Here our assumptions are that $\alpha$, $\beta$, and $\gamma$ are Gaussian Integers and ($\alpha$, $\beta$) = $\mathbb{G}$.

To prove this, I let $$\gamma = \alpha\delta$$ and $$\gamma = \beta\phi,$$ where $\delta, \phi\in\mathbb{G}$. This is from the definition of divisibility. I then considered $$\gamma\gamma = \alpha\delta\beta\phi = (\alpha\beta)\rho,$$ with $\rho = \delta\phi$. I.e., $\gamma\gamma|\alpha\beta$

I then showed that $\gamma|\gamma\gamma$, and via the transitive property $\gamma|\alpha\beta$.

However, my concern is that I did not use the assumption that ($\alpha, \beta$)=$\mathbb{G}$. I think that needs to be included, but I'm not sure how/where. Any help would be greatly appreciated!

  • $\begingroup$ First you write $\,(\alpha,\beta)=\Bbb G\,$ , then you write $\,(\alpha,\beta)\in\Bbb G\,$ , so which one is it? And what in the world is $\,\Bbb G\,$ , anyway?? $\endgroup$ – DonAntonio Sep 27 '12 at 3:20
  • $\begingroup$ To address what I think is wrong with your attempt, why should $\rho=\alpha\beta$? And $\gamma\gamma=(\alpha\beta)\rho$ implies $(\alpha\beta)$ divides $\gamma\gamma$, not the other way around. And lastly, in your title, you want to prove that $\alpha\beta$ divides $\gamma$, but the conclusion of your steps is the other way round? $\endgroup$ – alex.jordan Sep 27 '12 at 3:35
  • $\begingroup$ @DJM: Proof not right. If by $(\alpha,\beta)=\mathbb{G}$ you mean that $\alpha$ and $\beta$ generate $\mathbb{G}$, there is a very simple proof. For then there are objects $x$ and $y$ such that $x\alpha+y\beta=1$. Multiply by $\gamma$, and then it's all downhill. $\endgroup$ – André Nicolas Sep 27 '12 at 3:59
  • $\begingroup$ Sorry about the confusion! By $\mathbb{G}$ I mean the Gaussian Integers. $\endgroup$ – madisonfly Sep 27 '12 at 5:00
  • $\begingroup$ @alex.jordan my mistake! I meant that $\rho$ = $\delta\phi$. I did this in an attempt to make my point clearer. It would help if I actually wrote it right, eh? $\endgroup$ – madisonfly Sep 27 '12 at 5:05

Hint Note that $\rm\,\ a,b\:|\:c\:\Rightarrow\:ab\:|\:ca,cb\:\Rightarrow\:ab\:|\:(ca,cb) = c(a,b) = c\ $ by the GCD distributive law.

Or, Bezoutified $\rm\, \rm\,\ a,b\:|\:c\:\Rightarrow\:ab\:|\:ca,cb\:\Rightarrow\:ab\:|\:car\!+\!cbs = c(ar\!+\!bs) = c\ $ since, by Bezout's Identity, $\rm\:(a,b)= 1$ implies that there are $\rm\:r,s\in \Bbb G\:$ such that $\rm\:ar+bs = 1.$

The first proof works in any domain where GCDs exist, so in any UFD, e.g. Gaussian integers. The second proof also works in $\Bbb G$ since it is a Euclidean domain, i.e. it has a Euclidean algorithm for Division with Remainder, so, just as in $\Bbb Z$, the extended Euclidean algorithm yields the familiar Bezout linear representation of the gcd.

The first proof shows that if $\rm\,c\,$ is a common multiple of $\rm\,a,b\:$ then $\rm\:ab\:|\:c(a,b),\:$ or, equivalently, $\rm\:m = ab/(a,b)\:|\:c,\:$ i.e. $\rm\,m\,$ divides every common multiple of $\rm\,a,b.\:$ Further $\rm\:a,b\:|\:m\:$ since e.g. $\rm\:m/a = b/(a,b)\in\Bbb Z.\:$ Therefore, being a common multiple of $\rm\,a,b\,$ that divides every common multiple, $\rm\,m\,$ is the least common multiple of $\rm\,a,b.\:$ Hence we've proved the GCD * LCM law $\rm\:gcd(a,b)\, lcm(a,b) = ab.\:$ For a slicker proof of this basic law see here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.