The following dynamical system on polynomials comes mostly from idle curiosity, but I hope it is of some interest.


**Background** Fix some natural number $n$. Let $P$ be the quotient of the polynomial ring $\mathbb{Z}[x\_1,\ldots,x\_n]$ by the ideals $\lbrace(x\_j^2=x\_j)~|~1\leq j \leq n\rbrace$.


Let $\Sigma(n)$ denote the symmetric group on $n$ elements and note that it acts on $P$ by permuting the variables. The polynomials fixed by all $\sigma \in \Sigma(n)$ are sums and products of the symmetric polynomials and we can enumerate them by monomial degree. That is, the subring $P' \subset P$ of those polynomials is generated by the [elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) $$S\_1 = x\_1 + \ldots + x\_n,\ S\_2 = \sum\_{i<j}s\_is\_j,\ \dots\ ,\ S\_n = x\_1x\_2\ldots x\_n.$$ 


Consider the function $s:P \to P'$ defined as follows: $s$ takes a polynomial $p \in P$ to $$ \prod\_{\sigma \in \Sigma(n)}\sigma(p).$$ Also consider a function $e:P' \to P$ defined by sending the elementary symmetric polynomial $S\_j$ to its corresponding $x\_j$ and extending the domain to all of $P'$ via $\mathbb{Z}$-linearity.


The "dynamical system" I'm interested in comes from the self-map $f = e\circ s: P \to P$. 


**Example** Consider $n = 2$ and $p = x\_1$. Then, $s(p) = x\_1x\_2 = S\_2$ and so $f(p) = e(S\_2) = x\_2$. Then, $f(f(p)) = e \circ s(x\_2) = e(x\_1x\_2) = x\_2 = f(p)$. Thus, $p$ lies in the basin of attraction of the fixed point $x\_2$. On the other hand, if you start with $p = 2x\_1+x\_2$ then the orbit of $p$ is $$2x\_1 + x\_2 \to 2x\_1 + 5x\_2 \to 10x\_1 + 29x\_2 \to \ldots $$



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> **Question** What can be said about the behavior of $f:P \to P$ in terms of fixed points, periodic orbits, etc. for arbitrary $n$?
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**Update** Contrary to what was stated in the prior versions of this question, neither $s$ nor $e$ is a ring homomorphism.



Symmetric group action on squarefree polynomials
Interesting things happen if you reduce this dynamical system modulo some integer $N$.


This makes sense because your map $s$ is compatible with reduction modulo $N$, and your map $e$ is $\mathbf{Z}$-linear, so $f$ induces a map $f\_N : P\_N \to P\_N$ with $P\_N = P \otimes \mathbf{Z}/N\mathbf{Z}$.


In the case $N=2$, what is the map $f\_2$? Note that $P\_2 = \mathbf{F}\_2[x\_1,\ldots,x\_n]/(x\_1^2-x\_1,\ldots,x\_n^2-x\_n)$ can be identified with the ring of functions $\mathbf{F}\_2^n \to \mathbf{F}\_2$ (every such function is polynomial!). This ring is in bijection with the powerset of $\mathbf{F}\_2^n$ : we can map any subset of $\mathbf{F}\_2^n$ to its characteristic function. Now, the map $s\_2: P\_2 \to P\_2$ has a very natural interpretation : it maps a subset $A$ of $\mathbf{F}\_2^n$ to the subset $\cap\_{\sigma \in \Sigma(n)} \sigma(A)$. Also, by definition, the image of $e\_2 = e \otimes \mathbf{F}\_2$ is contained in the set of affine hyperplanes of $\mathbf{F}\_2^n$ (together with $\emptyset$ and $\mathbf{F}\_2^n$).


So we want to determine which affine hyperplanes are fixed by $f\_2$. This is not obvious at first sight, since the map $e\_2$ doesn't seem to have a nice interpretation like $s\_2$. But luckily, some linear algebra over $\mathbf{F}\_2$ shows that there are only $6$ possibilities for $s\_2(H)$ where $H$ is an affine hyperplane of $\mathbf{F}\_2^n$. Namely $s\_2(H)=H$ when $H$ is symmetric, and $s\_2(H)$ can only be $\emptyset$, $\{0\}$, $\{1\}$ or $\{0,1\}$ otherwise. Here $0$ resp. $1$ denotes the constant vector $(0,\ldots,0)$ resp. $(1,\ldots,1)$. It's possible to compute the image by $e\_2$ of these subsets, and after some computations we find that the only fixed points of $f\_2$ are the polynomials $0$, $1$, $x\_n$, together with $1+x\_1+\cdots+x\_{n-1}$ when $n$ is odd.


