AAR1

From Exampleproblems

In the ring $R=\mathbb{Z}\,$ , prove the following:

(a) $(m) \supset (n,m)$ if and only if $m\big| n\,$

(b) $(m,n) = (m)+(n)=(d)\,$ where $d=\gcd(m,n)\,$ . You need to show both equalities. Note that this implies that any ideal in the ring $\mathbb{Z}\,$ is principal.

Proof:

(a) $(\Rightarrow) (m)\supset(n) \Rightarrow kn \isin (m) \forall k \isin \mathbb{Z}\,$

Let $k=1\,$ . Then $n\isin(n)\,$ which means n is a multple of $m\,$ so $m\big| n\,$ .

$(\Leftarrow)\,$ If $m\big|\,$ then $n=km\,$ for some $k \isin \mathbb{Z}\,$ , so $n\isin (m)\,$ so $m\big| n\,$ .

(b) $(m,n)=\left\{am+bn\big| a,b\isin\mathbb{Z}\right\}\,$ $=\left\{am\big| a\isin \mathbb{Z} \right\}\,$ $+ \left\{bn\big| b\isin \mathbb{Z}\right\} = (m)+(n)\,$ .

Let $d=\gcd(m,n)\,$ . Then $(m,n)=\left\{am+bn\big| a,b\isin \mathbb{Z}\right\}\,$ $=\left\{(am'+bn')\big| a,b \isin\mathbb{Z}\right\}\,$ where $m'=m/d\,$ and $n'=n/d\,$ . Since $(m',n')=1\,$ , it is possible to write $am'+bn'=1\,$ for some $a,b\isin\mathbb{Z}\,$ and so any integer $c\,$ can be written as $c(am'+bn')=(a'm'+b'n')\,$ where $a'=ac\,$ and $b'=bc\,$ . This shows that $\left\{d(am'+bn')\big| a,b\isin\mathbb{Z}\right\} = \left\{dk\big| k\isin\mathbb{Z}\right\}=(d)\,$ .