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 $j<i$ 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)?


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 : $$ \begin{CD} F^2(\Omega^*_{P/A}):@.{}@.{}@.Pdx_1\wedge dx_2@>>> Pdx_1\wedge dx_2\wedge dx_3\\ @.@.@.@VVV@VVV\\ F^1(\Omega^*_{P/A}):@.{}@.Pdx_1\oplus Pdx_2@>>>Pdx_1\wedge dx_2\oplus Pdx_1\wedge dx_3\oplus Pdx_2\wedge dx_3@>>> Pdx_1\wedge dx_2\wedge dx_3\\ @.@.@VVV@VVV@VVV\\ F^0(\Omega^*_{P/A}):@.P@>>>Pdx_1\oplus Pdx_2\oplus Pdx_3@>>>Pdx_1\wedge dx_2\oplus Pdx_1\wedge dx_3\oplus Pdx_2\wedge dx_3@>>> Pdx_1\wedge dx_2\wedge dx_3 \end{CD} $$


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