What is the Proof that Every Ideal of Zn is Principal?

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SUMMARY

Every ideal of the ring Z_n is principal, as demonstrated through a proof involving the least positive integer in the ideal. By assuming an integer d' exists in the ideal that is not of the form md, the division algorithm is applied to express d' as d' = qd + r, where 0 < r < d. This leads to a contradiction by showing that r must also belong to the ideal, confirming that all elements in the ideal can be expressed as multiples of the least positive integer.

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Homework Statement


1.Prove that every ideal of Zn is principal

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The Attempt at a Solution



In 1-I've proved that if K is an ideal of Zn that contains an element k then it contains all the elements of the form mk (m in Zn)...But how can I prove that there are no elements that are not of the form mk?


Thanks in advance...
 
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Let d be the least positive integer in your ideal K. You know all integers of the form md where m is an integer is in your ideal. To prove these are the only ones assume there is some integer d' in K that is not of the form md. Then we use the division algorithm to write it on the form:
d' = qd + r
for some integer q and 0<r<d. Now can you show that r is in K? If you can you will have a contradiction since r is positive and less than d.
(this general approach works for Euclidean domains in general, and shows that Euclidean domains are principal ideal domains)
 
Well... if d' is in our ideal K then d'-qd is also in our ideal (since an ideal is also a sub-ring)... But d'-qd=qd+r-qd=r... Hence r must also be in our ideal and then we get a contradiction :)

THanks a lot!
 

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