## 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

91.3.1.1
$$\label{defos-equation-morphism-thickenings} \vcenter { \xymatrix{ (X, \mathcal{O}_ X) \ar[r]_ i \ar[d]_ f & (X', \mathcal{O}_{X'}) \ar[d]^{f'} \\ (S, \mathcal{O}_ S) \ar[r]^ t & (S', \mathcal{O}_{S'}) } }$$

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 (91.3.1.1) is cartesian in the category of ringed spaces.

Proof. Omitted. $\square$

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).