Q:

If the homomorphic image of a polynomial is irreducible, does it imply polynomial is irreducible

If the homomorphic image of a polynomial is irreducible, then is the original polynomial irreducible?
let $\phi: R[x]\rightarrow S[x]$ be a ring homomorphism. Let $f(x)\in R[x]$ be a polynomial and $g(x)\in S[x]$ be the polynomial obtained by applying $\phi$ to $f(x)$.
If $g(x)$ is irreducible, does it imply $f(x)$ is irreducible in $R[x]$

A:

Not necessarily: consider the quotient projection $\pi:\Bbb Z[x]\to \Bbb F_2[x]$. Then $\pi((2x+1)(x^2+x+1))=x^2+x+1$
An example of a UFD $R$ and a map $f:R[x]\to R[x]$ that maps a reducible element to an irreducible one is $R=\Bbb Z[T_n\,:\, n\in\Bbb N]$ and $f(p(T;x))=p(T_0,T_0,T_1,T_2,\cdots; x)$. In fact, $f((T_0-T_1+1)(x^2+1))=x^2+1$.
However, if in addition to $R$ being a domain we require that $f:R[x]\to R[x]$ is a $R$-algebra homomorphism and $f(x)\notin R$, indeed we'll have that preimage of irreducible polynomials will be irreducible. This is because $R[x]^*=R^*$ and $f(p)=p\circ [f(x)]$, with $\deg f(p)=\deg p\deg f(x)\ge \deg p$. Therefore, if $p=qr$, then $f(p)=(q\circ [f(x)])\cdot (r\circ [f(x)])$. The only way one of these factors could be in $R^*$ is that one of $r$ or $q$ has degree $0$ in the first place (and thus that it is equal to its image by $f$).
The previous observation applies to ring-homomorphisms $f:\Bbb Z[x]\to \Bbb Z[x]$, since they must be $\Bbb Z$-algebra homomorphisms. Of course, if you allow $f(x)\in\Bbb Z$, there are some reducible polynomials that take prime values in a couple of integers.

