In Frayleigh the definition of an UFD (unique factorization domain) is the following:

An integral domain $D$ is a unique factorization domain if the following conditions are satisfied:

  1. Every element of $D$ that is neither $0$ nor a unit can be factored into a product of a finite number of irreducibles.
  2. If $p_1\ldots p_r$ and $q_1\ldots q_s$ are two factorizations of the same element of $D$ into irreducibles, then $r = s$ and $q_j$ can be renumbered so that $p_i$ and $q_i$ are associates.

Now suppose that $ab = c^3$ in an $D$ which is an UFD and suppose that the only common factors of $a,b$ are units. Considering the facotorization of $a,b,c$, I deduce that each irreducible factor in $a$ must have multiplicity $3$ (since non of its factors is associated to the ones in $b$), however, this leaves me with $$a = up_1^3\ldots p_k^3$$ with $u$ some unit in $D$. Can we say something about this unit? (More specifically: can I say that this unit must be a third power of some other unit?


1 Answer 1


No: you can't. For example, suppose you are working in $\Bbb{Z}[i]$ (which is known to be a UFD), and consider two distinct irreducibles $p,q \in \Bbb{Z}[i]$. Then $$(ip^3)(-iq^3)=(pq)^3$$ And neither $i$ or $-i$ are cubes in this ring.

  • $\begingroup$ Thanks for your counterexample. I asked this question because I was working in the ring $\mathbb{Z} + \omega\mathbb{Z}$, where $\omega$ is a primitive third root of unity. I was able to show that $a + b\sqrt{-3}$ and $a - b\sqrt{-3}$ only have units as common divisor and I wanted to use this to solve the equation $a^2 + 3b^2 = c^3$ (a,b,c integers). Does the fact that $\omega$ is a third root of unity change anything? (I also know the units in this ring are $1, \omega, \omega^2$ up to sign) $\endgroup$
    – Student
    Commented May 18, 2017 at 9:10
  • 1
    $\begingroup$ That isn't a good example: $i$ and $-i$ are cubes in $\mathbf Z[i]$: $i = (-i)^3$ and $-i = i^3$. $\endgroup$
    – KCd
    Commented Nov 20, 2020 at 8:28

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