Consider the usual de Rham CDGA $(\Omega^\* Sym^\*(V),d)$ for a free $\mathbb{Z}\_{(p)}$-module $V$. What is known about its cohomology?


It is easy to compute ranks of primary summands in $H^\*(\Omega^\* Sym^\*(V),d)$ through the Bockstein spectral sequence or by applying the Kunneth's formula. Meanwhile a truly functorial in $V$ description of the cohomology groups is available only in special cases and generally seems to be hopeless (according to the work of Roman Mikhailov).


What is known about the ring structure? Can we realize $H^\*(\Omega^\* Sym^\*(V))$ as a quotient algebra of something else? Are there any "cohomology operations" acting here?



The integral cohomology of the de Rham complex
Following David Speyer's suggestion, fix exponents $e\_1,\dots, e\_n$ and consider the subcomplex consisting of sums of those monomials of the form $$\prod\_{i=1}^n \left( x\_i^{e\_i - \epsilon\_i} (d x\_i)^{\epsilon\_i} \right)$$ for $\epsilon\_i \in \{0,1\}$. The point of doing this is that this subcomplex is stable under differentitation and the de Rham complex is the sum.


I claim that the cohomology of this complex, in degree $d$, is $$ \left(\mathbb Z/ p^{ \min\_{i=1}^n v\_p( e\_i)}\right)^{ \binom{m-1}{d-1}}$$ where $m$ is the number of $e\_i$s that are nonzero.


To do this, note that this subcomplex is clearly the tensor product from $i$ from $1$ to $n$ of the complex consisting of $x\_i^{e\_i}$ in degree $0$ and $x\_i^{e\_i-1} dx\_i$ in degree $1$ (unless $e\_i=0$, in which case the $i$th complex is only in degree $0$). Calculating the derivative, this subcomplex is equivalent $\mathbb Z\_{(p)} \to \mathbb Z\_{(p)}$ with the differential multiplication by $e\_i$.


Because this is a complex of projective modules, tensoring with it computes the derived tensor product. Now choose some $i$ minimizing $v\_p(e\_i)$, and observe that the $i$th complex is quasi-isomorphic to $ \mathbb Z/p^{ v\_p(e\_i)}$ in degree $1$.


We may thus replace the $i$th factor in the tensor product with $\mathbb Z/p^{ v\_p(e\_i)}$ in degree $1$. But having done that, our tensor product complex is now isomorphic to $ \left(\mathbb Z/ p^{ \min\_{i=1}^n v\_p( e\_i)}\right)^{ \binom{m-1}{d-1}}$ in each degree $d$, and the differentials vanish because they are divisible by $p^{ \min\_{i=1}^n v\_p( e\_i)}$, so the cohomology in degree $d$ is also $ \left(\mathbb Z/ p^{ \min\_{i=1}^n v\_p( e\_i)}\right)^{ \binom{m-1}{d-1}}$, as desired.


More concretely, we can represent the forms as follows. Say $e\_n$ has the least $p$-adic valuation, for simplicity.


Then every $k$-form in the variables $x\_1,\dots,x\_n$ of exponent $(e\_1,\dots, e\_n)$ has the form
$$ g x\_n^{e\_n} + f x\_n^{e\_n-1} dx\_n$$ where $f$ is a $k-1$-form and $g$ is a $k$-form in the variables $x\_{1},\dots, x\_{n-1}$, each of exponent $(e\_1,\dots, e\_{n-1})$. Differentiating, we get


$$d ( g x\_n^{e\_n} + f x\_n^{e\_n-1} dx\_n)= dg x\_n^{e\_n} + (-1)^k e\_n g x\_n^{e\_{n}-1} d x\_n + df x\_n^{e\_n-1} dx\_n .$$


So the form is closed if and only if $$dg=0$$ and $$e\_n g = (-1)^{k-1} df$$ but the second equation implies the first because $e\_n$ is a nonzero divisor in $\mathbb Z$ and $d^2 =0$.


So the closed forms are exactly those of the form
$$ \frac{df}{e\_n} x\_n^{e\_n} + (-1)^{k-1} f x\_n^{e\_n-1} dx\_n$$ for $f$ a $k-1$-form in $x\_1,\dots, x\_{n-1}$ of exponents $e\_1,\dots, e\_{n-1}$, and this always exists since $e\_n$ divides $e\_1,\dots, e\_{n-1}$ over the $p$-adic numbers.


If $f$ is divisibly by $e\_n$, then such a form is $d ( f x\_n^{e\_n})$ and hence is exact. Conversely, because all the exponents are divisible by $e\_n$, if such a form is exact then $f$ is divisible by $e\_n$.


So this gives an explicit basis for this module over $\mathbb Z/p^{ v\_p(e\_n)}$, that being the expressions $$ \frac{df}{e\_n} x\_n^{e\_n} + (-1)^{k-1} f x\_n^{e\_n-1} dx\_n$$ for monomial $f$.


Multiplying two such expressions, one can, in principle, extract the structure constants for the cup / wedge product, but I haven't done this.


