Distributive law for ideals

Question

Let $A,B,C\triangleleft R$ be Ideals. prove that: $$B\cap(A+C)\subseteq C+(A\cap B)$$ $$\Updownarrow$$ $$B\cap(A+C)=(B\cap A)+(B\cap C)$$ I managed to proved the upper part from the lower, but I am struggling to prove the second direction.

Answer 1

One direction is straighforward as $$(B\cap A)+(B\cap C)\subseteq (B\cap A)+C=C+(A\cap B)\,.$$

For the other direction, first of all, we always have $(B\cap A)+(B\cap C)\subseteq B\cap (A+C)$.

Now, assuming $B\cap(A+C)\subseteq C+(A\cap B)$, let $b\in B\cap(A+C)$, that means in particular $b\in B$, and by the assumption, also $b\in C+(A\cap B)$, so $b=c+a$ with some $c\in C,\, a\in A\cap B$.
Then, $c=b-a\in B$, too, so $c\in B\cap C$, thus $b=a+c\,\in(B\cap A)+(B\cap C)$.

Answer 2

So the implication you're missing is proving the upper part from the lower.

Assume $B\cap(A+C)=(B\cap A)+(B\cap C)$. We want to prove the inclusion $B\cap(A+C)\subseteq C+(A\cap B)$. By assumption, this is equivalent to proving

$(B\cap A)+(B\cap C)\subseteq C+(A\cap B).$

Now notice that $B\cap A=A\cap B$ and that $B\cap C\subseteq C$. Therefore $(B\cap A)+(B\cap C)\subseteq C+(A\cap B) $ as we wanted.