Question: > 
> Let $R$ be a ring. If $I\_1,\ldots, I\_k$ are ideals of $R$ and for all $i$:
> 
> 
> $$I\_i + \bigcap\_{j\neq i} I\_j = R$$
> 
> 
> then for all $a\_1,\ldots,a\_k \in R$, there exists some $a \in R$ such that for all $i$, $a \equiv a\_i \pmod{I\_i}$
> 
> 
> 

I could prove this for $k=2$: for $a\_1,a\_2 \in R$, $a\_1 - a\_2 \in I\_1 + I\_2$, so there are $i\_i \in I\_i$ such that $a\_1 - a\_2 = -i\_1 + i\_2$. Hence $a\_1 + i\_1 = a\_2 + i\_2 = a$, and $a \equiv a\_i \pmod{I\_i}$ for $i=1,2$.

I couldn't prove the general case. Any help?

Answer: 
**Hint for the inductive step:**

Suppose $I\_1,\dots, I\_k,I\_{k+1}$ are ideals in $R$ such that for any $1\le i \le k+1$, 
$$I\_i+\bigcap\_{\substack{1\le j \le k+1\\j\ne i}}I\_j=R. $$

Then, *a fortiori*, we have for any $1\le i \le k$, 
$$I\_i+\bigcap\_{\substack{1\le j \le k\\j\ne i}}I\_j=R. $$
So, given $a\_1, \dots, a\_k, a\_{k+1}\in R$, there exists some $a\in R$ such that
$$a\equiv a\_i\mod I\_i\quad(1\le i\le k)$$
and $a$ is unique mod. $\displaystyle\bigcap\_{j=1}^k I\_j$.

Can you end the inductive step?

*Some more details*:

Note that if $J=\displaystyle\bigcap\_{j=1}^k I\_j $, we have $J+I\_{k+1}=R$. 
So we have to find $A\in R$ such that \begin{cases}A\equiv a& \mkern-12mu\bmod J\\A\equiv a\_{k+1},&\mkern-12mu\bmod I\_{k+1}.\end{cases}
which is possible by the case $k=2$, and the solution is unique modulo $J\cap I\_{k+1} =\displaystyle\bigcap\_{j=1}^{k+1} I\_j$.
**Hint** $ $ It is easy if we view the solutions as "vectors" in the product ring $\, (R/I\_1,\ldots, R/I\_k).\,$ Then every such congruence is solvable iff the natural image of $R$ in the product ring is onto iff the image contains all unit vectors $e\_j$ with $1$ in the $j$'th component and $0$ in all others, since they *span*, e.g. $$(a\_1,a\_2,a\_3) = a\_1(1,0,0)+a\_2(0,1,0)+ a\_3(0,0,1) = a\_1 e\_1 + a\_2 e\_2+ a\_3 e\_3$$ 

But the hypotheses imply that unit vector systems are solveable, since by hypothes $\,a+b = 1\,$ for $\,a\in I\_j,\ b\in I\_i, \forall\, i\neq j$ so $\,e\_j := 1\!-\!a = b\,$ $\,\Rightarrow\,e\_j\equiv 1\pmod{I\_j},\ $ $e\_j\equiv 0\pmod{I\_i},\forall\,i\ne j$ 

**Remark** $ $ The classical CRT formula for integers is a special case, where the unit vectors are constructed by the formula

$\qquad e\_j := M\_j(M\_j^{-1} \bmod m\_j)\,$ where $M\_j$ is the product of all moduli except $m\_j$

Note $\,M\_j \equiv 1\pmod{m\_j}\,$ and $\,M\_j\equiv 0\,$ mod other moduli since they divide $\,M\_j$