# For which rings $R$ can the number of irreducible factors of degree at least one of $f\in R[X]$ be bounded in terms of $\deg f$?

Question: For which integral domains $$R$$ does there exist a function $$M:\mathbb{Z}_{\ge 0}\to\mathbb{Z}_{\ge 0}$$, such that for all $$f\in R[X]$$, we have that $$f$$ has at most $$M(\deg f)$$ irreducible divisors of degree at least one?

If two irreducible divisors are unit multiples of each other, we regard them as the same divisor

Own work: The most general example I've found so far, is that when $$R$$ is a GCD domain, we may take $$M(n)=2^n-1.$$

Proof for GCD domains: Let $$R$$ be a GCD domain and $$f\in R[X]$$. Let $$g_1,\ldots, g_m$$ be irreducible divisors of $$f$$ in $$R[X]$$ of degree at least one and with $$(g_1),\ldots, ,(g_m)$$ all pairwise distinct.

Let $$K$$ be the quotient field of $$R$$ and let $$f=p_1\cdot \ldots\cdot p_l$$ be the prime factorization in $$K[X]$$. It is clear that, in $$K[X]$$, for all $$1\le i\le l$$ there exists a non-empty $$A_i\subset\{1,\ldots,l\}$$ such that $$(g_i)=\left(\prod_{j\in A_i}p_j\right).$$ Assume that $$m>2^l-1$$. Since $$\{1,\ldots,l\}$$ has $$2^l-1$$ non-empty subsets, the pigeonhole principle gives that there exist $$i\neq j$$ with $$(g_i)=(g_j)$$ in $$K[X]$$. This means that there exist constants $$a,b\in R$$ with $$ag_i=bg_j$$ in $$R$$.

Since $$R$$ is a GCD domain we may assume that $$a$$ an $$b$$ are coprime and subsequently that $$a\mid g_j$$. Because $$g_j$$ is irreducible and has degree at least one, $$a$$ must now be a unit (because $$g_j/a$$ definitely isn't). It follows that $$b$$ is also a unit and $$(g_i)=(g_j)$$; a contradiction.

We conclude that $$m\le 2^l-1\le 2^{\deg f}-1$$.

Edit: Let $$R$$ be a GCD domain, $$f\in R[X]$$ non-zero. Then there exists some non-zero $$a\in R$$ and $$g_1,\ldots, g_m\in R[X]$$ prime and at of degree at least one, with $$f=a\prod_{i=1}^mg_i,$$ which gives the sharper bound $$M(n)=n$$ for GCD domains.