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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.

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