• Oct 27th 2008, 03:56 PM
Pn0yS0ld13r
I've been stuck on this problem for a while now.

Suppose $\displaystyle 32 = \alpha\beta$ for $\displaystyle \alpha,\beta$ relatively prime quadratic integers in $\displaystyle \mathbb{Q}[i]$. Show that $\displaystyle \alpha=\epsilon\gamma^{2}$ for some unit $\displaystyle \epsilon$ and some quadratic integer $\displaystyle \gamma$ in $\displaystyle \mathbb{Q}[i]$.

Any pointers on how to start?

Thanks.
• Oct 27th 2008, 06:48 PM
NonCommAlg
Quote:

Originally Posted by Pn0yS0ld13r
I've been stuck on this problem for a while now.

Suppose $\displaystyle 32 = \alpha\beta$ for $\displaystyle \alpha,\beta$ relatively prime quadratic integers in $\displaystyle \mathbb{Q}[i]$. Show that $\displaystyle \alpha=\epsilon\gamma^{2}$ for some unit $\displaystyle \epsilon$ and some quadratic integer $\displaystyle \gamma$ in $\displaystyle \mathbb{Q}[i]$.

Any pointers on how to start?

Thanks.

$\displaystyle \alpha \beta=32=2^5=(i(1-i)^2)^5=i(1-i)^{10}.$ we know that $\displaystyle i$ is a unit and $\displaystyle 1-i$ is irreducible in $\displaystyle \mathbb{Z}[i],$ the ring of integers of $\displaystyle \mathbb{Q}[i].$ now use this fact that $\displaystyle \mathbb{Z}[i]$ is a UFD to finish the proof.
• Oct 27th 2008, 06:59 PM
ThePerfectHacker
I had a different proof (NonCommAlg beat me (Angry)) that did not involve finding factorizations.
Note that $\displaystyle N(32) = 2^{10}$. Therefore, if $\displaystyle \alpha$ (a non-unit) is decomposed into irreducibles it must mean that norm of each irreducible is a divisor of $\displaystyle 2^{10}$. Since $\displaystyle 2$ is not a prime and $\displaystyle N(2) = 4$ it means all irreducibles must have norm $\displaystyle 2$. Note that $\displaystyle 1+i$ is the only irreducible of norm $\displaystyle 2$ up to associates. This immediately forces $\displaystyle \beta$ to be a unit for otherwise it would mean $\displaystyle \alpha,\beta$ are not relatively prime. Thus, $\displaystyle \alpha = u(1+i)^{10}$ for some unit $\displaystyle u$.
• Oct 27th 2008, 11:11 PM
Pn0yS0ld13r
Thank you NonCommAlg and ThePerfectHacker! I got it now. :)