TITLE: Sum of ideals in a ring
QUESTION [10 upvotes]: I have the following question:
If $I,J$ are ideals in $R$, then denote $I+J = \{ r \in R | r = x+y, x \in I, y \in J  \}$
It is not hard to show that $I+J$ is again an ideal in $R$
If we then assume that $I+J = R$ we have the identity $R/(I \cap J) \simeq R/I \times R/J$
I managed to show this by constructing a function $f: R \to R/I \times R/J$ and then using the first isomorphism theorem. The proof then basically follows how one would derive the second isomorphism theorem from the first, which leads me the question: is it possible to approach this problem using the 2nd/3rd isomorphism theories directly?

REPLY [2 votes]: It is possible to prove this with the help of the "third isomorphism theorem". However you need to prove the following fact (which of course can be obtained from your result but can also be proved directly).
Step 1: If $I$ is an ideal of $R$ with complement $J$ (i.e. $I+J=R$ and $I\cap J=0$), then $R \cong I\times J$
Step 2: Let use the above to prove your result. Writing $q:R\to R/(I\cap J)$ for the canonical ring homorphism and taking images we obtain $q(I)=I/(I\cap J) \cong (I+J)/J = R/J$ and $q(J)=J/(I\cap J)\cong (I+J)/J = R/J$ (using the third isomorphism theorem twice). Applying the result from Step 1 (since $q(I) + q(J)=R/(I\cap J)$, and $q(I)\cap q(J) = 0$ we obtain that $R/(I\cap J) \cong q(I) \times q(J) \cong R/J \times R/I$ as desired.
Proof of statement in Step 1: By the "third isomorphism theorem" $R/I =(I+J)/I \cong J/(I\cap J)= J$ and similarly $R/J \cong I$ we obtain a
homorphism $R \to I\times J$ (from the two quotient homorphisms). A straight forward calculation shows that this homorphism is an isomorphism.
As a final remark this result is true in any semi-abelian category. This means that it is true for groups, Lie algebras over a commutative ring, Heyting semi-lattices and many other related structures (and that one can give a single proof to cover all of these cases). As another example in the language of groups it would be: If $I$ and $J$ are normal subgroups of a group $G$ and $IJ=G$, then $G/(I\cap J) \cong G/I \times G/J$.