Suppose $\phi:A\twoheadrightarrow B$ is a surjective homomorphism of commutative rings with unit, with $B$ an integral domain (note that $A$ is not an integral domain in general). Suppose $g\in B[X]$ is a monic irreducible polynomial. Can I find a preimage of $g$ in $A[X]$ which is also monic irreducible?
Here an element is irreducible means that it is not a unit, and if it is a product of two elements, then one of the elements is a unit.
A related (maybe easier or harder) question: if $b\in B$ is an irreducible element, can I find a preimage of $b$ in $A$ which is also irreducible?
Background: we want to prove that if $K/A$ is an integral ring extension with $K$ a field, then for any irreducible polynomial $f\in K[X]$, should there exists a monic irreducible polynomial $g\in A[X]$ such that $f\mid g$ as polynomials in $K$. It turns out that our proof only works when $A$ is a domain: in this case let $x$ be a root of $f$ in $\overline K$, and let $g$ be the minimal polynomial of $x$ over $A$.
When $A$ is not a domain, let $B$ be the image of $A$ in $K$, then $B$ is a domain and the above proof works for $K/B$. So we can get a monic irreducible polynomial $g\in B[X]$. Now the remaining part is the question I asked here.
[EDIT] Combined with current two answers, we know that it's true for irreducible rings, but not for not connected rings. The in-between cases are rings which are connected, but not irreducible, for example $k[u,v]/(uv)$. Can we say anything about such rings?
Best Answer
A step towards an answer (?)
Claim: let $g\in B[X]$ be an irreducible monic polynomial of degree $d>0$ and assume that the nilradical of $A$ is a prime ideal. Then every monic preimage $G\in A[X]$ of degree $d$ is irreducible.
(1) Let $Q$ be the kernel of $\phi$ and let $N$ be the nilradical of $A$. Then $N\subseteq Q$ and one can factor the homomorphism $\phi$ into
$ A\rightarrow A/N\rightarrow B, $
where by assumption $A/N$ is a domain. The induced homomorphisms
$ A[X]\rightarrow A/N[X]\rightarrow B[X] $
are surjective.
Every monic preimage $h\in A/N[X]$ of degree $d$ of $g$ is irreducible, since $A/N$ is a domain. Hence it suffices to prove the claim in the case $\mathrm{kern}(\phi)=N$.
(2) Let $G=PQ$ be a factorization in $A[X]$ of a monic preimage of $g$ of degree $d$. Let $p,q\in B[X]$ be the images of $P$ and $Q$. Then $g=pq$ and since $g$ is irreducible one can assume $p\in B^\times$. Hence $P=a_0+P_{>0}$, $a_0\in A\setminus N$ and $P_{>0}\in N[X]$, where the latter consists of all non-constant monomials of $P$. Let $Q=Q_{>d}+Q_{\leq d}$ be the decomposition of $Q$ into polynomials consisting of monomials of degree $>d$ respectively $\leq d$ only. Then $Q_{>d}\in N[X]$, since $g$ has degree $d$.
Let $a_i$, $b_j$ and $c_k$ be the coefficents of $P$, $Q$ and $PQ$ respectively. Then
$ a_0b_d+a_1b_{d-1}+\ldots +a_db_0=c_d\not\in N, $
since $g$ has degree $d$. However $a_1,\ldots ,a_d\in N$ yields $a_1b_{d-1}+\ldots +a_db_0\in N$, hence $\phi(a_0)\phi(b_d)=1$ and therefore $a_0b_d=1-n$ for some $n\in N$. By assumption $n$ is nilpotent, which implies that $1-n$ and therefore $a_0$ are units of $A$. Consequently $P$ is a unit in $A[X]$ and the factorization $G=PQ$ is not proper.