Now let's go back to $\mathbf{Z}$. Obviously every fixed point $p$ of $f$ must reduce to a fixed point of $f\_2$ mod $2$. Conversely, we have a kind of Hensel lemma : every fixed point of $f\_2$ lifts to a unique fixed point of $f\_{\mathbf{Z}\_2} : P \otimes \mathbf{Z}\_2 \to P \otimes \mathbf{Z}\_2$. Here $f\_{\mathbf{Z}\_2}$ is defined by just replacing the base ring $\mathbf{Z}$ by $\mathbf{Z}\_2$ in the definition of $f$.


This lifting property is a consequence of the following fact : $f$ is contracting for the $2$-adic topology on every ball $p\_0+2P$ with $p\_0 \in P$. Indeed, assume $p=p\_0+2q$ with $q \in P$. Then


\begin{equation\*}
s(p) = \prod\_{\sigma \in \Sigma(n)} p\_0^\sigma + 2q^\sigma \equiv s(p\_0)+2\sum\_{\sigma} q^\sigma \cdot \prod\_{\tau \neq \sigma} p\_0^\tau \pmod{4}
\end{equation\*}


Now, observe that the stabilizer $H$ of $\overline{p\_0} \in P\_2$ in $\Sigma(n)$ is isomorphic to $\Sigma(m) \times \Sigma(n-m)$ for some $0 \leq m \leq n$, so in particular $\operatorname{card}(H) \geq 2$. This implies that $\prod\_{\tau \neq \sigma} \overline{p\_0}^\tau = s(\overline{p\_0})$ in $P\_2$. Furthermore $\sum\_{\sigma} \overline{q}^\sigma = 0$ since the stabilizer of $\overline{q} \in P\_2$ has even order. It follows that $s(p) \equiv s(p\_0) \pmod{4}$. The same argument shows that $p \equiv p\_0 \pmod{2^m}$ implies $s(p) \equiv s(p\_0) \pmod{2^{m+1}}$, so $s$, and thus $f$, is contracting.


It follows immediately that the only fixed points of $f$ in the balls $2P$, $1+2P$ and $x\_n+2P$ are $0$, $1$ and $x\_n$ respectively. It remains to consider the fixed point $1+x\_1+\cdots+x\_{n-1}$ with $n \geq 3$ odd. The corresponding fixed point of $f\_{\mathbf{Z}\_2}$ is obtained as the limit of $f^m(1+x\_1+\cdots+x\_{n-1})$ in $P \otimes \mathbf{Z}\_2$ when $m \to \infty$.


**EDIT** This question is still surprising me. I thought this fixed point would not belong to $P$, but it does. In fact $1-x\_1+x\_2 -\cdots +x\_{n-1} \in P$ is a fixed point of $f$ for any odd $n \geq 3$. This is because of the following identity in $P$


\begin{equation\*}
\prod\_{\sigma \in \Sigma(n)} 1-x\_{\sigma(1)}+x\_{\sigma(2)}- \cdots +x\_{\sigma(n-1)} = 1-S\_1+S\_2-\cdots+S\_{n-1}
\end{equation\*}
which is true because it's true for any choice of $x\_1,\ldots,x\_n \in \{0,1\} \subset \mathbf{Z}$ (both sides are $0$ except when $x\_1=\cdots=x\_n=0$).


To sum up, for $n \geq 2$ the only fixed points of $f$ are $0$, $1$, $x\_n$, together with $1-x\_1+x\_2-\ldots+x\_{n-1}$ for odd $n$.


By the same method, one might determine the periodic orbits of $f$. Since $f$ is contracting on each ball, each $f$-orbit reduces mod $2$ to a $f\_2$-orbit with the same period. The only non-trivial orbit of $f\_2$ is $\{1+x\_1,1+x\_1+\cdots+x\_n\}$. Again, this non-trivial orbit lifts to a unique orbit in $P \otimes \mathbf{Z}\_2$, but I don't know whether this orbit is in $P$. I guess the next step would be to determine all preperiodic points of $f$.


This method also suggests to study the analogous dynamical system on $P = \mathbf{Z}[x\_1,\ldots,x\_n]/(x\_1^p-x\_1,\ldots,x\_n^p-x\_n)$ for any prime $p$.


