Polynomial reversal

Question

For a polynomial $a = {a_0} + {a_1}x + \ldots + {a_{n - 1}}{x^{n - 1}} + {a_n}{x^n} \in R[x]$ of degree at most $n \in {\mathbb{Z}_{ \geqslant 0}}$ over a ring $R$, let the $n$-reversal of $a$ be the polynomial ${\text{re}}{{\text{v}}_n}a = {a_n} + {a_{n - 1}}x + \ldots + {a_0}{x^n} \in R[x]$. Let $a,b \in R[x]$ be polynomials with $b$ monic and $n = \deg a \geqslant \deg b = m$. Show that the quotient $q \in R[x]$ and the remainder $r \in R[x]$ with $a = qb + r$ and $\deg r \leqslant m - 1$ satisfy the reversal identity

$${\text{re}}{{\text{v}}_n}(a) = {\text{re}}{{\text{v}}_{n - m}}(q) \cdot {\text{re}}{{\text{v}}_m}(b) + {x^{n - m + 1}} \cdot {\text{re}}{{\text{v}}_{m - 1}}(r)$$

Hint: Observe that for all$\,\,i = 0,1, \ldots ,n - m{\text{ and }}j = 0,1, \ldots ,m$ we have ${x^{n - i - j}} = {x^{n - m - i}}{x^{m - j}}$.

My work:

The equality is equivalent to $${\text{re}}{{\text{v}}_n}(a) \equiv {\text{re}}{{\text{v}}_{n - m}}(q) \cdot {\text{re}}{{\text{v}}_m}(b)\,\bmod \,({x^{n - m + 1}})$$Since $b$ is monic, ${\text{re}}{{\text{v}}_m}(b)$ will have a constant coefficient equal to $1$ and hence it is invertible modulo ${x^{n - m + 1}}$. Since ${\text{re}}{{\text{v}}_m}{(b)^{ - 1}} \in R[x]$ we can express ${\text{re}}{{\text{v}}_{n - m}}(q)$ uniquely as $${\text{re}}{{\text{v}}_{n - m}}(q) \equiv {\text{re}}{{\text{v}}_n}(a) \cdot {\text{re}}{{\text{v}}_m}{(b)^{ - 1}}\,\bmod \,({x^{n - m + 1}})$$ It follows that \begin{equation} \notag \begin{split} {\text{re}}{{\text{v}}_n}(a)\,\bmod \,({x^{n - m + 1}}) &= ({\text{re}}{{\text{v}}_{n - m}}(q) \cdot {\text{re}}{{\text{v}}_m}(b)\,\bmod \,({x^{n - m + 1}}) \,\, + \\ &+{x^{n - m + 1}} \cdot {\text{re}}{{\text{v}}_{m - 1}}(r)\,\bmod \,({x^{n - m + 1}}))\,\bmod \,({x^{n - m + 1}}) \hfill \\ &= {\text{re}}{{\text{v}}_n}(a) \cdot {\text{re}}{{\text{v}}_m}{(b)^{ - 1}} \cdot {\text{re}}{{\text{v}}_m}(b)\,\,\bmod \,({x^{n - m + 1}}) + 0 \hfill \\ &= {\text{re}}{{\text{v}}_n}(a)\,\bmod \,({x^{n - m + 1}}) \hfill \\ \end{split} \end{equation}

I'm not sure if my proof is correct since i don't know how to use the hint.

Answer 1

Hint: $\;\operatorname{rev}_n \,a(x) = x^n \cdot a\left(\frac{1}{x}\right)\,$. Now substitute $x \mapsto \frac{1}{x}$ in the euclidian division equation:

$$ a(x) = q(x) b(x) + r(x) \quad \implies \quad a\left(\frac{1}{x}\right) = q\left(\frac{1}{x}\right) b\left(\frac{1}{x}\right) + r\left(\frac{1}{x}\right) $$

Multiply by $x^n$ then spot the $\operatorname{rev}$ expressions:

$$ x^n \cdot a\left(\frac{1}{x}\right) \,=\, x^{n-m} \cdot q\left(\frac{1}{x}\right) \cdot x^{m} \cdot b\left(\frac{1}{x}\right) \,+\, x^{n-m+1} \cdot x^{m-1}\cdot r\left(\frac{1}{x}\right) $$