1
$\begingroup$

Given two-sided ideals $B$ and $C$ of a ring $A$,

(a) show that $BC \subseteq B \cap C$.

(b) If the ring $A$ is commutative and $B + C = A$, show that $BC = B \cap C$.

Here's what i have but I am unsure if it is correct.

(a) Since $A,B$ are both two-sided ideals, $aba' \in B$ and $aca' \in C$.

So, $BC = aba'(aca') = ab(a'a)ca' = abca' \subseteq B \cap C$.

(b) $B+C = aba'+aca'=a(b+c)a'$ and I got stuck.

Any help will be much appreciated.

$\endgroup$
  • $\begingroup$ I think you need A to be with a unit, then you have $\endgroup$ – user65304 Feb 3 '16 at 2:02
1
$\begingroup$

For a ) it's correct up to being careful in the last step noting how you show it both in $B$ and $C $

For b) You need $A$ to have a unit, so there exist : $c+b=1$ Now pick $d $ in $ B\cap C $ And multiply by previous equation, what do you conclude?

$\endgroup$
  • $\begingroup$ $cd+bd = d \in B \cap C$ $\endgroup$ – johnbowen Feb 3 '16 at 2:07
  • $\begingroup$ are you saying let $ 1 \in A$ ? $\endgroup$ – johnbowen Feb 3 '16 at 2:08
  • $\begingroup$ @mathguy84 yes. I am not sure if this equality holds if we drop the condition the $A $ has a unit. $\endgroup$ – user65304 Feb 3 '16 at 2:11
  • $\begingroup$ ok, that makes sense. Also for the last part of a) I am a little confused on the last step. How can I show that final step? $\endgroup$ – johnbowen Feb 3 '16 at 2:12
0
$\begingroup$

(a) To show $BC \subseteq B \cap C,$ all you need to do is show $BC \subseteq B$.

So consider $i \in BC$. We're trying to show $i \in B$. Write $i=bc$ with $b \in B$ and $c \in C$. But since $b \in B$ and $B$ is an ideal, hence $bc \in B$. Thus $i \in B$, as required.

(b) Assume $A \subseteq B+C$. Then we can find $\beta \in B$ and $\gamma \in C$ satisfying: $$1 = \beta+\gamma.$$

Now we're trying to show that $B \cap C \subseteq BC$. So consider $i \in B \cap C$. Our goal is to show $i \in BC$. It's clear that $$\beta i + i\gamma \in BC.$$ But

$$\beta i + i\gamma = \beta i + \gamma i = (\beta+\gamma) i = 1i = i$$

So $i \in BC$, as required.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.