Let $\mathcal{R}$ be a commutative ring with unity. Consider the ring $\frac{\mathcal{R}[x]}{{\langle f(x) g(x) \rangle}}$
$$\frac{\mathcal{R}[x]}{{\langle f(x) g(x) \rangle}}\cong \frac{\mathcal{R}[X]}{{\langle f(x) \rangle}}\times \frac{\mathcal{R}[X]}{{\langle g(x) \rangle}}\tag{1}$$
If $\mathcal{R}$ is a field and $\gcd (f(x), g(x)) =1$
$$\begin{align}{{\langle f(x) g(x) \rangle}}&={{\langle \textrm{lcm} (f(x), g(x)) \rangle}}\\&={{\langle f(x) \rangle}}\cap{{\langle g(x) \rangle}}\end{align}$$
Hence $1) $ is true by Chinese Remainder Theorem.
Under what condition(s) $1) $ is true for a general integral domain $\mathcal{R}$ ?