Hello! Here I share an example where I am able to find the catenary degree of .
Suppose that the minimal pair of is , where and is not monic. It is not hard to see that is atomic and . Let , and ,. If for some then there exists given by
and . Also . Following this procedure we can create a finite sequence such that and satisfies that for all .
Now, assume that , such that , and for all . Then
where . It is clear that for all , but the coefficient of the monomial of least degree in the right side of (1) is a multiple of . This is a contradiction. Then there exists a unique factorization such that every atom appears in less than times. Also note that is the factorization of minimal length of . We have proved that for every and there exists a n-chain of factorizations that connects and the unique factorization of minimal length of . It follows that . Following the same reasoning as before, we can prove that there is no such that , so . Then .