# Filtration in crystalline Poincaré Lemma

I am trying to understand section 20 in https://stacks.math.columbia.edu/download/crystalline.pdf, especially the proof of Lemma 20.2.

If $$A\rightarrow B$$ is a map of rings and $$P=B[x_i]$$ is some polynomial algebra, then we can look at the map of De-Rham complexes $$\Omega^*_{B/A}\rightarrow \Omega^*_{P/A}$$. We have a filtration $$F^i$$ on $$\Omega^*_{B/A}$$ given by truncating $$\Omega^*_{B/A}$$, i.e. by setting $$F^i(\Omega^*_{B/A})$$ to be the complex which is 0 in degree $$j and $$\Omega^j_{B/A}$$ for $$j\geq 0$$.

Now I want to use this filtration to define a filtration on $$\Omega^*_{P/A}$$. In the pdf above they do it by setting $$F^i(\Omega^*_{P/A})=F^i(\Omega^*_{B/A})\wedge \Omega^*_{P/A}$$ but I fail to see how this gives me a filtration. How is the wedge product defined (as $$F^i(\Omega^*_{B/A})$$ and $$\Omega^*_{P/A}$$ live over different rings)?

## 1 Answer

As you said, there is a map $$\Omega^*_{B/A}\to\Omega^*_{P/A}$$. This induces $$F^i(\Omega^*_{B/A})\to \Omega^*_{P/A}$$. Then $$F^i(\Omega^*_{P/A})$$ are the classes that can be written as a product of an element of the image of this map and an element of $$\Omega^*_{P/A}$$. Note that this is not all of $$\Omega^*_{P/A}$$ since $$F^i(\Omega^*_{B/A})$$ contains only elements of degree $$\geq i$$. Intuitively $$F^i(\Omega^*_{P/A})$$ is then spanned by the classes $$a_1\wedge ...\wedge a_n$$ such that at least $$i$$ of the $$a_1,...,a_n$$ comes from $$\Omega^*_{B/A}$$.

As an example, let $$B=A[x_1,x_2]$$ and $$P=B[x_3]=A[x_1,x_2,x_3]$$. Then $$\Omega_{B/A}$$ is the following complex with the following filtration :

$$\require{AMScd} \begin{CD} F^2(\Omega^*_{B/A}):@.{}@.{}@.Bdx_1\wedge dx_2\\ @.@.@.@VVV\\ F^1(\Omega^*_{B/A}):@.{}@.Bdx_1\oplus Bdx_2@>>>Bdx_1\wedge dx_2\\ @.@.@VVV@VVV\\ F^0(\Omega^*_{B/A}):@.B@>>>Bdx_1\oplus Bdx_2@>>>Bdx_1\wedge dx_2 \end{CD}$$ Then $$\Omega_{P/A}$$ is the following complex with filtration : $$\begin{CD} F^2(\Omega^*_{P/A}):@.{}@.{}@.Pdx_1\wedge dx_2@>>> Pdx_1\wedge dx_2\wedge dx_3\\ @.@.@.@VVV@VVV\\ F^1(\Omega^*_{P/A}):@.{}@.Pdx_1\oplus Pdx_2@>>>Pdx_1\wedge dx_2\oplus Pdx_1\wedge dx_3\oplus Pdx_2\wedge dx_3@>>> Pdx_1\wedge dx_2\wedge dx_3\\ @.@.@VVV@VVV@VVV\\ F^0(\Omega^*_{P/A}):@.P@>>>Pdx_1\oplus Pdx_2\oplus Pdx_3@>>>Pdx_1\wedge dx_2\oplus Pdx_1\wedge dx_3\oplus Pdx_2\wedge dx_3@>>> Pdx_1\wedge dx_2\wedge dx_3 \end{CD}$$