most recent 30 from http://mathoverflow.net 2013-05-25T18:57:15Z http://mathoverflow.net/feeds/question/12538 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12538/do-n-th-witt-polynomials-generate-p-p-is-divisible-by-n darij grinberg 2010-01-21T13:34:24Z 2012-03-11T19:38:33Z

<p>EDIT: Proved it on my own. It easily follows from the Witt integrality theorem. Sorry for posting.</p> <p>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 $n\in\mathbb{N}$.</p> <p>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 $\displaystyle P=\sum_{d\mid n}dP_d^{n/d}$.</p> <p>A few remarks on this: The $\Longleftarrow$ direction is trivial. I can prove the $\Longrightarrow$ if $n$ is a prime power.</p> <p>PS. No, this does not help in proving the Witt integrality theorem, even if it is true.</p>

http://mathoverflow.net/questions/12538/do-n-th-witt-polynomials-generate-p-p-is-divisible-by-n/12555#12555 darij grinberg 2010-01-21T17:32:12Z 2012-03-11T19:38:33Z

<p>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 <a href="http://arxiv.org/ftp/arxiv/papers/0804/0804.3888.pdf" rel="nofollow">Hazewinkel's Witt vectors</a>).</p> <p>EDIT: Wrote up <a href="http://mit.edu/~darij/www/algebra/witt5a.pdf" rel="nofollow">a proof</a>.</p>