About Relatively Prime Ideals I am a physics Master student who has been studying abstract algebra by himself. 
I have two questions about relatively prime ideals.
My first question goes as follows: 

Let $R$ be a ring, and $I$, $J_{1}, J_{2}, \dots, J_{n}$ be ideals of $R$. If $I$ is relatively prime to any $J_{i}$ for $\forall i=1,2,\dots$, how do I prove that $I$ is coprime to the product $\prod_{i}J_{i}$?

My second question is: 

If $I$ and $J$ are coprime ideals of $R$, how to prove that it is equivalent to $I+J$ being contained in no prime ideal?

I found a proof from a Chinese algebra book. But I suspect that the given proof in my Chinese book is incorrect. 
I hope you would help me. Thanks a lot.
 A: I am assuming that $R$ has a multiplicative neutral element $1$ (below I show that the claim is false otherwise).
The definition of two ideals $I,J$ being coprime is (AFAIK) that for some elements $i\in I, j\in J$ we have $i+j=1$. This is equivalent to the requirement $I+J=R$, when $R$ is unital (see comments).
So your assumptions imply that to each index $k=1,2,\ldots,n$ there exists elements $i_k\in I, j_k\in J_k$ such that $i_k+j_k=1$.
Hint: Expand the product
$$
1=1^n=(i_1+j_1)(i_2+j_2)\cdots(i_n+j_n).
$$


*

*When you are done, can you see that one term is in the ideal
$J_1J_2\cdots J_n$, and all the other terms are in the ideal $I$?

*Why does this prove the claim of your first question?



If we don't assume that $R$ has a multiplicative neutral element, and attempt to define coprimeness using the condition $I+J=R$, then the claim is actually false. Consider the rng $R=2\Bbb{Z}$. Let $I=4\Bbb{Z}$, $J_1=6\Bbb{Z}$ and $J_2=10\Bbb{Z}$.
We easily see that $2\in I+J_1$ and $2\in I+J_2$ implying that $R=I+J_1$ and
$R=I+J_2$. But here $J_1J_2\subseteq I$, so $I+J_1J_2=I$ is not all of $R$.
