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Ideals, elements r^n

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Homework Statement


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



Homework Equations


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


The Attempt at a Solution



Base Case: Assume r1[itex]\in[/itex]I. Then r[itex]\in[/itex]I.

Inductive Case: Assume that if rn[itex]\in[/itex]I then r[itex]\in[/itex]I for all n<n+1.
Assume rn+1[itex]\in[/itex]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[itex]\in[/itex]I then r[itex]\in[/itex]I by the inductive hypothesis. (QED)


Is this correct?

thank you for your time.
 

Answers and Replies

  • #2
Dick
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That would certainly be true. r^n=r(r^(n-1)). If r is in I, then r^n is in I.
 
  • #3
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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
Dick
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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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  • #5
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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
Dick
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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.
 
  • #7
Hurkyl
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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.
 

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