# Generalised Chinese Reminder Theorem - How to compute the cokernel?

Let $$R$$ be a commutative ring of dimension one with minimal prime ideals $$P_1,\ldots,P_n$$. We have the canonical injective map $$\phi_n: R/(P_1 \cap \ldots \cap P_n) \to \prod_{i=1}^n R/P_i.$$

My question is: Is there a formula for the cokernel of $$\phi_n$$? For instance as a product of rings again.

The case $$n=2$$ is well known and here $$\operatorname{coker}(\phi_2) \cong R/(P_1+P_2)$$. That this does not generalize to $$\operatorname{coker}(\phi_n) \cong \prod_{i < j}^n R/(P_i+P_j)$$ can be seen by a counter-example, see this answer.

I am grateful for any kind of insights, references and proofs!

• Just make sure - what does your CRT stand for? – wonderich May 24 at 15:49
• @wonderich. I suppose "Chinese Remainder Theorem"! – Kapil May 24 at 16:08
• I windsheaf, I would say the better (more geometric) generalization of the CRT is to ask what goes in the kernel, not what the meaning of the cokernel is. – Karl Schwede May 27 at 20:13
• @KarlSchwede Sure, that's true! – windsheaf May 28 at 5:54
• cokernel as which structure? The image of $\phi_n$ consists of $(r+P_1,\ldots, r+P_n)$ for $r \in R$. In general this is no ideal of $R/P_1 \times \cdots \times R/P_n$ and the cokernel has no natural ring structure. – tj_ Aug 14 at 6:40

A more general version of your question was asked and answered on mathoverflow 9 years ago. When we specialize that answer to your setting, the family $$\mathcal{F}$$ is $$\{P_1,\dots,P_n\}$$, and the answer gives a sheaf-theoretic characterization of the cokernel of $$\phi$$,

$$O(\mathcal{F}) = (\prod_{i=1}^n \limits R/P_i)/\phi(R/(P_1\cap\dots\cap P_n))$$

This paper also gives examples where the cokernel vanishes (i.e. the CRT isomorphism holds) and where it doesn't.

• I have seen this question and the answer to it and I also looked into the mentioned paper. But it does not answer the question, at least from my point of view. I was hoping that the sheaf theoretic interpretation provides a non-trivial answer to what requirements I need to impose on the tuples such that they provide a preimage under $\phi$. Did I miss something here? – windsheaf May 28 at 6:00
• @windsheaf In terms of what to ask for, I assume its some seminormality type condition for the various unions (strata). – Karl Schwede May 31 at 22:13

Let us first observe that $$M \Doteq \text{coker}(\phi_n)$$ is naturally an $$R/\text{rad}(R)$$-module with $$\text{rad}(R) \Doteq P_1 \cap \cdots \cap P_n.$$ So, there is no actual loss in generality if we suppose that $$R$$ is reduced, i.e., $$\text{rad}(R) = \{0\}$$: if there is a formula for $$M$$ as an $$R/\text{rad}(R)$$-module, we should be able to derive a formula for $$M$$ taken as an $$R$$-module by re-injecting $$\text{rad}(R)$$.

Let us assume that $$R$$ is reduced and let us identify $$R$$ with its image by $$\phi_n$$.

There is of course a formula that expresses $$M \Doteq \text{coker}(\phi_n) = \left(\prod_{i = 1}^nR/P_i\right)/R$$ as a function of the ideals $$P_i$$. This is for instance the following presentation of $$M$$ over $$R$$

$$M = \left\langle e_1, \dots, e_n \, \vert \, \sum_{j = 1}^n e_j = P_ie_i = 0,\quad i = 1, \dots, n \right\rangle$$ where $$e_i$$ corresponds to the identity element of $$R/P_i$$.

You may object that it doesn't to tell much about the structure of $$M$$ and I would agree. Still, note that it makes clear that $$M$$ can be generated by $$n - 1$$ elements, so that if $$n = 2$$, we see almost immediately that $$M$$ is the cyclic $$R$$-module $$R/(P_1 + P_2)$$.

Now comes the bad news. In general, the $$R$$-module $$M$$ does not split as a direct sum of factor rings of $$R$$, or equivalently, cyclic submodules of $$M$$. Here is a counter-example:

Claim 1. Let $$R$$ be the integral group ring of $$C_4$$, the cyclic group with $$4$$ elements, that is, $$R = \mathbb{Z}[C_4] = \mathbb{Z}[X]/(X^4 - 1)$$. Let $$\tilde{R}$$ be the integral closure of $$R$$, that is, $$\tilde{R} = \mathbb{Z}[X]/(X - 1) \times \mathbb{Z}[X]/(X + 1) \times \mathbb{Z}[X]/(X^2 + 1) \simeq \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[i]$$. Then $$M \Doteq \text{coker}(\phi_3) = \tilde{R}/R$$ is a non-cyclic two-generated $$R$$-module which is indecomposable.

