• Oct 27th 2008, 04:56 PM
Pn0yS0ld13r
I've been stuck on this problem for a while now.

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

Any pointers on how to start?

Thanks.
• Oct 27th 2008, 07:48 PM
NonCommAlg
Quote:

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

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

Any pointers on how to start?

Thanks.

$\alpha \beta=32=2^5=(i(1-i)^2)^5=i(1-i)^{10}.$ we know that $i$ is a unit and $1-i$ is irreducible in $\mathbb{Z}[i],$ the ring of integers of $\mathbb{Q}[i].$ now use this fact that $\mathbb{Z}[i]$ is a UFD to finish the proof.
• Oct 27th 2008, 07:59 PM
ThePerfectHacker
I had a different proof (NonCommAlg beat me (Angry)) that did not involve finding factorizations.
Note that $N(32) = 2^{10}$. Therefore, if $\alpha$ (a non-unit) is decomposed into irreducibles it must mean that norm of each irreducible is a divisor of $2^{10}$. Since $2$ is not a prime and $N(2) = 4$ it means all irreducibles must have norm $2$. Note that $1+i$ is the only irreducible of norm $2$ up to associates. This immediately forces $\beta$ to be a unit for otherwise it would mean $\alpha,\beta$ are not relatively prime. Thus, $\alpha = u(1+i)^{10}$ for some unit $u$.
• Oct 28th 2008, 12:11 AM
Pn0yS0ld13r
Thank you NonCommAlg and ThePerfectHacker! I got it now. :)