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I am trying to follow a general version of the Eisenstein's criterion, which states that

Let $R$ be a commutative ring and $\mathfrak{p}$ a prime ideal of $R$. Let $$f = a_0 + a_1x + \dots + a_nx^n \in R[x] \, . $$ be such that

  1. $a_n \notin \mathfrak{p}$;
  2. $a_0, \dots, a_{n-1} \in \mathfrak{p}$;
  3. $a_0 \notin \mathfrak{p}^2$.

Then $f$ is not a product of polynomials of degree less then $n$ in $R[x]$.

A crucial step in the proof states that if $f = a_nx^n$ and $f = gh$, where $f, g, h \in R'[x]$ for some integral domain $R'$ (in this case, $R' = R / \mathfrak{p}$), then $g$ and $h$ must also be monomials. I do not see how it follows.

I do know that it holds if $R'[x]$ is a UFD, as then $R'[x]$ is a UFD and $f$ would have a unique factorization.

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    If either $g$ or $h$ is not a monomial, then that factor has a distinct lowest degree term and a highest degree term. Because $R'$ is an integral domain the product $gh$ will then also have distinct highest and lowest degree terms. A contradiction. – 2017-02-21
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    @JyrkiLahtonen I have thought of that. My concern would be that one of $g, h$'s lowest coefficient is zero while the other is not. In this way, the lowest coefficient of $gh$ is still zero. Did I make a stupid mistake here? – 2017-02-21
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    I meant: lowest/highest degree non-zero term (that is non-zero in $R'$). My understanding was that we are looking at factorizations in $R'[x]$ at this step, so "non-zero" means "not in $\mathfrak{p}$", and the claim is that $g$ and $h$ must be monomials in $R'[x]$, that is, after reduction modulo $\mathfrak{p}$. May be I misunderstood? – 2017-02-21
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    @JyrkiLahtonen I see it. Thanks. When looking at “lowest” I was thinking $0$-th degree. – 2017-02-21

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I made a stupid mistake. As @JyrkiLahtonen pointed out, a proof can go as follows.

Assume not, then we can assume without loss of generality $g = b_dx^d + r_g + b_mx^m$ and $h = c_ex^e + r_h + c_nx^n$, where $b_fx^f, c_nx^n$ are the lowest non-zero terms of $g, h$ and $r_g, r_h$ are remainders. Clearly, $b_mc_nx^{m + n}$ would be the lowest non-zero term of $f = gh$, and $m + n < \deg f$, hence $f$ is not a monomial.