# For each $\sigma \in \mathrm{S}_{m}$, $p \mapsto \sigma \cdot p$ is an automorphism of $R[X_{1}, \ldots, X_{m}]$

I am trying to solve Exercise 7 in Section Rings, Fields and Polynomials from textbook Analysis I by by Amann/Escher.

Please have check on my attempt! Does it look fine or contain gaps/errors?

Let $$R$$ be a commutative ring with unity and $$m \in \mathbb{N}$$ with $$m \geq 2$$. Let $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, x) \mapsto \sigma \cdot x := x \circ \sigma^{-1}$$ be the action of $$(\mathrm{S}_{m}, \circ)$$ on $$\mathbb{N}^{m}$$ where $$\mathrm{S}_{m}$$ is the permutation group of $$\{1,\ldots,m\}$$ and $$\circ$$ is function composition.

1. The equation $$\sigma \cdot \sum_{x} p_{x} X^{x} :=\sum_{x} p_{x} X^{\sigma \cdot x}$$ defines an action $$\mathrm{S}_{m} \times R[X_{1}, \ldots, X_{m}] \rightarrow R[X_{1}, \ldots, X_{m}], \quad(\sigma, p) \mapsto \sigma \cdot p$$ of $$\mathrm{S}_{m}$$ on the polynomial ring $$R[X_{1}, \ldots, X_{m}]$$.

2. For each $$\sigma \in \mathrm{S}_{m}$$, $$p \mapsto \sigma \cdot p$$ is an automorphism of $$R[X_{1}, \ldots, X_{m}]$$.

My attempt:

1. It is easy to verify that $$\sigma \cdot p = p'$$ where $$p' (x) = p (x \circ \sigma)$$ for all $$\sigma \in \mathrm{S}_{m}$$, $$p \in R[X_{1}, \ldots, X_{m}]$$, and $$x \in \mathbb{N}^{m}$$.

Because $$\operatorname{id_{\{1,\ldots,m\}}} \cdot x = x$$ for all $$x \in \mathbb{N}^{m}$$, $$\operatorname{id_{\{1,\ldots,m\}}} \cdot p = p$$ for all $$p \in R[X_{1}, \ldots, X_{m}]$$.

For $$\pi,\sigma \in \mathrm{S}_{m}$$ and $$p \in R[X_{1}, \ldots, X_{m}]$$, we have $$\pi \cdot (\sigma \cdot p) = p''$$ where $$p'' = \pi \cdot p'$$ and $$p' = \sigma \cdot p$$. Then $$p'' (x) = p' (x \circ \pi) = p ((x \circ \pi) \circ \sigma) = p (x \circ (\pi \circ \sigma))$$. Thus $$\pi \cdot (\sigma \cdot p) = p'' =(\pi \circ \sigma) \cdot p$$.

1. If $$p_1 , p_2 \in R[X_{1}, \ldots, X_{m}]$$ such that $$\sigma \cdot p_1 = \sigma \cdot p_2$$, then $$p_1 (x \circ \sigma) = p_2 (x \circ \sigma)$$ for all $$x \in \mathbb{N}^{m}$$ and thus $$p_1 (x) = p_2 (x)$$ for all $$x \in \mathbb{N}^{m}$$. So $$p_1 = p_2$$ and subsequently $$p \mapsto \sigma \cdot p$$ is injective.

For $$p' \in R[X_{1}, \ldots, X_{m}]$$, we define $$p \in R[X_{1}, \ldots, X_{m}]$$ by $$p (x) = p' (x \circ \sigma^{-1})$$ for all $$x \in \mathbb{N}^{m}$$. Then $$\sigma \cdot p = p'$$ and so $$p \mapsto \sigma \cdot p$$ is surjective.

For $$\sigma \in \mathrm{S}_{m}$$, $$p_1 , p_2 \in R[X_{1}, \ldots, X_{m}]$$, and $$x \in \mathbb{N}^{m}$$, we have

\begin{aligned} (\sigma \cdot (p_1 + p_2)) (x) &= (p_1 + p_2) (x \circ \sigma) \\ &= p_1 (x \circ \sigma) + p_2 (x \circ \sigma) \\ &= (\sigma \cdot p_1) (x) + (\sigma \cdot p_2) (x) \\ &= (\sigma \cdot p_1 + \sigma \cdot p_2) (x) \end{aligned}

and

\begin{aligned} (\sigma \cdot (p_1 \cdot p_2)) (x) &= (p_1 \cdot p_2) (x \circ \sigma) \\ &= \sum_{y \le x \circ \sigma} \left [p_1 (y) + p_2(x \circ \sigma - y) \right]\\ &= \sum_{z \circ \sigma \le x \circ \sigma} \left [p_1 (z \circ \sigma) + p_2(x \circ \sigma - z \circ \sigma) \right] \quad (z \circ \sigma := y) \\ &= \sum_{z \le x} \left [p_1 (z \circ \sigma) + p_2((x - z) \circ \sigma) \right] \\ &= \sum_{z \le x} \left [(\sigma \cdot p_1) (z) + (\sigma \cdot p_2) (x-z) \right] \\ &= ((\sigma \cdot p_1) \cdot (\sigma \cdot p_2)) (x) \end{aligned}

As a result, $$p \mapsto \sigma \cdot p$$ is an automorphism.