The Chinese remainder theorem and distributive lattices

Question

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 (Wayback Machine).

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?

Answer 1

This came up before on MO, and the sentiment was that Rota is referring to the Elementwise Chinese Remainder Theorem. This holds in a ring $R$ if for any ideals $I_1,\ldots,I_n$ of $R$ and elements $x_1,\ldots,x_n \in R$, the following are equivalent:

(i) $x_i - x_j \in I_i + I_j$.
(ii) There is $x \in R$ with $x \equiv x_i \pmod{I_i}$ for all $i$.

It is clear that (ii) $\implies$ (i) in any ring. It turns out that the domains in which (i) $\implies$ (ii) holds are precisely the Prüfer domains. You can read a little bit about Prüfer domains in $\S 21$ of my commutative algebra notes, but not as much as I would like: this is the point at which the notes begin to trail off. In particular, the above result appears in the notes but the proof does not! (I think it appears in Larsen and McCarthy's Multiplicative Ideal Theory, for instance.)

Added much later: The proof does appear there now.

Answer 2

When I was an MIT student in the $\rm MathLab$ computer algebra group we were working on early implementions of Spear / Zacharias/ Grobner basis algorithms in $\rm Macyma.\,$ Motivated by this (and other number theoretic endavors) I was very interested in exploring the generality of various ideal-theoretic algorithms. Rota was a walking encyclopedia, so I would occasionally pick his brain on matters. In one of our conversations (circa $1980$) he emphasized the lattice theoretic view of the the well-known CRT congruence solvability criterion, i.e. that a system of congruences $\,x\equiv x_j\ \!\pmod{\!I_j}\,$ is solvable $\!\iff\! x_j\equiv x_k\! \pmod{\! I_j + I_k}\,$ for all $\,j,k.\,$ Namely, this criterion turns out to be equivalent to the lattice of ideals is distributive (for any commutative ring). Rota also emphasized this in writing in various places, e.g. further context from his The Many Lives of Lattice Theory, NAMS, 1997 (my emphasis below)

Dedekind outlined the program of studying the ideals of a commutative ring by lattice-theoretic methods, but the relevance of lattice theory in commutative algebra was not appreciated by algebraists until the sixties, when Grothendieck demanded that the prime ideals of a ring should be granted equal rights with maximal ideals. Those mathematicians who knew some lattice theory watched with amazement as the algebraic geometers of the Grothendieck school clumsily reinvented the rudiments of lattice theory in their own language. To this day lattice theory has not made much of a dent in the sect of algebraic geometers; if ever it does, it will contribute new insights. One elementary instance: the Chinese remainder theorem. 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 integral domains satisfying this property are known as Pruefer domains. They can be viewed as non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. Gauss's Lemma for polynomial content ideals, or for ideals: $\rm\ A\cap (B + C) = A\cap B + A\cap C,\ $ or $\rm\ (A + B)\ (A \cap B) = A\:\!B\:$ $\rm (GCD * LCM$ law), or contain = divides: $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm A,\,$ etc. It has been estimated that there are close to $100$ such characterizations known, e.g. below are close to $30$ characterizations (excerpted from my 2008/11/19 sci.math post).

Theorem $\ $ The following are equivalent for an integral domain $\:\!\rm D$/

(1) $\rm\:D\:$ is a Prüfer domain, i.e. every nonzero f.g. (finitely generated) ideal is invertible.
(2) Every nonzero two-generated ideal of $\rm\:D\:$ is invertible.
(3) $\rm\:D_P\:$ is a Prüfer domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(4) $\rm\:D_P\:$ is a valuation domain for every prime ideal $\rm\:P\:$ of $\rm\:D.\:$
(5) $\rm\:D_P\:$ is a valuation domain for every maximal ideal $\rm\:P\:$ of $\rm\:D.\:$
(6) Every nonzero f.g. ideal $\rm\:I\:$ of $\rm\:D\:$ is cancellable, i.e. $\rm\:I\:J = I\:K\ \Rightarrow\ J = K\:$
(7) $\: $ (6) restricted to f.g. $\rm\:J,K.$
(8) $\rm\:D\:$ is integrally closed and there is an $\rm\:n > 1\:$ such that for all $\rm\: a,b \in D,\ (a,b)^n = (a^n,b^n).$
(9) $\rm\:D\:$ is integrally closed and there is an $\rm\: n > 1\:$ such that for all $\rm\:a,b \in D,\ a^{n-1} b \ \in\ (a^n, b^n).$
(10) Ideals $\rm\:I\:$ of $\rm\:D\:$ are complete: $\rm\:I = \cap\ I\: V_j\:$ as $\rm\:V_j\:$ run over all the valuation overrings of $\rm\:D.\:$
(11) Each f.g. ideal of $\rm\:D\:$ is an intersection of valuation ideals.
(12) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I \cap (J + K) = I\cap J + I\cap K.$
(13) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:I\ (J \cap K) = I\:J\cap I\:K.$
(14) If $\rm\:I,J\:$ are nonzero ideals of $\rm\:D,\:$ then $\rm\:(I + J)\ (I \cap J) = I\:J.\ $ ($\rm LCM\times GCD$ law)
(15) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D,\:$ with $\rm\:K\:$ f.g. then $\rm\:(I + J):K = I:K + J:K.$
(16) For any two elements $\rm\:a,b \in D,\ (a:b) + (b:a) = D.$
(17) If $\rm\:I,J,K\:$ are nonzero ideals of $\rm\:D\:$ with $\rm\:I,J\:$ f.g. then $\rm\:K:(I \cap J) = K:I + K:J.$
(18) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of localizations of $\rm\:D.\:$
(19) $\rm\:D\:$ is integrally closed and each overring of $\rm\:D\:$ is the intersection of quotient rings of $\rm\:D.\:$
(20) Each overring of $\rm\:D\:$ is integrally closed.
(21) Each overring of $\rm\:D\:$ is flat over $\rm\:D.\:$
(22) $\rm\:D\:$ is integrally closed and prime ideals of overrings of are extensions of prime ideals of $\rm\:D.$
(23) $\rm\:D\:$ is integrally closed and for each prime ideal $\rm\:P\:$ of $\rm\:D,\:$ and each overring $\rm\:S\:$ of $\rm\:D,\:$ there is at most one prime ideal of $\rm\:S\:$ lying over $\rm\:P.\:$
(24) For polynomials $\rm\:f,g \in D[x],\ c(fg) = c(f)\: c(g)\:$ where for a polynomial $\rm\:h \in D[x],\ c(h)\:$ denotes the "content" ideal of $\rm\:D\:$ generated by the coefficients of $\rm\:h.\:$ (Gauss' Lemma)
(25) Ideals in $\rm\:D\:$ are integrally closed.
(26) If $\rm\:I,J\:$ are ideals with $\rm\:I\:$ f.g. then $\rm\: I\supset J\ \Rightarrow\ I|J.$ (contains $\:\Rightarrow\:$ divides)
(27) the Chinese Remainder Theorem $\rm(CRT)$ holds true in $\rm\:D\:,\:$ i.e. a system of congruences $\rm\:x\equiv x_j\ (mod\ I_j)\:$ is solvable iff $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$

Answer 3

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.