Relatively prime quadratic integers

**Pn0yS0ld13r** · October 27th 2008, 03:56 PM

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.

**NonCommAlg** · October 27th 2008, 06:48 PM

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.

**ThePerfectHacker** · October 27th 2008, 06:59 PM

I had a different proof (NonCommAlg beat me

) 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 $u$ .

**Pn0yS0ld13r** · October 27th 2008, 11:11 PM

Thank you NonCommAlg and ThePerfectHacker! I got it now.

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