PID Math Problem: Every Element Generating Ideal?

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Homework Statement


If R is a PID then all its ideals can be generated by a single element. It dosen't imply that every element in R can generate an ideal does it?




The Attempt at a Solution


I have to agree that it dosen't but can't think of a proof.
 
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Of course in any ring, any element can generate an ideal, the ideal generated by that element.
 
You are right. Each element can always generate an ideal, namely a principle ideal. If R is a PID than it tells us that all the ideals in R are those that have been generated by a single element hence are all principle ideals.

However one can turn any principle ideal into a nonprinciple ideal. i.e. in Z. <3> is a Principle ideal. Instead of denoting it by <3> we denote this ideal by {0, 3} or {3, 6} and either of those two will be equivalent to <3>.
 
It's still the same ideal, and is still a principal ideal, which is just an ideal that can be generated by a single element, regardless of whichever generating set you choose to focus on.
 
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...
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