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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 $\alpha|\beta$ then $N(\alpha)|N(\beta)$, but the previous assertion is not necesseraly 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.

Bill Dubuque
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Larara
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  • 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 Jan 13 '15 at 01:48
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    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 Jan 13 '15 at 01:49
  • Another false implication: $\alpha$ is prime in $\mathbf Z[i] \Rightarrow {\rm N}(\alpha)$ is prime in $\mathbf Z$. (Counterexamples include 3 and 7, which are both prime in $\mathbf Z[i]$ but their norms are 9 and 49.) Taking norms gives up too much information to make such conclusions. – KCd Jan 13 '15 at 01:52
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    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 Jan 13 '15 at 01:54
  • @KCd, if $\gamma=\pi_1\cdot\ldots\pi_r$, where $\pi_i$ is a prime element, then $\alpha|\varepsilon\pi_1^n\cdot\ldots\cdot\pi_r^n$, and same for $\beta$. Then, from $\alpha$ and $\beta$ are relatively prime, the result follows? – Larara Jan 13 '15 at 02:03
  • You have to use the relative primality somehow. What you just wrote is too lacking in detail to make comments on, other than that it is not helpful to just write out a prime factorization as a string of primes: collect those that are unit multiples together with an extra unit factor. For instance, in $\mathbf Z[i]$ don't write $1+7i$ as $(1+i)(1+2i)(2-i)$, but note $1+2i = i(2-i)$ and therefore $1+7i$ is $(-1+i)(2-i)^2$. If you are sloppy about paying attention to different prime factors in $\gamma$ that are unit multiples there is no way to solve this problem. – KCd Jan 13 '15 at 02:13
  • @KCd, once $\mathbb{Z}[i]$ is a UFD, we can write (up to units) $\alpha=\pi_1^{n_1}\cdot\ldots\cdot\pi_s^{n_s}$ and $\beta=\pi_1^{\prime m_1}\cdot\ldots\cdot\pi_t^{\prime m_t}$. From $\alpha$ and $\beta$ relatively prime, we have $\pi_i\neq\pi_j^{\prime}$ for all $i$ and $j$. If $\gamma=\tilde{\pi}_1^{k_1}\cdot\ldots\cdot\tilde{\pi}_r^{k_r}$, from $\alpha\beta=\varepsilon\gamma^n=\varepsilon(\tilde{\pi}_1^{k_1}\cdot\ldots\cdot\tilde{\pi}_r^{k_r})^n$, we have that $s+t=r$, and we have, up to units $\alpha=\tilde{\pi}_1^{k_1n}\cdot\ldots\cdot\tilde{\pi}_s^{k_sn}$ and analougosly for $\beta$... – Larara Jan 13 '15 at 02:19
  • ... (from $s+1$ to $r$). The units $\varepsilon^{\prime}$ and $\varepsilon^{\prime\prime}$ come out from the fact that the $\pi_i$'s and $\pi_j^{\prime}$'s are associate (i.e. unit multiples) with the $\tilde{\pi}_l$'s. – Larara Jan 13 '15 at 02:25
  • Don't write equations up to missing units: include the units explicitly. I also would not write exponents as $n_1,\dots,n_s$ when the right side has an exponent called $n$. Let the exponents in the factorizations of $\alpha$ and $\beta$ be $e_i$ and $f_j$, for instance. – KCd Jan 13 '15 at 02:32
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    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 Jan 13 '15 at 02:33
  • You can't prove that the first prime $\widetilde{\pi}_1$ in your factorization of $\gamma$ is showing up in $\alpha$: there is no definite ordering imposed on the factors of $\gamma$. – KCd Jan 13 '15 at 02:34
  • Anyway, I think you basically get the idea up to not being as careful as you should with some bookkeeping issues related to where prime factors are appearing. – KCd Jan 13 '15 at 02:35
  • @KCd, $\tilde{\pi}_1|\pi_i$ for some i or $\tilde{\pi}_1|\pi_j^{\prime}$ for some j. Then $\tilde{\pi}_1$ is a unit multiple of $\pi_i$ or $\pi_j^{\prime}$ and I think I'm able to finish. Thank you very much for your help and attention! – Larara Jan 13 '15 at 02:51
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    @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. – orangeskid Jan 13 '15 at 03:34
  • @orangeskid, great suggestion! +1 – Larara Jan 13 '15 at 04:03

2 Answers2

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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.

Bill Dubuque
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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 $$\beta=\epsilon''\prod_k p_k^{na_k}=\epsilon''\left(\prod_k p_k^{a_k}\right)^n=\epsilon''\zeta^n$$

Maths fan123
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