1. ## Ideal Test

Let $R$ be a commutative ring with identity, and let $I$ be any ideal of $R$. Define the radical RADICAL of I be the set $N(I)=\{ r \in R: r^n \in I \}$ for some positive integer $n$.

Prove that $N(I)$ is an ideal of $R$.

My Attempt:
I think I should use the "ideal test"; that is to show that (i) $a-b \in N(I)$ whenever $a,b \in N(I)$, and (ii) show that $ar$ and $ra$ are in $N(I)$ whenever $a \in N(I)$ and $r \in R$.

(i) Let $a,b \in N(I)$, then that means $a,b \in R$ such that $a^n, a^m \in I$. Since $I$ is an ideal of $R$, $a^n-b^m \in I$, and I think this implies that $a-b \in N(I)$.

But first I think we must show that $a-b \in R$. How do we show this?

(ii) Since I is an ideal of R we know that for any $r \in R$ and $a \in N(I)$, $ar$ and $ra$ are both in $N(I)$. Is this correct?

2. Originally Posted by demode
Let $R$ be a commutative ring with identity, and let $I$ be any ideal of $R$. Define the radical RADICAL of I be the set $N(I)=\{ r \in R: r^n \in I \}$ for some positive integer $n$.

Prove that $N(I)$ is an ideal of $R$.

My Attempt:
I think I should use the "ideal test"; that is to show that (i) $a-b \in N(I)$ whenever $a,b \in N(I)$, and (ii) show that $ar$ and $ra$ are in $N(I)$ whenever $a \in N(I)$ and $r \in R$.

(i) Let $a,b \in N(I)$, then that means $a,b \in R$ such that $a^n, a^m \in I$. Since $I$ is an ideal of $R$, $a^n-b^m \in I$, and I think this implies that $a-b \in N(I)$.

But first I think we must show that $a-b \in R$. How do we show this?

(ii) Since I is an ideal of R we know that for any $r \in R$ and $a \in N(I)$, $ar$ and $ra$ are both in $N(I)$. Is this correct?
In both parts you need commutativity. So,

For (i), what you need to show is that if $a^i, b^j \in I$ then there exists $n$ such that $(a-b)^n \in I$. Can you see why this is what you want to do? Now, you will need to use the binomial theorem for this, and basically you want to pick $n$ to be sufficiently large. Can you think of an $n$ that would work, and can you prove that it works? (As a small help, remember that if $a^i \in I$ then $a^{i+k} \in I$ for all $k>0$ (assuming $i>0$)).

For (ii), you need to prove that if $r^n \in I$ then $(ar)^n \in I$ for all $a \in R$. This actually falls out pretty easily, if you remember that your ring is commutative, and that $r^n in I$...

3. Originally Posted by Swlabr
In both parts you need commutativity. So,

For (i), what you need to show is that if $a^i, b^j \in I$ then there exists $n$ such that $(a-b)^n \in I$. Can you see why this is what you want to do? Now, you will need to use the binomial theorem for this, and basically you want to pick $n$ to be sufficiently large. Can you think of an $n$ that would work, and can you prove that it works? (As a small help, remember that if $a^i \in I$ then $a^{i+k} \in I$ for all $k>0$ (assuming $i>0$)).
Frankly, I can't follow because I don't know much about binomial theorem or combinatorics and my knowledge in this area is very poor. I don't think this is how they expect us to solve this problem...

But... maybe pick j+k=n?

$(a-b)^n = xa^jb^k$

$(a-b)^{n+1} = xa^jb^k(a-b)$

For (ii), you need to prove that if $r^n \in I$ then $(ar)^n \in I$ for all $a \in R$. This actually falls out pretty easily, if you remember that your ring is commutative, and that $r^n in I$...
I see. But doesn't the method in my previous post work?

4. Originally Posted by demode
Frankly, I can't follow because I don't know much about binomial theorem or combinatorics and my knowledge in this area is very poor. I don't think this is how they expect us to solve this problem...

But... maybe pick j+k=n?

$(a-b)^n = xa^jb^k$

$(a-b)^{n+1} = xa^jb^k(a-b)$
No, that isn't true. You will need the Binomial theorem for this part of the proof. Look it up.

