# Thread: Proving (I:a) is an ideal of a ring

1. ## Proving (I:a) is an ideal of a ring

Suppose R is a commutative ring and I is an ideal of R and a∈R.

Define (I:a) = {r∈R : ar∈ I}

I want to prove that (I:a) is also an ideal of R.

I know i need to prove:
1) 0∈(I:a)

2) If b,c∈ (I:a) , then b + c ∈ (I:a)

3) If b∈ (I:a) and s∈ R, then bs∈(I:a) and sb∈(I:a)

I started by trying to prove 1):

I know 0∈I (by definition of an ideal) so ar=0 for some a, some r.
Hence, 0∈(I:a)

I am not sure if this is correct, or how to proceed with the rest of the proof.

Any help is appreciated.
Thanks

2. Originally Posted by Siknature
Suppose R is a commutative ring and I is an ideal of R and a∈R.

Define (I:a) = {r∈R : ar∈ I}

I want to prove that (I:a) is also an ideal of R.

I know i need to prove:
1) 0∈(I:a)

2) If b,c∈ (I:a) , then b + c ∈ (I:a)

3) If b∈ (I:a) and s∈ R, then bs∈(I:a) and sb∈(I:a)

I started by trying to prove 1):

I know 0∈I (by definition of an ideal) so ar=0 for some a, some r.
Hence, 0∈(I:a)

I am not sure if this is correct, or how to proceed with the rest of the proof.

Any help is appreciated.
Thanks
1) Is it true that $0\in (I:a)$ , i.e. is it true that $a\cdot 0=0\in I$ ?

2) is it true that $ab\in I\,,\,\,ac\in I\Longrightarrow a(b+c)=ab+ac\in I$ ?

3) Is it true that $b\in (I:a)\,,\,\,r\in R\Longrightarrow br\in (I:a)$ , i.e. $a(br)=abr\in I$ ?

Tonio

3. Originally Posted by tonio
1) Is it true that $0\in (I:a)$ , i.e. is it true that $a\cdot 0=0\in I$ ?

2) is it true that $ab\in I\,,\,\,ac\in I\Longrightarrow a(b+c)=ab+ac\in I$ ?

3) Is it true that $b\in (I:a)\,,\,\,r\in R\Longrightarrow br\in (I:a)$ , i.e. $a(br)=abr\in I$ ?

Tonio
Yes i think it is true in all 3 cases because we know I is an ideal and by the definition of ideal i think that all these cases must hold.

Thanks for your help in showing me this.

A further question i would appreciate your help with is this:

suppose bR is an associate of a.

How do i prove that (I:b) = (I:a) ?

I started off by saying that a,bR are associates if b=ua for some unit uR.

I tried substituting this in to (I:b)

So (I:b) = {rR : br R}
= {rR : uar R} for some unit uR

but i cant see how to proceed for here to try to make it equal (I:a).

Do you have any suggestions?

4. Originally Posted by Siknature
Yes i think it is true in all 3 cases because we know I is an ideal and by the definition of ideal i think that all these cases must hold.

Thanks for your help in showing me this.

A further question i would appreciate your help with is this:

suppose bR is an associate of a.

How do i prove that (I:b) = (I:a) ?

I started off by saying that a,bR are associates if b=ua for some unit uR.

I tried substituting this in to (I:b)

So (I:b) = {rR : br R}
= {rR : uar R} for some unit uR

but i cant see how to proceed for here to try to make it equal (I:a).

Do you have any suggestions?

Let $x\in(I:b)\Longrightarrow bx=i\in I$ , but $i=bx=uax\Longrightarrow$ if $uv=1$ , then $vi=vuax=ax\in I\Longrightarrow x\in (I:a)\Longrightarrow (I:b)\subset (I:a)$.

In a similar way can the other direction be proved.

Tonio

5. Originally Posted by tonio
Let $x\in(I:b)\Longrightarrow bx=i\in I$ , but $i=bx=uax\Longrightarrow$ if $uv=1$ , then $vi=vuax=ax\in I\Longrightarrow x\in (I:a)\Longrightarrow (I:b)\subset (I:a)$.

In a similar way can the other direction be proved.

Tonio
Is I⊆(I:a) ?

Is a∈(I:a) ?

6. Originally Posted by Siknature
Is I⊆(I:a) ?

Obviously (why?) yes...!

Is a∈(I:a) ?

Not necessarily. For example, $3\notin (4\mathbb{Z}:3)\,,\,\,R=\mathbb{Z}$

Tonio

7. Originally Posted by tonio
Is I⊆(I:a) ?

Obviously (why?) yes...!

Is a∈(I:a) ?
I am not sure why this is obvious.

Is it because every element of (I:a) must belong to I by definition of (I:a) ?

8. Originally Posted by Siknature
I am not sure why this is obvious.

Is it because every element of (I:a) must belong to I by definition of (I:a) ?

What happens if you take any element of I and multiply it by ANY element of R? So what happens if you multiply any element of I by a?!

Tonio

9. Originally Posted by tonio
What happens if you take any element of I and multiply it by ANY element of R? So what happens if you multiply any element of I by a?!

Tonio
Oh, I see....

If you take any element of I and multiply it by any element of R,the result is in I (by definition of an ideal).

So if you multiply any element by aR, the result will still be in I.

Hence, I⊆(I:a)

10. Originally Posted by Siknature
Oh, I see....

If you take any element of I and multiply it by any element of R,the result is in I (by definition of an ideal).

So if you multiply any element of I by aR, the result will still be in I.

Hence, I⊆(I:a)

Check the red addition to your post above!

Tonio

11. Originally Posted by tonio
Check the red addition to your post above!

Tonio
Oh yes!

Thankyou, I very much appreciate your help in this series of questions.