invert coefficients as monoid homomorphism of polynomial rings
1. Proposition
Let \(R\) be an Integral domain, \(R[X]\) the polynomial ring. Then
\begin{align*} \mathrm{inv}: R[X] \rightarrow& R[X] \\ g \mapsto& X^{\mathrm{deg}} \cdot g(\frac{1}{X}) \end{align*}is a monoid-homomorphism of the multiplicative monoid
2. Proof
2.1. unit
follows from
\begin{align*} \mathrm{inv}(1) =& X^{\mathrm{deg}(1)} \cdot 1(\frac{1}{X}) \\ =& X^0 \cdot 1 \\ =& 1 \end{align*}2.2. multiplicative
Note that
\begin{align*} \mathrm{inv}(\sum_{i=0}^{n} \alpha_{i} X^i) = \sum_{i=0}^{n} \alpha_{n-i} X^i \end{align*}Suppose
\begin{align*} f =& \sum_{i=0}^{n} \alpha_{i} X^i \\ =& (\sum_{i=0}^{k} \beta_i X^i) \cdot \sum_{i=0}^{l} \gamma_i X^i) =& \sum_{i \leq k, j \leq l}^{} \beta_{i} \cdot \gamma_{j} X^{i + j} \end{align*}Then
\begin{align*} \mathrm{inv}(f) =& \sum_{i=0}^{n} \alpha_{n - i} X^i \end{align*}and
\begin{align*} \mathrm{inv}((\sum_{i=0}^{k} \beta_i X^i)) \cdot \mathrm{inv}(\sum_{i=0}^{l} \gamma_i X^i)) =& \sum_{i=0}^{k} \beta_{k-i} X^i \cdot \sum_{i=0}^{l} \gamma_{l-i} X^i \\ =& \sum_{i=0}^{k} \beta_{i} X^{k-i} \cdot \sum_{i=0}^{l} \gamma_{i} X^{l-i} \\ =& \sum_{i \leq k, j \leq l} \beta_i \gamma_{j} \cdot X^{k + l - i - j} \\ =& \sum_{i=0}^{n} \alpha_{n-i} X^i \end{align*}