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I am familiar with the Chinese Remainder Theorem:

Suppose that $R$ is a ring and $I_1, \dots, I_n$ are two-sided ideals of $R$. If any of the following conditions holds, then $$ R \Bigm{/} \bigcap_{i=1}^n I_i \cong \bigoplus_{i=1}^n R/I_i $$ For $1 \leq i,j \leq n$,

  1. $\forall i \neq j,\; I_i + I_j = R$
  2. $\forall i,\; I_i + \bigcap_{j \neq i} I_j = R$
  3. $\forall i,\; I_i + \bigcap_{j<i} I_j = R$.

So, for example, $\mathbb{R}[x,y] / (x^2-1) \cong \mathbb R[x,y] / (x-1) \oplus \mathbb{R}[x,y] / (x+1) \cong \mathbb{R}[x] \oplus \mathbb{R}[x]$ since $(x+1) - (x-1) = 2$ generate $\mathbb{R}[x,y]$.

But $\mathbb{R}[x,y] / (xy) \ncong \mathbb{R}[x,y] / (x) \oplus \mathbb{R}[x,y] / (y) \cong \mathbb{R}[x] \oplus \mathbb{R}[x]$ (it is evident from the geometry interpretation). And $(x) + (y) \neq \mathbb{R}[x,y]$, it naturally motivates the converse of CRT:

If $I,J$ are the ideals of $R$ and $R / (I \cap J) \cong R / I \oplus R / J$, then $I + J = R$. In fact, you can ask this question for all kinds of rings.

Maybe I have known that the converse holds for finite rings: the ring of integers, and matrix rings over rings with identity. Simple explanation:

  1. $|I + J| \, |I \cap J| = |I| \, |J|, \frac{|R|}{|I \cap J|} = \frac{|R|^2}{|I| \, |J|} \implies |I+J| = |R|$
  2. If $\mathbb{Z} / \operatorname{lcm}(m,n)\mathbb Z \cong \mathbb{Z} / (m) \oplus \mathbb{Z} / (n),\; \operatorname{lcm}(m,n) = mn \implies \gcd(m,n) = 1$
  3. the ideals of $M_n(R)$ have the form $M_n(I)$ where $I$ is the ideal of $R$.

However, there exist counterexamples like $\bigoplus_{i=1}^{\infty}\mathbb R$ and $\prod_{i=1}^{\infty} \mathbb{Z}$. In addition, a claimed but unproven result states that the converse holds for Dedekind domains. But I encounter a difficulty in applying the isomorphism condition (it's not always canonical). Therefore, is there a good way to determine this problem for polynomial rings?

Sammy Black
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PengHe Chen
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1 Answers1

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The implication "Isomorphism of $R$-algebras $\Rightarrow$ ideals coprime", that is, $$R/I\cap J \overset{\sim}{\to}R/I \times R/J \quad \implies \quad I+J=R$$ holds true in any commutative ring, see for example the answers to this question or [Atiyah--Macdonald, Proposition 1.10.ii)]. Note that there is only one map of $R$-algebras from left to right mapping $r$ to $(r,r)$ modulo the respective ideals.

Leobeth
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