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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!

Martin Sleziak
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madisonfly
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  • 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?? – DonAntonio Sep 27 '12 at 03:20
  • 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? – 2'5 9'2 Sep 27 '12 at 03:35
  • @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. – André Nicolas Sep 27 '12 at 03:59
  • Sorry about the confusion! By $\mathbb{G}$ I mean the Gaussian Integers. – madisonfly Sep 27 '12 at 05:00
  • @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? – madisonfly Sep 27 '12 at 05:05
  • @DJM: Does $(\alpha,\beta)=\mathbb{G}$ mean, in your course, that $\mathbb{G}$ is generated by the set ${\alpha,\beta}$? If so, you should be able to write a proof using my suggestion. If needed, I can do it, if pinged. – André Nicolas Sep 27 '12 at 06:04

1 Answers1

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Hint $ $ note that $\rm\,\ a,b\:|\:c\:\Rightarrow\:ab\:|\:ca,cb\:\Rightarrow\:ab\:|\:(ca,cb)\! =\! c(a,b)\! =\! c\ $ by 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 duality-based proof of this basic law see here.

Bill Dubuque
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