# Ideals, elements r^n

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?

Homework Helper
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?

Homework Helper
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?

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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?

Homework Helper
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.

Staff Emeritus