Let $R$ be an integral domain. I am trying to find a $f \in R[x]$, such that $\deg(f) \geq 1$, and $f$ does not have any irreducible factors in $R[x]$.
Does such $f$ exist?
Though I haven't been able to find an example, I have some thoughts on the required properties of $R$:
For this to work, $R$ must not be ACCP:
$f$ in particular is not irreducible. If $f$ factors into 2 non-constant polynomials, then each must have the same property. Since the degree cannot descend infinitely, we may assume that $f$ only factors into $f(x) = c \cdot g(x)$. Then $c$ is non-zero, non-unit, and cannot have irreducible factors, so $R$ is not ACCP.
$R$ contains several elements $r_0, r_1, \dots, r_m$ with no maximal common divisors:
The $c$ in the above factorization is a common divisor of the coefficients of $f$. Take $r_i$ to be these coefficients, then these have no maximal common divisor, for otherwise, factoring out a maximal common divisor of the $r_i$'s would leave $g$ behind irreducible.
In particular, the set of common divisors is infinite and has no maximal elements. This is stronger than requiring that their GCD does not exist.
Background: I came across this problem here as a detour during attempts of weakening a condition in a homework problem, regarding a generalization of the Eisenstein criterion, which has the following formulation:
Let $R$ be a UFD. Suppose $f(x) = a_0 + a_1\cdot x + \dots + a_n \cdot x^n$, with a prime $\pi \in R$ and $k \in \{1, \dots, n\}$, where $\pi$ is divisor of $a_0,\dots,a_{k-1}$, while not of $a_k$, and $\pi^2$ is not a divisor of $a_0$, then $f$ has a irreducible factor of degree $\geq k$.
I have managed to show that any factorization of $f$ contains a factor of degree $\geq k$, and the UFD condition seems only to provide a factorization into irreducibles. So the criterion might generalize over all ACCP domains, or whenever $f$ is "primitive", in the sense that the coefficients have no common divisors other than units.
