Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
He probably means the version where condition (1) is omitted and $A$ is the zero ideal. –  Zhen Lin May 14 '12 at 21:35
    
@ZhenLin I'm not sure I understand. Does it mean that for a ring with a distributive lattice of ideals $(2)$ implies $(1)$, when $A=(0)?$ But that's not true, is it? If the ideals $A_1,\ldots,A_k$ are pairwise comaximal, then their intersection doesn't have to be trivial. For example, $\mathbb Z\ni 6\in (2)\cap (3),$ and $(2)+(3)=\mathbb Z$ (and the lattice of ideals is distributive in $\mathbb Z$.) –  user23211 May 14 '12 at 22:30
1  
There's a pretty big error in that article, with the definition of $2$-distributive lattice being horribly wrong. The correct condition is: $$a\lor(x\land y\land z) = (a\lor (x\land y)) \land (a\lor (x\land z))\land (a\lor (y\land z))$$ –  Thomas Andrews May 14 '12 at 23:08
    
@ymar: Sorry, I wasn't thinking. –  Zhen Lin May 14 '12 at 23:19
add comment

3 Answers

up vote 6 down vote accepted

I discussed this with Rota, so I can assure you that he refers to Prüfer domains. They are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. For example, they are those domains satisfying either the Chinese Remainder Theorem for ideals, or 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\:,\ $ or $\rm\ A\supset B\ \Rightarrow\ A\:|\:B\ $ for fin. gen. $\rm\:A\:$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my 2008/11/19 sci.math post for 30 odd characterizations. Below is an excerpt:

THEOREM $\ \ $ Let $\rm\:D\:$ be a domain. The following are equivalent:

(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 $\rm\:x_j\equiv x_k\ (mod\ I_j + I_k).$

share|improve this answer
    
Thanks! That's a lot of conditions, some of which I've been trying to think about lately. May I ask how much of this carries over to non-commutative rings, if anything? –  user23211 May 14 '12 at 23:42
add comment

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.

share|improve this answer
    
Thank you very much! I think I've found the MO question you are talking about: link. I see that Prufer domains are commutative rings. But the condition that the lattice of ideals is distributive (which I see in your notes is equivalent for integral domains to being a Prufer domain) doesn't need commutativity to make sense. Can anything be said about this for non-commutative rings? –  user23211 May 14 '12 at 22:53
1  
@ymar: "Can anything be said about this for non-commutative rings?" Probably, but not by me at this time. (Like many mathematicians, I am somewhat "non-commutatively challenged".) –  Pete L. Clark May 14 '12 at 23:11
    
OK, maybe someone else will answer this one. :) I will try to find that book. Before I do, I'd like to ask one thing out of curiosity. In the theorem you give in your notes (about the equivalence "prufer domain iff satisfies the ECRT", can we drop the domain requirement? I think commutative rings with distributive lattices of ideals are called arithmetical. Is it also true that a ring is arithmetical iff it satisfies the ECRT? –  user23211 May 14 '12 at 23:32
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.