TITLE: The Chinese remainder theorem and distributive lattices
QUESTION [20 upvotes]: In The Many Lives of Lattice Theory Gian-Carlo Rota says the following.

Necessary and sufficient conditions on a commutative
ring are known that insure the validity
of the Chinese remainder theorem. There is, however,
one necessary and sufficient condition that
places the theorem in proper perspective. It states
that the Chinese remainder theorem holds in a
commutative ring if and only if the lattice of ideals
of the ring is distributive.

The essay can be found here, and the quotation comes from the third page in the file.
I know the following version of the Chinese remainder theorem for rings (not necessarily commutative).

Suppose $R$ is a ring and $A,A_1, \ldots,A_k$ are ideals of $R.$ If
$(1)$ $A_1 \cap \ldots \cap A_k = A,$ and
$(2)$ $A_i + A_j = R$ for all $1 \leq i < j \leq k,$
then $R/A \cong R/A_1\times\ldots\times R/A_k$ via an isomorphism which is both a ring isomorphism and an $R$-module isomorphism.

This version of the theorem comes from these lecture notes.
Clearly, there are some lattice-theoretic conditions on the ideals here, but I don't understand what G.C. Rota means by "the Chinese remainder theorem". He cannot mean this version because it holds for any rings. Could you give me the exact wording of the theorem he mentions? Also, can I find its proof anywhere? And if it's possible, could you explain to me why (or if) commutativity is important in the theorem he mentions?

REPLY [10 votes]: In fact, commutativity is not needed at all for this completely elementary equivalence. The proof is below.
(ECRT)$\Rightarrow$ (Prüfer). Assume that $R$ satisfies the ECRT, let $I,J,K$ be three ideals of $R$. The inclusion $(I\cap J)+(I\cap K) \subseteq I\cap (J+K)$ is obvious. Conversely, let $i\in I \cap (J+K)$. Then, there is a $j\in J$ and a $k\in K$ with $i=j+k$. By ECRT, there is an $x\in R$ such that $x\equiv 0 \ ({\sf mod } I)$, $x\equiv j \ ({\sf mod } J)$ and
$x\equiv i \ ({\sf mod } K)$. If we set $y=i-x$, we have $x\in I \cap J, y\in I \cap K$
so $i\in (I\cap J)+(I\cap K)$. Finally $(I\cap J)+(I\cap K) = I\cap (J+K)$ and $R$ is a  Prüfer ring.
(Prüfer)$\Rightarrow$ (ECRT). Let $R$ be a  Prüfer ring. It will suffice to show the ECRT for three ideals instead of $n$. So let $I_1,I_2,I_3$ be three ideals of $R$, and $x_1,x_2,x_3$ in $R$ such that
$x_j-x_i \in I_i+I_j$ for any $i<j$. Then we have $a_i,b_i \in I_i (1 \leq i \leq 3)$ such that 
$$
x_2-x_1=a_1+a_2, \ \  x_3-x_1=b_1+a_3, \ \  x_3-x_2=b_2+b_3
$$
Then we have
$$
b_3-a_3=(b_1-a_1)-(b_2-a_2)
$$
So the element $b_3-a_3$ is in $I_3 \cap (I_1+i_2)$. Since $R$ is Prüfer, we deduce that there are constants $c_{13}\in I_1\cap I_3, c_{23}\in I_2\cap I_3$ such that
$b_3-a_3=c_{13}+c_{23}$. Then
$$
b_2+a_2+c_{23}=b_1-a_1-c_{13}
$$
and the RHS above is in $I_1$, while the LHS is in $I_2$. So we have found an element 
$c_{12}$ of $I_1 \cap I_2$. Now I claim that $x=x_1+a_1+c_{12}$ satisfies the congruences we want ; indeed, for each $k \in \lbrace 1,2,3 \rbrace$ we have
$$
x-x_k=a_k+c_{1k} \in I_k
$$ 
qed.