**Proof plan.** The key ingredient is that the product of two primitive polynomials is primitive (Lemma A). Using this, we show reducibility over $R$ implies reducibility over $F$ (trivially), and reducibility over $F$ can be pulled back to $R$ by clearing denominators and comparing contents. **Step 1: Product of primitives is primitive (Lemma A).** [claim: Primitive Product] If $f, g \in R[X]$ are primitive, then $fg$ is primitive. [/claim] [proof] Suppose for contradiction that $fg$ is not primitive. Then some irreducible (hence prime, since $R$ is a UFD) element $p \in R$ divides all coefficients of $fg$. Since $f$ is primitive, $p \nmid f$; let $a_k$ be the first coefficient of $f$ not divisible by $p$. Similarly, let $b_\ell$ be the first coefficient of $g$ not divisible by $p$. The coefficient of $X^{k+\ell}$ in $fg$ is \begin{align*} \sum_{i+j=k+\ell} a_i b_j = a_k b_\ell + \sum_{\substack{i+j=k+\ell \\ i < k}} a_i b_j + \sum_{\substack{i+j=k+\ell \\ j < \ell}} a_i b_j. \end{align*} The sums on the right are divisible by $p$ (by choice of $k$ and $\ell$). Since $p$ divides the whole sum, $p \mid a_k b_\ell$. Since $p$ is prime and $p \nmid a_k$, we get $p \mid b_\ell$, a contradiction. [/proof] **Step 2: Content is multiplicative (Lemma B).** [claim: Content Multiplicativity] For $f, g \in R[X]$, $c(fg)$ is an associate of $c(f)c(g)$. [/claim] [proof] Write $f = c(f)f_1$ and $g = c(g)g_1$ with $f_1, g_1$ primitive. Then $fg = c(f)c(g)f_1g_1$. By Lemma A, $f_1g_1$ is primitive, so $c(f)c(g)$ is a gcd of the coefficients of $fg$, which equals $c(fg)$ up to a unit. [/proof] **Step 3: ($\Rightarrow$) Reducible over $R$ implies reducible over $F$.** If $f = gh$ in $R[X]$ with $g, h$ not units in $R[X]$, then since $f$ is primitive, neither $g$ nor $h$ is constant (a constant $r$ dividing $f$ would divide $c(f) = 1$, making $r$ a unit). So $g, h$ have positive degree and are non-units in $F[X]$. Thus $f$ is reducible in $F[X]$. **Step 4: ($\Leftarrow$) Reducible over $F$ implies reducible over $R$.** Suppose $f = gh$ in $F[X]$ with $g, h$ having positive degree. Clear denominators: for each coefficient of $g$, multiply by the product of denominators to get $a \in R$ with $ag \in R[X]$; similarly $b \in R$ with $bh \in R[X]$. Then \begin{align*} ab \cdot f = (ag)(bh) \in R[X]. \end{align*} Write $ag = c(ag)g_1$ and $bh = c(bh)h_1$ with $g_1, h_1$ primitive. By Lemma B, $c(abf) = c(ab \cdot f)$ is an associate of $c(ag)c(bh)$. Since $f$ is primitive, $c(abf) = ab$ (up to a unit). So \begin{align*} ab \cdot f = c(ag)c(bh) \cdot g_1 h_1 \implies f = u \cdot g_1 h_1 \end{align*} for some unit $u \in R$. Since $\deg g_1 = \deg g > 0$ and $\deg h_1 = \deg h > 0$, neither $g_1$ nor $uh_1$ is a unit in $R[X]$, so $f = g_1 \cdot (uh_1)$ is a non-trivial factorization in $R[X]$. $\square$