Abstract Algebra Hello Experts: Proving Theorems About Ideals and Radicals

[DukeSteve]

Hello Experts,

I can't find the proof of this theorems please help me:

Given that there is a commutative ring R and 2 ideals I and J, also given that I is included in J

I need to prove
1) radical of I is in radical of J
2) radical of radical of ideal I = radical of ideal I.


Please give me a detailed proof, I want to understand the proof and your way.

 

[micromass]

Is this a homework problem? If so, then this belongs in the homework forums...

As for the problem: what did you try already to solve the problem?

 

[DukeSteve]

It's not a hw, I just want to understand why it's right. If you don't have the answer just don't answer, let others answer it.

 

[spamiam]

I'm sure micromass knows the answer to this: he's a very clever guy. It's just against PF rules to give help on homework without seeing the poster's attempt.

1) This seems straightforward: Given $x \in \sqrt{I}$ then $x^n \in I$ for some positive integer n. Since $I \subseteq J$ then $x^n \in J$ as well.

2)Suppose $x \in \sqrt{\sqrt{I}}$. Then there is some positive integer n s.t. $x^n \in \sqrt{I}$. By the definition of $\sqrt{I}$, there is some pos. int. m s.t. $(x^n)^m \in I$. Then $x^{nm} \in I$, so $x \in \sqrt{I}$.

The converse is trivial: given $x \in \sqrt{I}$ then $x^1 = x \in \sqrt{I}$, so $x \in \sqrt{\sqrt{I}}$.

 

[DukeSteve]

Thanks a lot for this great explanation. It's really not from HW.

Could you please do me a favor and explain why is this right:

If I have D ring with division but not a field and it's char !=2. If I know that F is centralizer of D (that means F is a field) and I have x in D such that x^2 is in F but x is not in F.

What does this mean that x is not in F? Why?

 

[spamiam]

I'm not sure I understand the question. I'm pretty sure you can have x in D with x² in F but still have $x \notin F$. Take the quaternions for example (http://en.wikipedia.org/wiki/Quaternion). They form a division ring but the element j isn't in F since ij = k but ji = -k. However j² = -1, which does commute with every element and thus is in F.

 

[mathwonk]

F = center(D).