Proving r is in ideal I if r^n is in I

[EV33]

Homework Statement

In the process of trying to prove something else I found it would be helpful if rⁿ$\in$I, where I is an ideal, n$\in$N, and r$\in$R and R is a ring, then r is in I.

Homework Equations

I is an ideal if a$\in$I and b$\in$I then a+b$\in$I, a$\in$I and r$\in$R then ar$\in$I, and I is not the empty set.

The Attempt at a Solution

Base Case: Assume r¹$\in$I. Then r$\in$I.

Inductive Case: Assume that if rⁿ$\in$I then r$\in$I for all n<n+1.
Assume rⁿ⁺¹$\in$I. Since rⁿ⁺¹= rⁿ*r then either rⁿ or r is in I. We only need to show the first case works since the second is trivial. If rⁿ$\in$I then r$\in$I by the inductive hypothesis. (QED)Is this correct?

thank you for your time.

 

[Dick]

That would certainly be true. r^n=r(r^(n-1)). If r is in I, then r^n is in I.

 

[EV33]

By what you wrote makes me think you were thinking of the converse of my statement. Were you just showing that it, in fact, is an iff statement?

 

[Dick]

Yes, I was thinking of the converse, sorry. Take the example of the ring of integers Z. 9Z is an ideal. Pick r=3 and n=2. What do you say now?

 

[EV33]

Ok. So I get 3²=9=0 mod 9. So 3² is an element of 9z. By definition 3 is an element of 9z. So I get that it holds for this case. That's what I was supposed to get right?

 

[Dick]

Ok. So I get 3²=9=0 mod 9. So 3² is an element of 9z. By definition 3 is an element of 9z. So I get that it holds for this case. That's what I was supposed to get right?

You were supposed to get that it is a counterexample. 9Z is the ideal of integers that are divisible by 9. 3 is NOT in 9Z. 3^2 is in 9Z. So?

I think what you are missing is that the definition of ideal does NOT say that if ab is in I, then a or b is in I. And it doesn't follow from the definition either.

 

[Hurkyl]

I found it would be helpful if rn∈I, where I is an ideal, n∈N, and r∈R and R is a ring, then r is in I.

This is the definition of a "radical ideal".A prime ideal is one where rs in I implies r in I or s in I. Your proof:

Since rⁿ⁺¹= rⁿ*r then either rⁿ or r is in I.

only works for prime ideals. It is true that every prime ideal is radical.