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Consider the symmetric power series $$f = \prod_{i \in I}\left(1+x_i+x_i^2+x_i^4+x_i^8 + x_i^{16} +\ldots \right)$$ in variables $(x_i)_{i \in I}$ over $\mathbb F_2$. Fix some degree $r$, smaller than the number of variables, and denote the degree $r$ part of $f$ by $f^{(r)}$.

From looking at small-dimensional examples, I got the impression that the following might be true: If $r$ is odd, then $f^{(r)}$ is divisible by $\sigma_1 = \sum_{i \in I} x_i$. Could somebody provide a proof or counterexample of that statement? It would be interesting to me knowing the answer, either way.

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Let $I=\{i_0\} \cup J$ and $g(x)=1+x+x^2+x^4+\dots$.

We show that adding the relation $\sum_{i\in I}x_i=0$, i.e. $x_{i_0}=\sum_{i\ne i_0}x_i$ leads to an even degree power series:

$$ f=g(x_{i_0})\prod_{j\in J} g(x_j) =\left(1+\sum_{i\in J} (g(x_i)-1)\right)\prod_{j\in J} g(x_j)\\ = \prod_{j\in J} g(x_j) + \sum_{i\in J}\left(g(x_i)-1\right)g(x_i)\prod_{j\in J\setminus \{i\}}g(x_j)\\ = \prod_{j\in J} g(x_j) + \sum_{i\in J}x_i\prod_{j\in J\setminus \{i\}}g(x_j)= \prod_{j\in J} g(x_j) + \sum_{i\in J}x_i\frac{\partial}{\partial x_i}\prod_{j\in J}g(x_j)$$ Now this is an even degree power series: check that each monomial with an odd number of $x_j^1$ cancels.

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    $\begingroup$ Wonderful proof! Let me just spell out a few of the steps in more detail. (1) You are using the fact that $g\left(x_{i_0}\right) = 1 + \sum\limits_{i\in J}\left(g\left(x_i\right)-1\right)$. This follows from the observation that the power series $h\left(x\right) := g\left(x\right) - 1$ is additive (i.e., it satisfies $h\left(x+y\right) = h\left(x\right) + h\left(y\right)$) and thus satisfies $h\left(\sum\limits_{i \in J} x_i\right) = \sum\limits_{i \in J} h\left(x_i\right)$. Of course, the additivity of $h\left(x\right)$ follows from the additivity of $x^{2^i}$ for each $i > 0$. $\endgroup$
    – darij grinberg
    Commented Aug 2, 2016 at 14:19
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    $\begingroup$ (2) You are using the fact that $\left(g\left(x\right)-1\right) g\left(x\right) = x$. This follows from the computation $\left(g\left(x\right)\right)^2 = \left(1+x+x^2+x^4+\cdots\right)^2 = 1+x^2+x^4+x^8+\cdots = g\left(x\right) - x = g\left(x\right) + x$. $\endgroup$
    – darij grinberg
    Commented Aug 2, 2016 at 14:20
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    $\begingroup$ (3) At the very end, you are using the fact that if $p$ is any power series in $\mathbb{F}_2\left[\left[x_i \mid i \in J\right]\right]$, then $p + \sum\limits_{i \in J} x_i \dfrac{\partial}{\partial x_i} p$ is a power series whose all monomials have even degree. This is easy to check by linearity. $\endgroup$
    – darij grinberg
    Commented Aug 2, 2016 at 14:22

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