Irreducibility of a polynomial linked to his reduction mod $m$

Question

Assume to have a domain $D $, consider a maximal ideal $m$, take $f \in D[X]$ and take $\overline{f}\in (D/m)[X]$ (which is a field) s.t. $\overline{f}$ is irreducible. In this situation in which hypothesis $f$ is irreducible? Thanks for the help!

Answer 1

This is true (with the additional assumption that the leading coefficient of $f$ is not in $m$). It is a special case of the result

Let $I$ be a proper ideal in the integral domain $R$ and let $p(x)$ be a nonconstant monic polynomial in $R[x]$. If the image of $p(x)$ in $(R/I)[x]$ is irreducible, then $p(x)$ is irreducible.

Let's modify the standard proof to your case:

Take $f \in D[X]$ with the leading coefficient of $f$ not in $m$. Suppose $f$ is reducible, but $\bar{f} \in (D/m)[x]$ is irreducible. Then $f=gh$ with $g,h$ not units. Now the leading coefficient of both $g$ and $h$ cannot be in $m$ (otherwise their product would be in $m$, which is not true by assumption). So $\bar{g}$ and $\bar{h}$ both have degree greater than one. Then we have a nontrivial factorization $\bar{f}= \bar{g} \bar{h}$. Contradiction. So $f$ was irreducible.

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Notice that we need the assumption that the leading coefficient of $f$ is not $m$, or else it can easily be the case that either $f$ or $g$ is a unit in $(D/m)[x]$ and our proof doesn't work.

For an example where we don't enforce the leading coefficient condition: Take $f = 2x^2+x-1$. Then $\bar{f} = x+1$ mod $2$ is irreducible in $(\mathbb{Z}/2\mathbb{Z})[x]$. But $f$ was reducible in $\mathbb{Z}$: $f = (2x-1)(x+1)$.