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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.