TITLE: $R$ with an upper bound for degrees of irreducibles in $R[x]$
QUESTION [10 upvotes]: One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as 

If $f\in\mathbb{R}[x]$ has degree larger than $2$, then $f$ is reducible.

However, the proof as I know depends highly on the fact that $\mathbb{C}$ is algebraically closed, and the very nice property: $z\in\mathbb{C}$ is in $\mathbb{R}$ if and only if $z=\bar{z}$.
This makes a generalization to other integral domains rather difficult. So the problem is 

For what kind of integral domain $R$, we have a finite upper bound on the degree of irreducible elements in $R[x]$?

Some most familiar examples are ruled out: in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{F}_{p}[x]$, there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge.
Another result might help is Eisenstein's criterion and its generalized form, which says if we can find a prime ideal $\mathfrak{p}$ in $R$ such that $\mathfrak{p}^2\neq\mathfrak{p}$, then by picking $a\in\mathfrak{p}^2\backslash\mathfrak{p}$ we have an irreducible $a+x^d$, where $d$ can be arbitrary, and hence the upper bound is not possible.
So we only need to focus on domains where $\mathfrak{p}^2=\mathfrak{p}$ for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it.
Can someone give a hint?
Thanks!

REPLY [6 votes]: Case 1: $R$ is a field
I will refer to these notes by Keith Conrad on the Artin-Schreier theorem. 
We will proceed by condition on whether or not $R$ is a perfect field. If $R$ is an imperfect field then there always exist irreducible polynomials of arbitrary large degree. This follows from Lemma 2.1 in Keith's notes which tell us that if $\alpha \in F \setminus F^{p^l}$, which is non-empty because $R$ is imperfect, then $p(t)=t^{p^l}-\alpha$ is irreducible. 
If $R$ is perfect then we can deduce it has finite index in its algebraic closure. Let $A$ be the algebraic closure of $R$ then if $[A:R]=\infty$ we can construct arbitrarily large extensions by adjoining elements of a basis of $A$ to $R$. Then using the primitive element theorem we can find elements of arbitrarily large degree. So if every polynomial of degree larger than $n$ is reducible then the index of $R$ in $A$ is finite.
Now if $R=A$ we are done otherwise
the Artin-Schreier theorem tells us that if $1<[A:F] < \infty$ we have that $[A:F]=2$, $F$ is formally real closed and $A=F(i)$. 
Case 2: R is not a field
Note that $R$ cannot be a UFD. A UFD is a field if and only if it has no prime elements. So if $R$ was a UFD that wasn't a field we could find some prime $p \in R$ and by the OP's observations we would have to have $(p)^2=(p)$. Any finitely generated idempotent ideal is generated by an idempotent, the only idempotents of an integral domain are $0$ and $1$. So $(p)=(0),R$ which is impossible, thereby $R$ cannot be a UFD if its not a field. This observation is probably not that important in the end.
Let $R$ be an integral domain such that every polynomial over $R[x]$ of degree greater than $n$ is reducible. If $K$ is the fraction field of $R$ then $K[x]$ necessarily satisfies the same condition, in particular $K$ is algebraically closed or satisfies the conclusion of Artin-Schreier. One thing to notice is that if $K$ is algebraically closed then every polynomial splits over $R$ since any $a/b \in K$ satisfies $bx-a\in R[x]$. If $K$ is not algebraically closed then every polynomial splits over $R[i]$ for much the same reason.  This MO question gives some details on how to construct these sorts of rings with valuation theory.