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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.

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You should post your result as an answer and accept it. That way, if someone were to ask a similar question, this question will show up as answered. The npeople who ask the same question wll be able to find it more easily. –  Harry Gindi Jan 21 '10 at 17:27

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up vote 4 down vote accepted

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).

EDIT: Wrote up a proof.

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