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This is the second problem in Neukirch's Algebraic Number Theory. I did the proof but it feels a bit too slick and I feel I may be missing some subtlety, can someone check it over real quick?

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 $\alpha =\varepsilon '\xi ^n$ and $\beta =\varepsilon ''\eta ^n$, with $\varepsilon '$,$\varepsilon ''$ units.

So basically, because the Gaussian integers are a unique factorization domain and alpha and beta are relatively prime, I have the prime decomposition:

$\alpha = \varepsilon' p_1^{e_1}...p_r^{e_r}$

$\beta = \varepsilon'' p_s^{e_s}...p_y^{e_y}$

$\varepsilon\gamma^n = \varepsilon q_1^{nf_1}...q_k^{nf_k}$

And so $\alpha\beta = p_1^{e_1}...p_r^{e_r}p_s^{e_s}...p_y^{e_y}$. Where we have a one-to-one correspondence between the $p_i^{e_i}$ and the $q_i^{nf_i}$ and thus setting $p_i^{e_i} = q_i^{f_i}$, in accordance with this correspondence, we obtain our desired xi and eta.

Does this make sense? I never used anything specific to the Gaussian integers so if this is right then it holds for all UFDs. Thanks.

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oops! I'll fix that, thanks for checking this over =]. –  Ron Jeremy Nov 10 '12 at 22:14
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3 Answers 3

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The proof looks OK, but you might want to make the role of the fact that $\alpha$ and $\beta$ are relatively prime more apparent. You probably mean the right thing, but the current wording sounds a bit like you're using that to get the prime decompositions, not for concluding that $p_1$ through $p_r$ and $p_s$ through $p_y$ are disjoint sets of primes; you never mention that crucial fact explicitly.

Also I'd continue the numbering with $p_{r+1}$; the additional index $s$ appears somewhat unmotivated and adds to the unclarity whether there might be an overlap between the two sets of primes.

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you're right $p_{r+1}$ is a more suggestive notation. –  Ron Jeremy Nov 11 '12 at 18:46
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The trouble with Gaussian integers is that unique factorization is only unique up to units. For example consider $-i \cdot (1+i)(2+i)(1+2i)^4 = -1 \cdot (1+i)(-1+2i)(2-i)^4$: The primes $2+i$ and $-1+2i$ are just associates rather than equal. Prime ideals factor out associates and free us from caring about choosing associates correctly: The ideal $(2+i)$ is equal to $(-1+2i)$ and so the factorization into ideals $(1+i)(2+i)(1+2i)^4$ is unique (even if we have a few different ways to write it).

With that in mind take the prime ideal factorization of $\gamma$ so consider $$(\alpha)(\beta) = (\pi_1)^{n r_1} (\pi_2)^{n r_2} \cdots (\pi_k)^{n r_k}$$ so in particular $(\pi_1)|(\alpha)(\beta)$ and by the characterization of a prime ideal this implies that $(\pi_1)|(\alpha)$ or $(\pi_1)|(\beta)$, suppose $(\pi_1)|(\alpha)$ then by coprimality $(\pi_1)\not|(\beta)$ so by unique factorization $(\pi_1)^{n r_1}|(\alpha)$. Just doing the same thing for each prime factor of $(\gamma)$ gives the result that both $(\alpha)$ and $(\beta)$ are products of $n$th powers of prime ideals.

Travelling back to the Gaussian integers themselves, this result requires us to bring units back in so we conclude that $\alpha = \varepsilon' \xi^n$, $\beta = \varepsilon'' \eta^n$.

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That's an interesting point. I did find dealing with the units to be the most tedious part of this proof. So in algebraic number theory is that the general procedure to stop factoring things into primes and instead into prime ideals? –  Ron Jeremy Nov 11 '12 at 18:52
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Yes, it seems to hold in every UFD.

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ah my suspicions confirmed! –  Ron Jeremy Nov 11 '12 at 18:52
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