Past exam of the rings and modules course of the University of Oxford / 2021 / 1 / a / i

We prove the stated Eisenstein criterion directly. Assume $n\ge 1$. 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 $r=\mathrm{deg}g>0$ and $s=\mathrm{deg}h>0$.

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 giveThe 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