Proof of chinese remainder theorem, showing that Isomorphism is surjective

Question

So I was trying to proof the chinese remainder theorem, which (in its general form) says:
Let $R$ be a Ring (with unity) and $I_1,...,I_n \subset R$ Ideals such that $I_i+I_j=R$ for $i\not=j$. Then: $$R\diagup (I_1\cap...\cap I_n) \cong \bigoplus_{i=1}^n R\diagup I_i$$ Now the general way of going about this, is to show that $$\phi: R\diagup (I_1\cap...\cap I_n)\rightarrow \bigoplus_{i=1}^n R\diagup I_i \quad ,\; a+(I_1\cap...\cap I_n)\mapsto (a+I_1,...,a+I_n)$$ is an isomorphism.
So far I could show that $\phi$ is well-defined, linear and injective. I was however not able to show surjectivity . Here´s my approach for $n=2$:
Let $\;(p+I_1, q+I_2)\in (R\diagup I_1) \oplus (R\diagup I_2)$ than since $I_1$ and $I_2$ are coprime there exist $\;p_1, q_1 \in I_1$ and $\;p_2, q_2 \in I_2$ such that $p=p_1 +p_2,\; q=q_1+q_2$ and hence: $$\phi (p_2 +q_1+(I_1\cap I_2))=(p_2 +q_1+I_1,p_2 +q_1+I_2)=(p_2 +I_1, q_1+I_2)=(p+I_1, q+I_2)$$ so $\phi$ has to be surjective. Any hints on how to go about the general case?

Answer 1

Hint:

Proceed by induction on $n$. All you need to show is that if $I_1,\dots,I_n$ are pairwise coprime, then $I_1\cap\dots\cap I_{n-1}$ and $I_n$ are coprime.