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
Best Answer
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.