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Finding ideals of a ring

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all ideals I of Z mod 18Z. Then find what (Z mod 18Z)/I is isomorphic to for every ideal I.

    3. The attempt at a solution
    We know that the whole ring and {0} are ideals. since Z/18Z is not a field there are more. So are Z/nZ where n is a divisor of 18, all of them?
     
  2. jcsd
  3. May 14, 2009 #2
    Not quite. The correspondence theorem guarantees that there is a bijection between ideals of [tex] \mathbb{Z}/18\mathbb{Z} [/tex] and ideals of [tex] \mathbb{Z} [/tex] containing [tex] 18 \mathbb{Z} [/tex], which are of the form [tex] n \mathbb{Z} [/tex], where [tex] n \mid 18 [/tex], as you said. However, [tex] \mathbb{Z}/n\mathbb{Z} [/tex] isn't an ideal of [tex] \mathbb{Z}/18 \mathbb{Z} [/tex]. (It will turn out that this is isomorphic to the quotient ring.)
     
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