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.

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

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?

  • $\begingroup$ $cd+bd = d \in B \cap C$ $\endgroup$ – 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

(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.


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