According to [1, pages 19--20], the integral group ring $$R = \mathbb{Z}[G]$$ of a finite Abelian group $$G$$ is always Gorenstein, but it is a Bass ring if and only if the cardinality $$\vert G \vert$$ of $$G$$ is square-free. By a Bass ring, we mean a Noetherian reduced commutative unital ring $$R$$ of Krull dimension one and such that its integral closure $$\tilde{R}$$ is a cyclic $$R$$-module.

Thus the fact that the module $$M$$ of Claim 1 is not cyclic is already predicted by Bass's theorem, see [2, Theorem 2.1] and the rings $$\mathbb{Z}[G]$$ may yield further counter-examples.

Proof of Claim 1. Let us denote by $$x$$ the image of $$X$$ in $$R$$ and set $$P_1 = R(x - 1), P_2 = R(x + 1)$$ and $$P_3 = R(x^2 + 1)$$. We denote by $$\epsilon_i$$ the identity element of $$R/P_i$$ and by $$e_i$$ its image in $$M$$ for $$i = 1, 2, 3$$. It is easily checked that $$M \simeq \left(R/P_1 \times R/P_2 \right) /P_3(\epsilon_1 + \epsilon_2)$$ and that $$P_3(\epsilon_1 + \epsilon_2) = \mathbb{Z} \cdot 2(\epsilon_1 + \epsilon_2) + \mathbb{Z} \cdot 2(\epsilon_1 - \epsilon_2)$$. As a result, $$M$$ has the following $$R$$-module presentation: $$M = \langle e_1, e_2 \, \vert \, 4e_1 = 4e_2 = 2(e_1 + e_2) = 0 \rangle.$$ In particular $$M$$ is isomorphic to $$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$$ as an Abelian group and we can take $$e \Doteq e_1 + e_2$$ as the canonical generator of $$\mathbb{Z}/2\mathbb{Z}$$ and any of $$e_1$$ or $$e_2$$ as the canonical generator of $$\mathbb{Z}/4\mathbb{Z}$$.

Now we claim that:

• $$M$$ is not a cyclic $$R$$-module.
• $$Re_1, Re$$ and $$Re_2$$ are the only $$R$$-submodules of $$M$$ with $$4$$ elements.
• $$R \cdot 2e_1 = R \cdot 2e_2$$ is the only $$R$$-submodule of $$M$$ with $$2$$ elements.

The fact that $$R \cdot 2e_1$$ is the unique minimal non-zero $$R$$-submodule of $$M$$ proves instantly that $$M$$ is indecomposable. For the first assertion, let us reason by contradiction, assuming that $$M = Rf$$ with $$f = ae + be_1, a, b \in \mathbb{Z}$$. Then $$M$$ is generated as a $$\mathbb{Z}$$-module by $$f$$ and $$xf = -ae + (2a + b)e_1$$, so that the $$\mathbb{Z}$$-module $$M/(R \cdot 2e_1) \simeq_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$$ is generated by the image of $$f$$, a contradiction.

The last two assertions are routine. $$\square$$

Remark. In the answer to your MSE post, the following example was considered. Let $$R = \mathbb{C}[X, Y]/XY(X - Y)$$, with minimal primes $$P_1 = Rx, P_2 = Ry$$ and $$P_3 = R(x - y)$$ where $$x, y$$ are the images of $$X$$ and $$Y$$ in $$R$$. Then the integral closure of $$R$$ is $$\tilde{R} = R/P_1 \times R/P_2 \times R/P_3$$ and it is not difficult to show that $$\tilde{R}/R \simeq (R/(P_1 + P_2))^2.$$ Indeed, we have $$\tilde{R}/R \simeq \left(R/P_1 \times R/P_2 \right) /P_3(\epsilon_1 + \epsilon_2)$$ and $$P_3(\epsilon_1 + \epsilon_2) = P_2\epsilon_1 + P_1\epsilon_2$$. Hence this $$M \Doteq \tilde{R}/R$$ does split as a direct sum of two indecomposable cyclic $$R$$-modules.

On the positive side, we have:

Claim 2. Assume that $$M \Doteq \text{coker}(\phi_n)$$ can be generated by $$d \le n - 1$$ over $$R$$. Then we have $$\text{Fitt}_0(M) \subseteq \text{ann}(M)$$ and $$\text{ann}(M)^{d} \subseteq \text{Fitt}_0(M)$$ where the Fitting ideal $$\text{Fitt}_0(M)$$ is given by $$\text{Fitt}_0(M) = \sum_{1 \le i_1 < \cdots < i_{n - 1} \le n} P_{i_1} \cdots P_{i_{n - 1}}.$$ In particular, $$\text{ann}(M) = \text{Fitt}_0(M)$$ if $$M$$ can be generated by one element over $$R$$.

Proof. This is [3, Proposition 20.7].

[1] H. Bass, "On the ubiquity of Gorenstein rings", 1963.
[2] L. Levy, R. Wiegand, "Dedekind-like behaviors of rings with $$2$$-generated ideals", 1985.
[3] D. Eisenbud, "Commutative Algebra with a View Toward Algebraic Geometry", 1995.