Wednesday, July 26, 2017

abstract algebra - Greatest common divisor in the Gaussian Integers



Let $a$ and $b$ be integers. Prove that their greatest common divisor in the ring of integers is the as their greatest common divisor in the ring of Gaussian Integers.




Ring of Gaussian Integers is:



$\mathbb{Z}[i]=\{a+ib: a,b\in \mathbb{Z}\} $



Attempt at proof:



Suppose that the GCD (greatest common divisor) of $a$ and $b$ is $d$. Then for any other common divisor of $a$ and $b$, say $e$, we must have that $e$ divides $d$. This is the definition of greatest common divisor.



Extend to Gaussian Integers. Suppose that $x+iy$ divides $a$ and $b$. That is:




$a=(x+iy)(n_0+in_1)$



$b=(x+iy)(m_0+im_1)$



Then I need to show that $x+iy$ divides $d$. From here I don't know where to go, I could write:
$a=Nd=(x+iy)(n_0+in_1)$ for some integer $N$.
But then I don't know that $N$ will divide $n_0+in_1$



Thanks for any help!



Answer



Remember that $d$ is a greatest common divisor of $a$ and $b$ if and only if:




  1. $d|a$ and $d|b$; and

  2. If $c|a$ and $c|b$, then $c|d$.



Let $a$ and $b$ be integers, and let $d$ be their greatest common divisor as integers. Then we know that $d$ satisfies the two properties above, where "divides" means "divides in $\mathbb{Z}$; and $c$ in point 2 is an arbitrary integer.




The thing that makes this very easy is that in the integers, we can express a gcd as a linear combination: we know that there exist integers $\alpha$ and $\beta$ such that $d=\alpha a + \beta b$.



Thus, if $\mathbf{x}$ is a Gaussian integer that divides $a$ and divides $b$ (in the Gaussian integers), then it also divides $\alpha a$, it also divides $\beta b$, and hence also divides $\alpha a + \beta b$. Thus, if $\mathbf{x}$ divides $a$ and divides $b$ in the Gaussian integers, then it divides $d$ in the Gaussian integers.



Thus, $d$ is a greatest common divisor for $a$ and $b$ in the Gaussian integers, since it satisfies the two requirements to be a greatest common divisor:




  1. $d|a$ and $d|b$ (in the Gaussian integers); and

  2. If $c$ is a Gaussian integer that divides both $a$ and $b$, then it divides $d$ (in the Gaussian integers).




Thus, $d$ is a greatest common divisor for $a$ and $b$ in $\mathbb{Z}[i]$.



No need to muck about with norms or with products.



P.S. Note that in the integers we can talk about "the" greatest common divisor because, even though every pair of numbers (except for $0$ and $0$) has two greatest common divisors ($d$ and $-d$), there is a natural way to "choose" one of them. In the Gaussian integer each pair of numbers (again, except $0$ and $0$) has four greatest common divisors (if $d$ is one, then so are $-d$, $id$, and $=id$); there is no universally accepted way of deciding which one would be "the" greatest common divisor, so generally you should talk about a greatest common divisor rather than the greatest common divisor.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...