Ideals and commutative rings with unity

[STEMucator]

Homework Statement

From contemporary abstract algebra :

http://gyazo.com/08def13b62b0512a23505811bcc1e37e

Homework 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

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.

 

[STEMucator]

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 ABCase : 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 :).

 

[Dick]

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 :).

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?

 

[STEMucator]

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?

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?

 

[Dick]

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?

No, it all looks good to me.