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?

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.
