So given $I_1,I_2,\dots, I_n$ are ideals in a commutative ring $R$ with unity, suppose that
$$I_1+I_2+\dots+I_n=R$$
I want to show that given arbitrary $c_1, c_2, \dots, c_n\geq1$, we have
$$I_1^{c_1}+I_2^{c_2}+\dots + I_n^{c_n}=R.$$
$\textbf{Here's what I know so far:}$
The arbitrary sum of ideals is an ideal by a simple lemma, products of ideals are ideals by another lemma, and if an ideal contains 1 then it is equal to the whole ring. Furthermore, if an ideal is equal to the whole ring, then it's sum or product with another ideal is also the whole ring.
Hence, there should be a "smallest ideal sum" $I_1 + \dots + I_j$ such that it contains 1, but that for all indices $1\leq k<j$, none of $I_1+\dots + I_k$ contains unity. Then there are elements $i_1 \in I_1, \dots ,i_j \in I_j$ such that $$i_1+\dots + i_j=1.$$
Now, I know that $i_k^{e_k}\in I_k^{e_k}$, but this doesn't actually seem relevant since I have no reason to believe they would sum to unity.
The whole sum-of-ideals-equals-the-ring thing screams Chinese Remainder theorem, but I'm not sure how quotients of $R$ by $I_1, \dots, I_n$ are supposed to help me find elements in $I_1^{e_1},\dots,I_k^{e_k}$ that sum to unity. I feel like I have a few of the major pieces of a solution, but it's not clear how I should combine them. Can anyone point me in the right direction?
Best Answer
If $I_1+\cdots+I_n=R$ then there are $a_j\in I_j$ with $$a_1+\cdots+a_n=1.\tag{1}$$ Write $C=c_1+\cdots+c_n$. Take the $C$-th power of $(1)$, and we get $$(a_1+\cdots+a_n)^C=1^C=1.\tag{2}$$ Expanding out the left side of $(2)$, each term has a factor of $a_j^{c_j}$ for some $j$, so that $1\in I_1^{c_1}+\cdots+I_n^{c_n}$.