Sum of Powers of Coprime Ideals Let $R$ be a commutative unital ring and let $I$, $J$ be ideals in $R$ such that $I+J=R$.
Show that $II+JJ=R$.
What I've tried: 
Clearly $II+JJ \subseteq I+J =R$ as $I$ and $J$ are closed under addition and multiplication. So it suffices to show that $R \subseteq II+JJ$. Since $1\in R$, there exist $i_1 \in I, j_1 \in J$ such that $i_1+j_1 = 1$. Similarly, for each $r\in R$, there exist $i_r, j_r$ such that $i_r + j_r = r$. Now $r = r1= (i_r+j_r)(i_1+j_1)= i_ri_1+i_rj_1+j_ri_1+j_rj_1$. From here I would like to show $ i_rj_1+j_ri_1 \in II+JJ$. 
Any suggestions?
 A: If $I^2+J^2$ is not equal to $R$ then it is contained in a maximal ideal, call it $M$. Then we have that $I^2\in M$ and $J^2\in M$. 
Claim: If $P$ is a prime ideal and $AB\subset P$ where $A$ and $B$ are ideals, then we must have either $A\subset P$ or $B\subset P$. 
Proof: You try it :) 
Use the claim ($M$ is prime, why?) to show that since $I^2\subset M$, then we have $I\subset M$. Similarly $J\subset M$, how does this create a contradiction?
A: $i_r j_1 + j_r i_1$ = $(i_r j_1 + j_r i_1) \cdot 1$ = $(i_r j_1 + j_r i_1) \cdot (i_1 + j_1)$
Now if you expand the right side you see that every term has either 2 $i$'s or two $j$'s in it, hence it is in $II+JJ$.
A: How about this:
Since $I + J = R$, there exist $e_I \in I$ and $e_J \in J$ with $e_I + e_J =1 \in R$.  Then $1 = 1^3 = (e_I + e_J)^3 = e_I^3 + 3e_I^2e_J + 3e_Ie_ J^2 + e_J^3$.  Observe that $e_I^3 + 3e_I^2 = e_I^2(e_I + 3e_J) \in II$, and that, similarly, $3e_Ie_J^2 + e_J^3 = e_J^2(3 e_I + e_J)  \in JJ$.  Thus $1 \in II + JJ$, so $II + JJ = R$.  QED
Note Added in Edit:  Apparently the same technique can be used to show that $I^n + J^n =R$; simply consider $1 = 1^m = (e_I + e_J)^m$ for sufficiently large $m$; I believe that $m \ge 2n - 1$ will suffice; then each term in the binomial expansion of $(e_I + e_J)^m$ has at least one of $e_I$, $e_J$ to the power $n$.
Hope this helps.  Cheers, 
and as always, 
Fiat Lux!!!
A: $I+J = R \implies (I+J)^n = R$ for any natural $n$.
Using the fact for any ideal $H$, $H + H = H$, a binomial expansion will sum up terms without "coefficients", for example $(I+J)^2 = I^2 + IJ + J$. Then
$R=(I + J)^4 = I^2 (I+J)^2 + J^2(I+J)^2 = I^2 +J^2$
You can generalize this to $I^n + J^m = R$ for any natural $n,m$.
