1. ## Ring theory - radicals.

Hello, thank you very much for reading.

I have two homework assignments which I struggle with for quite some time.
1) Let R be a commutative, unitary (R has a unit) Ring, and let I be an Ideal to R.
I was given the definition of the radical of I:
Rad(I) = {x in R| there exists n in N such that x^n is in I}
I needed to prove that Rad(I) is an ideal in R, that Rad(Rad(I))=Rad(I), and that the set of all nilpotent elements N(R) (such elements x that have n in N such that x^n is 0) is an ideal.
So I've proved all of those. Rad(I) is an ideal by definition, and it took some time. Rad(Rad(I))=Rad(I) was easy. The third claim is also easy, noticing that this particular set N(R) is actually Rad({0}), which I've just proved is an ideal (since {0} is an ideal).
Now - the last question is: assuming that R is not commutative - which of the claims is correct?
However, I can't see if either the first or the third claims are correct here. First of all, if the first is still correct, the third would also be.
That's why I tend to think the first isn't correct, and the third is. However, I didn't succeed in proving the third (that N(R) is an ideal even though R isn't commutative), and I can't find a counter-example for the first claim (that Rad(I) is not an ideal).

So that's one question

2) (much shorter!)
I'm given a ring R, a sub-ring S of R, and an ideal I of R.
I proved that the intersection of I and S is an ideal in S.
Is every ideal in S an intersection of S with an ideal in R?
Cannot show either it's correct or false :-\
It reminds me of the definition of a closure of a set in a subset in topology, but unfortunately I make any analogies here...

THANK YOU VERY MUCH FOR READING!!!
Bless you all

Love,
Tomer.

2. Originally Posted by aurora
2) (much shorter!)
I'm given a ring R, a sub-ring S of R, and an ideal I of R.
I proved that the intersection of I and S is an ideal in S.
Is every ideal in S an intersection of S with an ideal in R?
Cannot show either it's correct or false :-\
It reminds me of the definition of a closure of a set in a subset in topology, but unfortunately I make any analogies here...
This is false because $S$ is its own ideal and what we are saying is that we can find $I$, ideal of $R$, so that $S = S\cap I$.
This forces $I = S$. But that means then $S$ must be an ideal of $R$. Not all subrings are ideals, like if $R$ are $2\times 2$ matrices over $\mathbb{R}$ and $S$ are the invertible matrices. Then $S$ is a subring but not an ideal.

EDIT: This is wrong.
Love,
I love you too, .

3. Hacker, thanks
At first I thought you solved it (so I posted something else), but now I realize you're probably wrong...
First of all, S = I (intersection) S doesn't force I to be S at all.... it could be that I=R. That way I is an ideal in R, and every sub-ring S, that is also it's own ideal, can be written in the form R (intersection) S. (and R is an ideal of itself).
Secondly, you're example of a sub-ring isn't right... it doesn't contain the zero element. for example, unit minus unit is zero, and that's not an element in the ring.
But thanks
So.... Still stuck on this one (unless I'm really wrong about saying you are wrong, in which case I would be glad to read an explenation...)
(Greedy but: no clue about quetion 1?)

Oh, and I love you for being such a great help all the time. What do you have to love me for, just giving you hard riddles with no pay...

4. Originally Posted by aurora
assuming that R is not commutative - which of the claims is correct?
one counter-example is enough to disprove both first and third claims: let $R=M_2(\mathbb{Z}).$ let $r=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \ a=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}.$ then $a^2 = 0$ but $(ra)^n = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \neq 0, \ \forall n \in \mathbb{N}. \ \Box$

2) Is every ideal in S an intersection of S with an ideal in R?
a simple counter-example: let $R=\mathbb{Q}, \ S=\mathbb{Z}, \ I=2\mathbb{Z}.$ now $I$ is an ideal of $S$ and since $R$ is a field, the only ideals of $R$ are $(0)$ and $R.$ we have $(0) \cap S = (0) \neq I, \ R \cap S = S \neq I. \ \Box$

5. NoncommAlg, I love you .