# Ideals and commutative rings with unity

1. Jan 28, 2013

### Zondrina

1. The problem statement, all variables and given/known data

From contemporary abstract algebra :

http://gyazo.com/08def13b62b0512a23505811bcc1e37e

2. Relevant equations

"A subring A of a ring R is called a (two-sided) ideal of R if for every r in R and every a in A both ra and ar are in A."

So I know that since A and B are ideals of a ring R, $ar, ra \in A$ and $br, rb \in B$ for all $a \in A, \space b \in B, \space r \in R$

3. The attempt at a solution

So my guess is to argue the double inclusion for this.

Case : $A \cap B \subseteq AB$

Suppose $k \in A \cap B$, then $k \in A$ and $k \in B$. We want to show $k \in AB$

I'm having trouble seeing how the given facts are supposed to steer the argument from here. Help would be much appreciated.

Last edited: Jan 28, 2013
2. Jan 28, 2013

### Zondrina

Hmm I thought about this for an hour or so... and I THINK I may be on to something. I'll continue my flow below :

Case : $A \cap B \subseteq AB$

Suppose $k \in A \cap B$, then $k \in A$ and $k \in B$. We want to show $k \in AB$.

Since we know R is a commutative ring with unity, let e denote the unity in R.

Now since $R = A + B$, we know that $e = a + b$ for some $a \in A$ and $b \in B$. Now :

e = a + b
ke = k(a + b)
k = ka + kb

Hence $ka \in A$ and $kb \in B$ hence $k \in A + B$

We can take $a \in A$ and we can take $k \in B$. We can also take $b \in B$ and we can take $k \in A$. Putting these together we have $ka \in AB$ and $kb \in AB$ so that $k \in AB$ since $a \in A$ and $b \in B$ as desired.

$∴ A \cap B \subseteq AB$

Case : $A \cap B \supseteq AB$

Let $ab \in AB$

Now, $ab \in A$ since $a \in A$, $b \in R$ and A is an ideal of R. Also, $ab \in B$ since $b \in B$, $a \in R$ and B is an ideal of R. Hence $ab \in A \cap B$.

$∴ A \cap B \supseteq AB$

$∴ A \cap B = AB$

I think that's it... If someone could look over it for me and tell me maybe if I missed something or if it's completely wrong that would be great :).

3. Jan 28, 2013

### Dick

Writing e=a+b is a great idea. I think you could write it read a little better. Like did you really need to say "Hence $ka \in A$ and $kb \in B$ hence $k \in A + B$"? Did you ever use that k is in A+B?

4. Jan 28, 2013

### Zondrina

Yes good point. I never used it at all so I don't need it there. I wrote it down just to have it around incase I MAY have needed it yknow?

Does anything else not make sense or has early morning thinking saved me again?

5. Jan 28, 2013

### Dick

No, it all looks good to me.