# Finding ideals of a ring

1. May 13, 2009

### curiousmuch

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. May 14, 2009

### VKint

Not quite. The correspondence theorem guarantees that there is a bijection between ideals of $$\mathbb{Z}/18\mathbb{Z}$$ and ideals of $$\mathbb{Z}$$ containing $$18 \mathbb{Z}$$, which are of the form $$n \mathbb{Z}$$, where $$n \mid 18$$, as you said. However, $$\mathbb{Z}/n\mathbb{Z}$$ isn't an ideal of $$\mathbb{Z}/18 \mathbb{Z}$$. (It will turn out that this is isomorphic to the quotient ring.)

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