# Coprime cofactors of n'th powers are n'th powers, up to associates, for Gaussian integers & UFD

I'm trying to do the following exercise from Neukirch's Algebraic Number Theory (exercise 2, $$\S 1$$, chapter 1):

Show that, in the ring $$\mathbb{Z}[i]$$, the relation $$\alpha\beta=\varepsilon\gamma^n$$, for $$\alpha,\beta$$ relatively prime numbers and $$\varepsilon$$ a unit, implies $$\alpha=\varepsilon^{\prime}\xi^n$$ and $$\beta=\varepsilon^{\prime\prime}\eta^n$$, with $$\varepsilon^{\prime},\varepsilon^{\prime\prime}$$ units.

My attempt: once $$\alpha$$ and $$\beta$$ are relatively prime, $$\alpha\nmid\beta$$ and $$\beta\nmid\alpha$$. Therefore $$N(\alpha)$$ and $$N(\beta)$$ are relatively prime too (where $$N(a+ib)=a^2+b^2$$). Therefore, $$N(\alpha\beta)=N(\alpha)N(\beta)=N(\varepsilon)N(\gamma^n)=N(\gamma)^n=(p_1\cdot\ldots\cdot p_r)^n=p_1^n\cdot\ldots\cdot p_r^n$$

And once $$N(\alpha)$$ and $$N(\beta)$$ are relatively prime, we have (if necessary, we reorder the $$p_i$$'s):

$$N(\alpha)=p_1^n\cdot\ldots\cdot p_k^n=(p_1\cdot\ldots\cdot p_k)^n=N(\xi)^n=N(\xi^n)$$ and $$N(\beta)=p_{k+1}^n\cdot\ldots\cdot p_r^n=(p_{k+1}\cdot\ldots\cdot p_r)^n=N(\eta)^n=N(\eta^n)$$

And the result follows.

But I'm not sure about one step. It is $$\alpha$$ and $$\beta$$ relatively prime $$\Rightarrow$$ $$N(\alpha)$$ and $$N(\beta)$$ relatively prime. I know that if $$\alpha\mid\beta$$ then $$N(\alpha)\mid N(\beta)$$, but the previous assertion is not necessarily true, and if it is, and don't know how to prove it. If it is true, the result follows, unless I made something wrong in the rest of the proof.

That's it. If you know another way of doing the exercise, please show me.

• Being relatively prime does not imply neither $\alpha$ nor $\beta$ divides the other: maybe $\alpha$ or $\beta$ is a unit too! And it is false that relatively prime elements have relatively prime norm. Try $1+2i$ and $1-2i$. Their norms are both 5 but they are relatively prime. So the whole strategy is just wrong.
– KCd
Commented Jan 13, 2015 at 1:48
• This exercise really is not so much about $\mathbf Z[i]$ specifically, but is a property of any ring with unique factorization (like $\mathbf Z$, $\mathbf Z[i]$, $\mathbf R[x]$, and $\mathbf R[x,y]$): when relatively prime elements in a unique factorization domain have a product that is an $n$th power times a unit, then both factors are $n$th powers times units.
– KCd
Commented Jan 13, 2015 at 1:49
• You should focus attention on the multiplicity of each prime in the numbers $\alpha$, $\beta$, and $\gamma$ because a nonzero element of $\mathbf Z[i]$ is determined up to unit multiple by the multiplicity of each prime in its prime factorization. That is, two nonzero elements that are divisible equally often by every prime are unit multiples, and conversely.
– KCd
Commented Jan 13, 2015 at 1:54
• From $\alpha$ and $\beta$ relatively prime you have more than $\pi_i \not= \pi_j'$: $\pi_i$ and $\pi_j'$ are not unit multiples of each other.
– KCd
Commented Jan 13, 2015 at 2:33
• @Larara: Try to solve the similar statement for the ring $\mathbb{Z}$. Once you understand that, the proof for $\mathbb{Z}[i]$ will be a piece of cake. Commented Jan 13, 2015 at 3:34

The proof for naturals (or integers) via prime factorization immediately generalizes to any UFD, but we need to account for unit (invertible) factors, so we work up to associates (unit multiples). [ENT readers:  in $$R = \Bbb Z,\,$$ $$\,u\,$$ is unit $$\!\iff\! u=\pm1,\,$$ so $$\,m,n\,$$ are associate $$\!\iff\! m = \pm n$$]
Theorem $$\$$If $$R\,$$ is a UFD and coprime $$\,a,b\in R\,$$ satisfy $$\,ab=c^n$$ for some $$\,0\ne c\in R,\ n \ge 1,\,$$ then $$\,a=u\,r^n$$ and $$\,b=u^{-1}s^n$$ for some $$\,r,s\in R\,$$ and for some unit (i.e. invertible) $$u\in R.\,$$ Therefore both factors $$\,a\,$$ and $$\,b\,$$ are ― like $$\,c^n$$ ― associates of $$\,n$$'th powers in $$R$$.
Proof $$\$$ We induct on $$\,k =\,$$ number of prime factors of $$\,c.\,$$ If $$\,k=0\,$$ then $$\,c\,$$ is a unit, so $$\,a,b\,$$ are units, so $$\,a = a\cdot 1^n,\ b = a^{-1}c^n$$ works. Else $$\,k\ge 1,\,$$ so a prime $$\,p\mid c,\,$$ so $$\,p^n\mid c^n\! = ab\,$$ hence $$\,p^n\mid a\ {\rm or}\ b\,$$ by $$\,a,b$$ coprime, $$R\,$$ UFD. Wlog $$\,p^n\!\mid b\,$$ so canceling $$\,p^n$$ we obtain $$\,a(b/p^n) = (c/p)^n.\,$$ $$\,c/p\,$$ has fewer prime factors than $$c\,$$ so induction $$\Rightarrow a = ur^n,\ b/p^n\! = u^{-1} s^n,\,$$ so $$\,b = u^{-1}(ps)^n$$.
Remark $$\$$ For generalizations, see here for a proof using gcds (or ideals), and see here for Weil's remarks on the relationship with Fermat's method of infinite descent.
Assume for now that $$\gamma$$ is a non zero, non unit Gaussian integer ($$\gamma=0$$ is trivial and I would be repeating the comment above for when $$\gamma$$ is a unit). Since $$\mathbb{Z}[i]$$ is a UFD, $$\epsilon\gamma=\epsilon\prod_{i}p_i^{a_i}$$. Therefore $$\alpha\beta=\epsilon^n\prod_{i}p_i^{na_i}$$. Now we reorder the primes (they share no common factors up to units so we are safe to do this since none of the primes in the products clash. I hope this makes sense, but basically we can do this because they are coprime) and use the fact that any unit$$^n$$ is still a unit in the Gaussian integers to obtain the following unique factorisations: $$\alpha=\epsilon'\prod_j p_j^{na_j}=\epsilon'\left(\prod_j p_j^{a_j}\right)^n=\epsilon'\xi^n$$ and $$\beta=\epsilon''\prod_k p_k^{na_k}=\epsilon''\left(\prod_k p_k^{a_k}\right)^n=\epsilon''\zeta^n$$