In a UFD, why does $ab | c^n$ implies $ab | c$?

Question

In a UFD, if I have two irreducible elemnts, $a$ and $b$ such that $ab | c^n$, why does this imply that $ab | c$?

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Here $a$ and $b$ are different and coprime.

Answer 1

The above is not true unless $a,b$ are coprime, else we may take the example $a=-b=c=2$ in the UFD $\mathbb Z$, where $ab =- 4 | 2^2 = 4$ but $ab =-4\nmid 2$.

HINTS :

• Every element in a UFD can be written as a product of irreducible elements. Use this to conclude that an element in a UFD is irreducible if and only if it is prime.

• By unique factorization, the factorization of $c^n$ can be determined from the factorization of $c$.

• If one of $a$ or $b$ don't appear in the factorization of $c$, then ...

Answer 2

In a UFD irred's are prime so $\,a,b\mid c^n\,\Rightarrow\, a,b\mid c,\,$ which implies $\,ab\mid c\,$ if $\,a,b\,$ are coprime $(\!\!\iff\! a\nmid b,\,$ by $\,a\,$ prime). Else it may fail, e.g. for any unit $u\!:\, $ $\, a(au)\mid a^2\,$ but not $\, a(au)\mid a.\,$

To prove it you can either directly use existence and uniqueness of irreducible factorizations, or else employ well-known consequences such as Euclids Lemma $\,(a,c)=1,\, a\mid cd\,\Rightarrow\, a\mid d$