$\prod I_i = \bigcap I_i$ for ideals $I_1,\dots, I_n$ with $I_i+I_j=R$ for $i\neq j$ Consider a rng $R$ and ideals $I_1,\dots , I_n$ such that $I_i+I_j=R$ for $i\neq j$. Show that the inclusion $\prod I_i\subseteq \bigcap I_i$ is an equality (Exercise 1.1.4, Bosch's Algebraic Geometry and Commutative Algebra)
I already  proved  Chinese remainder theorem (in the previous exercise), but I don't know how to approach this one, and I've been thinking for a while. Even if I show $R/\prod (I_i)\cong R/ \bigcap (I_i)$, the thesis is not proven, and I can't think of anything else. Can you only give a hint to start? Thank you
 A: First assume $n=2$. If $I+J=R$ then we can write $1=x+y$ where $x\in I, y\in J$. So if $z\in I\cap J$ then we have:
$z=z(x+y)=zx+zy\in IJ$
Now try to apply induction in order to prove the general case.
A: Below is a sketch of a simple inductive proof.
Theorem $\ I_n \cap \cdots I_2\cap I_i\, =\, I_n\cdots I_2 I_1\ $ if $\ I_j + I_k = 1\, $ for $j\neq k$.
Proof $ $ (Sketch) $\ $ We proceed by (complete) induction on $n$.
Base case: $\,n = 2,\,$ i.e. $\, I\cap J = I J\ $ if $\ \color{#c00}{I+J = 1}.\,$  Direction $\, I\cap J \supseteq IJ$ is clear. Conversely
$\, I\cap J = (\color{#c00}{I+J})(I\cap J) = I(I\cap J) + J(I\cap J) \subseteq IJ.\,$
Induction step: $ $ assume it is true for all $\,k< n.\,$ Then
$$\begin{align}
&I_n \cap (I_{n-1}\cap \cdots\cap I_1)\\
=\ &I_n \cap (I_{n-1}\times \cdots\times I_1)\ \ {\rm by\ induction} \\
=\ &I_n \times (I_{n-1}\times \cdots\times I_1)\ \ \text{by $\,k=2\,$ case & Lemma below }
\end{align}$$
Lemma $\ \forall i\!:  \color{#0a0}{I + I_i = 1}\Rightarrow I + I_1\cdots I_{n-1} = 1\,\ $ [Euclid's Lemma in ideal form]
Proof $ $ Induct using $\,I + I_i J = I + I_i J + I J = I + (\color{#0a0}{I_i+I})J = I+J$
Or $\bmod I\!:\ I_k\equiv 1\Rightarrow \prod I_k \equiv 1^n = 1\ $ [ideal form of coprimes are closed under product]
