What is an example of ideals $I$ and $J$ such that the sequence $0\to I\cap J\to I\oplus J\to I+J\to 0$ is not split? I have a problem  

Prove there is an exact sequence of $R$-modules
  $$0\to I\cap J\to I\oplus J\to I+J\to 0$$ 
  given that $I,J$ are ideals of $R$.  

I solved it with homomorphisms $f:I\cap J \to I\oplus J$ and $g:I\oplus J\to I+J$ s.t. $f(a)=(a,-a)$ and $g(a,b)=a+b$.
My teacher claimed that it's generally not split, though I can't find a counterexample. Help me with this, thank you.
 A: Let $k$ be a field (or actually any commutative ring) and $R=k[x,y]$ and let $I=(x)$ and $J=(y)$.  Then $I\cap J=(xy)$.  Suppose your map $f:I\cap J\to I\oplus J$ had a splitting $h:I\oplus J\to I\cap J$. Let $h(x,0)=s$ and $h(0,y)=t$.  Then we must have $$xy=h(f(xy))=h(xy,-xy)=h(y\cdot(x,0)-x\cdot(0,y))=ys-xt.$$
But $s$ and $t$ must both be elements of $I\cap J=(xy)$.  In particular, every term of $s$ and $t$ must have degree at least $2$.  It follows that every term of $ys-xt$ has degree at least $3$, so $ys-xt$ cannot be equal to to $xy$.  Thus no such splitting exists.
In fact, by a bit of magic this shows that for this choice of $I$ and $J$, there cannot exist any split exact sequence $$0\to I\cap J\to I\oplus J\to I+J\to 0,$$ even if you use maps different from your $f$ and $g$.  For the fact that the sequence with your $f$ and $g$ does not split witnesses that the module $I+J$ is not projective.  But if such a split exact sequence existed with different maps, $I+J$ would be a direct summand of $I\oplus J$.  The modules $I$ and $J$ are both projective (they are freely generated by $x$ and $y$ respectively), so $I\oplus J$ is projective.  A direct summand of a projective module is product, so this would imply $I+J$ is projective, which is a contradiction.
