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Suppose we have some $a \in \mathbb{Z}$ with prime factorization $$a = \prod_{i=1}^kp_i^{n_i}.$$ If we write the multiset for this prime factorization as $$X_a=\\{p_1, ..., p_1, p_2, ..., p_2, ..., p_k\\},$$ where each $p_i$ appears $n_i$ times, then a divisor of $a$ is any integer $b$ such that the multiset for the prime factorization of $b$ ,$X_b$, is a subset of the multiset for the prime factorization for $a$, $X_a$.

My question is: is there a name for a divisor $b$ of $a$ such that for each $p_i \in X_b$, the multiplicity of $p_i$ in $X_b$ is equal to the multiplicity of $p_i$ in $X_a$? I am also wondering if such a divisor is of importance over a generic UFD, or really anywhere in commutative algebra. Thanks!

categorically_stupid
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  • This holds iff $,b,$ is coprime to its cofactor $,b' = a/b,,$ so such divisors $,b,$ correspond to factorizations of $a$ into coprime factors $,a = bb',, \gcd(b,b')=1.,$ I don't recall any widely used name for such factors. Factorizations into (pairwise) coprimes share many of the properties of prime factorizations (e.g. uniqueness) and can prove very fruitful in number theory (e.g. see gcd-free basis here) – Bill Dubuque Oct 15 '22 at 00:18

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If $n=\prod p_i^{n_i}$ with $p_i$ distinct, then it is sometimes written in number theory that $p_i^{n_i}\vert\!\vert n$ and said that $p_i^{n_i}$ exactly divides $n$. So I guess you can say that such a number $b$ in your question is a product of numbers that exactly divide $a$.

YiFan Tey
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