X in ideal if x^n is in ideal?

[pivoxa15]

Homework Statement

In a ring with multiplicative identity, If x^n is in an ideal then is x also in the ideal? with n a natural number.

The Attempt at a Solution

I can't find a proof. Which is likely to mean the statement is false?

 

[radou]

Let x^n be in the ideal I of a ring R. It can be represented as x^n = x^(n-1) x. Now, either both x^(n-1) and x are in the ideal (in that case the statement if proved), or one of them is in the ideal. If x is in the ideal, the proof is finished. If x^(n-1) is in the ideal, then you can apply the same argument again.

I hope this works, since I'm a bit new to rings.

 

[Dick]

The statement that a*b in an ideal implies a and/or b in the ideal is only true for prime ideals. Not any ideal. The statement is false.

 

[Dick]

Eg. consider the ideal generated by 4 over the integers.

 

[radou]

[This is a sign I should stop learning from various lecture notes, and turn to books. :uhh:]

 

[matt grime]

No. It is a sign you should think about what you're reading.

 

[matt grime]

I can't find a proof. Which is likely to mean the statement is false?

Just think about an example for a second. Like the simplest ring there is, the integers. Why didn't you do some examples to see if it was true?

 

[pivoxa15]

Right, I should have done that. So the ideal generated by <x^2> in a ring certainly wouldn't have x in it. So the statement is false.

 

[matt grime]

It is perfectly possible for the ideal generated by x^2 to have x in it. It is just not *certain* to have x in it.

 

[pivoxa15]

It is perfectly possible for the ideal generated by x^2 to have x in it. It is just not *certain* to have x in it.

Good point. An example might be the ideal <2^2> in N which contains no element 2.