0

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

edit

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

roi_saumon
  • 4,196
  • It suffices to show that every irreducible element is prime in a UFD – D_S Jan 29 '19 at 14:48
  • 1
    you should probably add 'distinct' irreducible elements ($a\neq b$) – user120527 Jan 29 '19 at 14:52
  • So if I show that every irreducible is prime in a UFD, then since $ab | c^n$ then $a | c^n$ and $b|c^n$ so by induction (and the fact that $a$ and $c$ are prime) $a|c$ and $b|c$ but then... – roi_saumon Jan 29 '19 at 14:57

2 Answers2

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 ...

Sarvesh Ravichandran Iyer
  • 76,105
1

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 $\,(a,c)=1,\, a\mid cd\,\Rightarrow\, a\mid d$

Bill Dubuque
  • 272,048
  • How to show ab | c using existence and uniqueness of irreducible factorization? Do you decompose $a$ , $b$ and $c$ into irreducible factors? $a=u_a a_1...a_m, b=u_b b_1...b_n, c=u_c c_1...c_l$? – roi_saumon Jan 29 '19 at 18:39
  • @roi_saumon $,a,b,$ are already prime, being irred's in a UFD. Consider the complete (unique) prime factorizations of both sides of $,c^n = abd,$ The primes $a$,$b$ occur in the RHS factorization, so they mist also occur on the RHS (up to unit factors),. But the primes on the RHS are the same as those in $c$, again using uniqueness, since a prime factorization of $c^n$ arises from that of $c$ by duplicating each prime $n$ times. Uniqueness (and existence) is used many times in the proof so be sure to explicitly mention each use to be completely rigorous. – Bill Dubuque Jan 29 '19 at 19:00