What are the ideals of Z mod 18Z?

  • Thread starter Thread starter curiousmuch
  • Start date Start date
  • Tags Tags
    Ring
curiousmuch
Messages
7
Reaction score
0

Homework Statement


Find all ideals I of Z mod 18Z. Then find what (Z mod 18Z)/I is isomorphic to for every ideal I.

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?
 
Physics news on Phys.org
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.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top