Originally Posted by demode
I see. But doesn't the method in my previous post work?
No, as all you have done is re-state the question. You need to prove that $ar \in N(i)$, not just state that it is.

5. Originally Posted by Swlabr
No, that isn't true. You will need the Binomial theorem for this part of the proof. Look it up.
The binomial theorem is about expansion of powers of a binomial, so when you have $(a-b)^n$, depending on what the value of "n" is, you get a sum (negative in this case) of terms of the form $xa^ib^j$. In each of these terms the sum of the exponents of a and b must be n. That's why I said n=i+j, if that's wrong... I really don't know

No, as all you have done is re-state the question. You need to prove that $ar \in N(i)$, not just state that it is.
$a \in N(I) =\{ r \in R: r^n \in I \}$ and $r \in R$ since the ring is commutative ar=ra, and they belong to R as it is closed under multipication. And if ar is in R, how would you know if there is an integer for this particular member of R such that $(ar)^n \in I$?

6. Originally Posted by demode
The binomial theorem is about expansion of powers of a binomial, so when you have $(a-b)^n$, depending on what the value of "n" is, you get a sum (negative in this case) of terms of the form $xa^ib^j$. In each of these terms the sum of the exponents of a and b must be n. That's why I said n=i+j, if that's wrong... I really don't know
What you just typed there is correct, and yes n=i+j works. However, $(a-b)^{n} = \sum x_ka^k(-b)^{n-k}$ (or something of that form). Which isn't what you said in your above post...

$a \in N(I) =\{ r \in R: r^n \in I \}$ and $r \in R$ since the ring is commutative ar=ra, and they belong to R as it is closed under multipication. And if ar is in R, how would you know if there is an integer for this particular member of R such that $(ar)^n \in I$?[/quote]

If $a^n \in I$ then $(ar)^n \in I$. This holds because $R$ is commutative, and because $a^n \in I$. What do you get when you expand $(ar)^n$?

7. This is what I've done:

$a^n \in I$ and $b^m \in I$ for some $m,n \in \mathbb{Z}$.

$(a+(-b))^k = \sum {k \choose i} a^ib^j$

where $a^i \in I$

$k=m+n$

$i+j = m+n$

$\implies j \geq m$ or $j \geq n$

So $j \geq m \implies b^j \in I$

Therefore ${k \choose i} a^ib^j \in I$.

----------
And for the multipication condition:

$a \in N(I)$ so $a^n \in I$, and $r\in R$.

$ra \in N(I)$ because $(ra)^n = r^na^n \in I$.

Are these correct? (I hope so because I must complete these by tomorrow)

8. Originally Posted by demode
This is what I've done:

$a^n \in I$ and $b^m \in I$ for some $m,n \in \mathbb{Z}$.

$(a+(-b))^k = \sum {k \choose i} a^ib^j$

where $a^i \in I$

$k=m+n$

$i+j = m+n$

$\implies j \geq m$ or $j \geq n$

So $j \geq m \implies b^j \in I$

Therefore ${k \choose i} a^ib^j \in I$.

----------
And for the multipication condition:

$a \in N(I)$ so $a^n \in I$, and $r\in R$.

$ra \in N(I)$ because $(ra)^n = r^na^n \in I$.

Are these correct? (I hope so because I must complete these by tomorrow)
You second part is fine. Your first one contains a few minor errors though (mostly that you state that $i+j=m+n \Rightarrow j \geq m$ or $j \geq n$ which isn't true as $j$ can be zero...This may just be a typo though!).

9. Originally Posted by Swlabr
You second part is fine. Your first one contains a few minor errors though (mostly that you state that $i+j=m+n \Rightarrow j \geq m$ or $j \geq n$ which isn't true as $j$ can be zero...This may just be a typo though!).
Thank you. About the minor error, what can I change it to so that it would be alright?

10. Originally Posted by demode
Thank you. About the minor error, what can I change it to so that it would be alright?
Ah, well, I'll leave you to think about that. This is work you're to hand in after all...

11. Originally Posted by Swlabr
Ah, well, I'll leave you to think about that. This is work you're to hand in after all...
I meant $j \geq m$ or $i \geq n$, if that's the typo you meant...