A funny Proof For Sunzi’s Remainder Theorem

This post will assume familiarity with some basic notions from commutative algebra such as localization for modules and exact sequences. Here we will mean “ring” to mean a commutative ring with unity.

This post is about to the  “Chinese Remainder Theorem.) however, I prefer a renaming— while Sunzi’s identity is unknown, I do not think “Sunzi’s remainder theorem” will cause any great confusion.

Anyhow, the theorem is best remembered by the informal statement “you can simultaneously solve a system of modular conditions, given that the moduli are coprime.” We’ll consider a mild generalization to arbitrary rings, granted that we can make sense of “coprime” outside of principal ideal domains, leading us to the following definition:

Definition 1 Two ideals I,J of a ring R are said to be coprime if I+J=R.

One can easily verify that this definition is “correct” since it is both well defined, and also agrees with our usual notion of coprime numbers, since the condition I+J=R is equivalent to the existence of i \in I and j \in J so that i+j=1.

Now, we can state the theorem:

 Main Theorem the map \phi: R \to \prod_{i=1}^{n} R/I_{i} given by

    \[x \mapsto (x+I_1,...,x+I_n)\]

is surjective when all I_i, I_j are pairwise coprime. (It is also the case that

    \[\ker \phi =\bigcap_{I=1}^{n} I_i=I_1\dots I_n,\]

but this is much less fun to prove.)

If we take R = \mathbb Z, then this essentially says that the map x \mapsto (x \mod a_1,...,x \mod a_n) is surjective when a_1,...,a_n are coprime. This ensures some solution to the set of modular conditions.

There is a standard proof for the main theorem, discoverable by any cursory google search. It is essentially the same as the special case for R=\mathbb Z,  using Bezout’s identity. However, I consistently forget this proof since I did not think of it myself, and  no part of my being wishes to be reminded of this fact. Consequently, I want to present a more interesting one using a hint from Eisenbud’s “Commutative Algebra With a View Towards Algebraic Geometry.” We will need the notion of localization for modules, and some of its basic properties. In particular, we will need to use the fact that localization is “exact,” and that surjectivity is a “local property,” statements that will be made precise immediately:

given any R-module homomorphism \phi: M \to N, localization at S induces a new S^{-1}R-module homomorphism S^{-1}\phi: S^{-1}M \to S^{-1}N, defined in the most natural way one can imagine: m/s \mapsto \phi(m)/s. Presently, we only care about localization at prime ideals, so in the case where S=R \backslash P for a prime ideal P, we will denote the map by \phi_{P} and the localization of M by M_P.

If I were a better person I would have made clear how the localization construction works with modules, but one must have faith that all is good in the world. A nice reference would be Atiyah-Macdonald’s classical text, from which we will borrow the statement and proof of the following proposition:

Proposition 2: if

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is exact at M, then

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is exact as well.

proof: The first inclusion \mathrm{Im} (S^{-1}f) \subseteq \ker (S^{-1}g) is immediate since S^{-1}g \circ S^{-1}f=S^{-1}(g \circ f)=S^{-1}(0)=0. For the other direction, let m/s \in \ker(S^{-1}g). Then there is some t \in S so that 0=t\cdot g(m)=g(t \cdot m), so tm \in \ker (g)=\mathrm{Im}( f) by our hypothesis. Now, we simply take m^{\prime} \in M^{\prime} so that tm=f(m^{\prime}). But then m/s=f(m^{\prime})/st which was our definition for S^{-1}f(m^{\prime}/st) \in \mathrm{Im}(S^{-1} f) , proving the proposition.

With all of this sorted out we can finally prove that surjectivity is a local property, meaning that if it holds for localizations at every prime ideal, it holds for the entire ring! We will need one corollary from the previous proposition: quotients “commute” with taking localizations, or rather S^{-1}(M/N) \cong S^{-1}M/S^{-1}N, which can be shown by considering the sequence 0 \to N \to M \to M/N \to 0.

Here is the final lemma needed before our proof of the main theorem:

Lemma 3: As before, consider a  sequence of modules

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We claim that if

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is exact for every maximal ideal P of R, then the first sequence is exact as well.

Proof: Again the first inclusion \mathrm{Im} f \subseteq \ker g is clear. This means that \ker g /\mathrm{im} f is well defined, and is 0 at the localization, since (\ker g/ \mathrm{f})_P \cong (\ker g)_P/(\mathrm{Im}f)_P=\ker g_P / \mathrm{Im} f_P=0 by proposition 2 (applied twice) and the hypothesis of local exactness.

Thus, it will suffice to show that if M_P=0 for all maximal ideals, then M=0. This this will show that \ker g= \mathrm{Im} f.

I will not lie, this is an annoying verification, so the following is hasty: assume that 0 \neq m \in M exists so its annihilator, \mathrm{ann} (m), is contained in some maximal ideal P since \mathrm{ann}(m) \neq (1). If we localize at this maximal ideal, m/1 will be zero by hypothesis. But then m gets eaten by something outside of P, a contradiction.

Sigh! the deed is done, we have all that we require to show that surjectivity is local, since one need only remark that if the sequence

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is exact at every maximal ideal, so is the “non-localized version.”

We proceed now to the allegedly interesting proof of the main theorem:

Main Theorem: the map \phi: R \to \prod_{i=1}^{n} R/I_{i} given by

    \[x \mapsto (x+I_1,...,x+I_n)\]

is surjective whenever all I_i, I_j are pairwise coprime.

proof: localize at an arbitrary maximal ideal P. We now note that if none among the I_i belong to P, the map \phi_P is trivially surjective. On the other hand suppose that I_i \subseteq P, meaning that no other I_j among the family can also belong to P since this would contradict coprimality. Then, (\prod_{i=1}^{n} R/I_{i})_P \cong (R/I_i)_P \cong R_P/(I_i)_P. But then \phi_P is really just the natural surjection \phi_P: R_P \to R_P/(I_i)_P. Since P was arbitrary we have by lemma 3, that \phi is a surjective map.