$R$ is a commutive ring, and $A, B, C$ are ideals in $R$. If $A + B = A + C$ and $A ∩ B = A ∩ C$, then it is necessary that $B = C$? My question is: $R$ is a commutive ring, and $A, B, C$ are ideals in $R$.  If $A + B = A + C$ and $A ∩ B = A ∩ C$, then it is necessary that $B = C$?
I come up with this question because when $ R $ is P.I.D, $A + B$ is generated by the gcd of the generators of $ A $ and $ B $ and $ A ∩ B $ is generated by the lcm of the generators of $ A $ and $ B $. In this case, the above statement is clearly true. I was wondering whether we need such strong condition (e.g., $ R $ is P.I.D) to make the statement true.
 A: This doesn't hold in general, though it does hold for arithmetical rings (in particular Prüfer domains), which are more general than PIDs.
For the counterexample, take $R=k[x,y]/(x^2,xy,y^2)$ for a field $k$, and denote by $\delta=x+(x^2,xy,y^2)$ resp. $\varepsilon=y+(x^2,xy,y^2)$ the classes of $x$ resp. $y$ in $R$. Then $1,\delta,\varepsilon$ form a $k$-basis of $R$, and we have the realtions $\varepsilon^2=\varepsilon\delta=\delta^2=0$.
Now take $A=(\delta+\varepsilon)$, $B=(\delta)$ and $C=(\varepsilon)$. Then using the description above, one may verify that $A+B=(\delta,\varepsilon)=A+C$, and $A\cap B=(0)=A\cap C$. However, $B\neq C$.
On the other hand, suppose that $R$ is arithmetical: a common definition for this is that for all ideals $I,J,K$ of $R$ we have $I\cap(J+K)=I\cap J+I\cap K$. Now take ideals $A,B,C$ of $R$ satisfying $A+B=A+C$ and $A\cap B=A\cap C$. In particular, we have $A\cap (B+C)=A\cap B+A\cap C=A\cap B=A\cap C$.
Now take any element $b\in B$. Then $b\in A+C$, so there exist $a\in A$ and $c\in C$ such that $b=a+c$. Thus $b-c\in A\cap (B+C)=A\cap C$ by the above argument, and hence also $b\in C$. So we proved $B\subseteq C$, and by symmetry we obtain $B=C$.
It would be interesting to know whether the condition
$$
\forall\text{ ideals }A,B,C\subseteq R:\left((A+B=A+C)\ \land\ (A\cap B=A\cap C)\right)\implies B=C
$$
implies the condition
$$
\forall\text{ ideals }I,J,K\subseteq R: I\cap(J+K)=I\cap J+I\cap K,
$$
but intuitively I don't think this is the case.
Edit: This last question is settled here by Jacob Manaker: contrary to what I expected, the only rings where $A+B=A+C$ and $A\cap B=A\cap C$ always implies $B=C$ are precisely the arithmetical rings.
