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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\in (m)\forall k\in {\mathbb  {Z}}\,

Let k=1\,. Then n\in (m)\, which means n is a multiple of m\, so m{\big |}n\,.

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


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

Let d=\gcd(m,n)\,. Then (m,n)=\left\{am+bn{\big |}a,b\in {\mathbb  {Z}}\right\}\, =\left\{(am'+bn'){\big |}a,b\in {\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\in {\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\in {\mathbb  {Z}}\right\}=\left\{dk{\big |}k\in {\mathbb  {Z}}\right\}=(d)\,.

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