[proofplan] We prove existence and uniqueness of the quotient-remainder pair by two separate arguments. Existence is established by strong induction on $\deg f$: if $\deg f < \deg g$ the division is trivial, and otherwise we subtract a suitable multiple of $g$ to lower the degree and apply the induction hypothesis. Uniqueness follows from the fact that a monic polynomial of degree $m$ cannot divide a nonzero polynomial of degree less than $m$. [/proofplan] [step:Establish existence by strong induction on $\deg f$] We adopt the convention $\deg 0 = -\infty$. Let $m = \deg g$. We induct on $\deg f$. **Base case.** If $\deg f < m$ (including $f = 0$), set $h = 0$ and $r = f$. Then $f = 0 \cdot g + f$, with $\deg h = -\infty \leq \deg f - m$ and $\deg r = \deg f < m$. **Inductive step.** Suppose $\deg f = d \geq m$ and the result holds for all polynomials of degree less than $d$. Write $f = aX^d + (\text{lower terms})$ where $a \in R$ is the leading coefficient of $f$. Since $g$ is monic of degree $m$, we have $g = X^m + (\text{lower terms})$. Define \begin{align*} f_1 := f - aX^{d - m} g. \end{align*} The leading term of $aX^{d-m}g$ is $aX^d$, which cancels the leading term of $f$, so $\deg f_1 < d$. By the induction hypothesis applied to $f_1$ and $g$, there exist $h_1, r \in R[X]$ with $f_1 = h_1 g + r$, $\deg h_1 \leq \deg f_1 - m < d - m$, and $\deg r < m$. Setting $h = aX^{d-m} + h_1$, we obtain \begin{align*} f = aX^{d-m}g + f_1 = aX^{d-m}g + h_1 g + r = (aX^{d-m} + h_1)g + r = hg + r. \end{align*} The degree bound on $h$ holds: $\deg h \leq \max(d - m, \deg h_1) = d - m = \deg f - \deg g$. [guided] We want to divide $f$ by the monic polynomial $g$ of degree $m$, producing a quotient $h$ and a remainder $r$ with $\deg r < m$. The idea is polynomial long division: at each step, we cancel the leading term of the dividend by subtracting an appropriate multiple of $g$. **Base case.** If $\deg f < m$, then $f$ is already of degree less than $\deg g$, so it is its own remainder: set $h = 0$ and $r = f$. **Inductive step.** Suppose $\deg f = d \geq m$. Let $a \in R$ be the leading coefficient of $f$. Since $g$ is monic, the polynomial $aX^{d-m}g$ has leading term $aX^d$, the same as $f$. Why is the monic hypothesis essential here? Because we need $aX^{d-m} \cdot (\text{leading coefficient of } g)$ to equal $a$. If the leading coefficient of $g$ were a general element $b \in R$, we would need $b$ to be a unit in $R$ to solve for the correct multiple, and in a general commutative ring not every nonzero element is a unit. The monic hypothesis ($b = 1$) sidesteps this entirely. Define $f_1 := f - aX^{d-m}g$. By construction, $\deg f_1 < d = \deg f$, so the induction hypothesis applies to the pair $(f_1, g)$. We obtain $f_1 = h_1 g + r$ with $\deg r < m$. Setting $h = aX^{d-m} + h_1$: \begin{align*} f = aX^{d-m}g + f_1 = aX^{d-m}g + h_1 g + r = hg + r. \end{align*} For the degree bound: $\deg(aX^{d-m}) = d - m$ and $\deg h_1 \leq \deg f_1 - m < d - m$, so $\deg h = d - m = \deg f - \deg g$. [/guided] [/step] [step:Prove uniqueness by analysing the difference of two representations] Suppose $f = h g + r = h' g + r'$ with $\deg r < m$ and $\deg r' < m$. Then \begin{align*} (h - h')g = r' - r. \end{align*} The right-hand side has degree at most $\max(\deg r', \deg r) < m = \deg g$. If $h - h' \neq 0$, then the left-hand side has degree $\deg(h - h') + \deg g \geq \deg g = m$, since $g$ is monic (so its leading coefficient is $1_R$, which is not a zero divisor, and multiplication by $g$ does not reduce degree). This contradicts $\deg((h - h')g) < m$. Therefore $h - h' = 0$, so $h = h'$, and consequently $r = r'$. [guided] To see why the monic hypothesis is crucial for uniqueness, suppose $f = hg + r = h'g + r'$ with both remainders of degree less than $m$. Rearranging: \begin{align*} (h - h')g = r' - r. \end{align*} The right-hand side satisfies $\deg(r' - r) \leq \max(\deg r', \deg r) < m$. Now consider the left-hand side. If $h - h' \neq 0$, say $h - h'$ has leading term $cX^k$ with $c \neq 0$. The product $(h - h')g$ then has a term $cX^{k+m}$ (since the leading term of $g$ is $X^m$ and $c \cdot 1 = c$). Can this term cancel with lower-degree terms? In a general ring, two nonzero elements can multiply to zero (zero divisors), which would make the leading term vanish. But here the leading coefficient of $g$ is $1$, so the leading term of $(h - h')g$ is $c \cdot 1 \cdot X^{k+m} = cX^{k+m} \neq 0$. Therefore $\deg((h - h')g) = k + m \geq m$, contradicting $\deg(r' - r) < m$. Hence $h - h' = 0$, which gives $h = h'$ and $r = f - hg = f - h'g = r'$. Both the quotient and remainder are unique. [/guided] [/step]