= Solution
{c}
We prove the stated <Eisenstein criterion> directly. Assume $n\geq1$. Suppose $f=gh$ in the <polynomial ring> $R[X]$. Since the <principal ideal domain> $R$ is an <integral domain>, <polynomial degree> is additive under <polynomial multiplication>, and the product of the <leading coefficients> of $g,h$ is the <leading coefficient> $1$ of $f$. These two <coefficients> are therefore <units>. Multiplying $g$ and $h$ by inverse <units>, without changing their product, makes both factors <monic polynomials>.
If either factor is a <constant polynomial>, that factor is now $1$, so the original factor was a <unit>. It remains to rule out $r=\deg g>0$ and $s=\deg h>0$.
The <prime element> $p$ generates a <maximal ideal> in the <principal ideal domain> $R$. Indeed, if $(p)\subseteq(d)$, then $p=de$, and the <irreducible element> property of $p$ forces either $(d)=R$ or $(d)=(p)$. Consequently $k=R/(p)$ is a <field>. Apply <reduction modulo an ideal> to every <coefficient>. The <monic polynomial> factors retain their <polynomial degrees>, and the hypotheses give
$$
\overline g\,\overline h=\overline f=X^n\quad\text{in }k[X].
$$
The <polynomial ring> over a <field> is a <unique factorization domain>, and $X$ is an <irreducible polynomial>. Thus its only <monic polynomial> divisors in $X^n$ are powers of $X$. Hence $\overline g=X^r$ and $\overline h=X^s$. Both have zero <constant term>, so $p\mid g(0)$ and $p\mid h(0)$. This gives $p^2\mid g(0)h(0)=a_0$, a contradiction.
The <monic polynomial> $f$ has positive <polynomial degree>, so it is a nonzero <nonunit>. Every factorization therefore has a <unit> factor, proving
$$
\boxed{f\text{ is irreducible in }R[X]}.
$$
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