EDIT: Proved it on my own. It easily follows from the Witt integrality theorem. Sorry for posting.


Let $P\in\mathbb{Z}\left[\Xi\right]$ be a polynomial (where $\Xi$ is a family of symbols that we use as indeterminates, for instance $\Xi=\left(X\_1,X\_2,X\_3,...\right)$). Let $n\in\mathbb{N}$.


Prove or disprove that $\displaystyle\frac{\delta}{\delta\xi}P\in n\mathbb{Z}\left[\Xi\right]$ for every $\xi\in\Xi$ if and only if there exist polynomials $P\_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum\_{d\mid n}dP\_d^{n/d}$.


A few remarks on this: The $\Longleftarrow$ direction is trivial. I can prove the $\Longrightarrow$ if $n$ is a prime power.


PS. No, this does not help in proving the Witt integrality theorem, even if it is true.



Do n-th Witt polynomials generate {P | P' is divisible by n} ?
Solved. The key is that the set of polynomials $P$ for which there exist polynomials $P\_d\in\mathbb{Z}\left[\Xi\right]$ for all divisors $d$ of $n$ such that $\displaystyle P=\sum\_{d\mid n}dP\_d^{n/d}$ is a subring of $\mathbb{Z}\left[\Xi\right]$ (by the Witt integrality theorem, which is 9.73 in [Hazewinkel's Witt vectors](https://arxiv.org/abs/0804.3888v1)).


EDIT: Wrote up [a proof](http://www.cip.ifi.lmu.de/~grinberg/algebra/witt5a.pdf).


