If I?{0} , then m?I ?-m?I.
So I contains positive integer.
Let n be the small positive integer in I.
Since I is an ideal , so r.n , n.r ?I, ?r?Z.
Hence (n)?I …(1)
Now , any k?I can be written as k=qn+r where 0?r?n , q,r?Z.
Since k,qn? I ,so k-qn=r?I.
By def; of n we obtain r=0.
And so k=qn ?(n)
? I?(n) …(2)
Therefore I=(n) [from (1) and (2)].
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Def:-If a1.a2….an are common multiple of a1,…,an the element m is called the least common multiple of a1,…,an if:-
1)m is a common multiple of a1,…,an.
2)any other common multiple of a1,…,an is a multiple of m.
Theorem:- If {(Ii,+,.)} is the collection of ideals of the ring (R,+,.).Then (?Ii,+,.) is also an ideal of R .
Proof:-since ?i 0?Ii ,so 0??Ii ,hence ?Ii??.
Let a,b??Ii , r?R.
a,b ?Ii ,?I so a-b?Ii ?i.
?? a-b??Ii.
Let a??Ii , r?R
?a?Ii , ?i ,r?R.
So a.r ?Ii , ?i
? a.r,r.a ??Ii
?(?Ii,+,.) is an ideal of (R,+,.).
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