# Homework Help: Ideals, elements r^n

1. Sep 4, 2012

### EV33

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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?

2. Sep 4, 2012

### Dick

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

3. Sep 4, 2012

### 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?

4. Sep 4, 2012

### 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?

Last edited: Sep 5, 2012
5. Sep 5, 2012

### 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?

6. Sep 5, 2012

### Dick

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.

7. Sep 5, 2012

### Hurkyl

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