Sum of ideals in a ring 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?
 A: In general, you always have that $R/(I\cap J)$ embeds into $R/I \times R/J$. This simply follows because if $f$ and $g$ are any two ring homomorphism $f\colon R\to S$ and $g\colon R\to T$, then letting $f\times g\colon R\to S\times T$ by the map $(f\times g)(r) = (f(r),g(r))$, you always have $\mathrm{ker}(f\times g) = \mathrm{ker}(f)\cap\mathrm{ker}(g)$. Thus, $R/(I\cap J)$ is isomorphic to its image in $R/I\times R/J$ by the First Isomorphism Theorem. This always holds, without any conditions on $I$ or on $J$. 
The only issue here is showing that this map is onto, and for this you use the fact that $I+J=R$: given $x,y\in R$ and $(x+I,y+J)\in R/I\times R/J$, there exists $a_1,a_2\in I$, $b_1,b_2\in J$ such that $x=a_1+b_1$, $y=a_2+b_2$. Then $(b_1+a_2)+I = b_1+I = x+I$ and $(b_1+a_2)+J = a_2+J = y+J$, so $b_1+a_2$ maps to $(x+I,y+J)$. 
As far as I can tell, this only uses the first isomorphism theorem, and
I don't see any use or reconstruction of the second isomorphism theorem (usually, that there is an inclusion preserving bijection between the subrings of $R/I$ and the subrings of $R$ that contains $I$, and the bijection maps ideals to ideals) nor the third (usually, that for a subring $R'$ and an ideal $I$, $(R'+I)/I \cong R'/(R'\cap I)$). Perhaps you have a different numbering, but I still don't see the application of anything other than the first isomorphism theorem and some basic properties about products in the proof above.
A: 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 $G/(I\cap J) \cong G/I \times G/J$.
