcal{I}$ is surjective.

The following lemma in particular shows that a morphism $(f, f') : (X \subset X') \to (S \subset S')$ of thickenings of schemes is strict if and only if $X = S \times _{S'} X'$.

Lemma 91.3.3. In Situation 91.3.1 the morphism $(f, f')$ is a strict morphism of thickenings if and only if ( is cartesian in the category of ringed spaces.

Proof. Omitted. $\square$

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 08KY. Beware of the difference between the letter 'O' and the digit '0'.