I was wondering whether there are is a nice characterization of commutative rings $R$ where for ideals $I,J,K$ we always have: $$(I+K)\cap (J+K) = I\cap J +K $$


There is an extended version of the classical chinese remainder theorem in number theory which goes as follows:

Generalized Chinese Remainder theorem. If $a_1,a_2,...,a_n$ and $m_1,m_2,..,m_n$ are integers such that $a_i \equiv a_j \,\text{mod} \gcd(a_i,a_j)$ for all $i,j\in \{1,\dots,n\}$ then the system of equations $$ \begin{cases} x \equiv a_1 \mod m_1 \\ x \equiv a_2 \mod m_2 \\ \dots \\ x \equiv a_n \mod m_n \\ \end{cases}$$ Has a solution which is unique $\text{mod}\, \text{lcm}(a_1,\dots,a_n)$

The usual CRT has a general version in a commutative ring setting, and we can do the same for the generalized CRT:

Generalized CRT: Commutative Rings. Let $R$ be a commutative ring (with identity) and $I_1,\dots,I_n$ ideals of $R$. Then the canonical map $$\varphi : R \rightarrow R/I_1\times R/I_2\times \cdots R/I_n $$ has kernel $I_1\cap\dots\cap I_n$ and has image precisely all the $(\overline{a_1},\dots,\overline{a_n})$ with $a_k = a_l \,\text{mod}\, (I_k+I_l)$ for all $k,l \in \{1,\dots,n\}$

The case where $n=2$ is proven quite easily in the general case but when translating the proof of $\mathbb{Z}$ for $n\geq 3$ we make use of the fact that $$(I_1+I_n)\cap(I_2+I_n)\cap\dots\cap(I_{n-1}+I_n) = I_1\cap\dots\cap I_{n-1} + I_n$$ Which is true for ideals in $\mathbb{Z}$ but not in general! In fact, I noted that this is also a necessary condition: for a commutative ring $R$ the generalized CRT holds if and only if for all ideals $I,J,K$ of $R$ we have $$(I+K)\cap (J+K) = I\cap J +K$$

So knowing how to handle this last condition might be interesting.

Thoughts so far

The proof in $\mathbb{Z}$ boils down to the fact that $$\text{lcm}(\gcd(a,c),\gcd(b,c)) = \gcd(\text{lcm}(a,b),c)$$ for all integers $a,b,c$ which can be checked by looking at the exponents of the prime powers. Hence, since every PID is a UFD the equality holds for every PID with the same proof.

However, I know from this question that there exists counterexamples, even in the UFD $\mathbb{Z}[X]$ which is already quite disturbing. Does the equality hold in Dedekind rings, local rings?

Thanks in advance!

  • 3
    $\begingroup$ You should search Prüfer domain. (Or see [Bourbaki (1972), pp.558-559].) $\endgroup$ – Orat Feb 19 '17 at 12:40
  • 2
    $\begingroup$ This distributive law is one of many known laws that characterize Prufer domains, e.g.. see condition $(12)$ in this answer.. See also here for other examples with non-distributive lattices of ideals. See also this answer for the proof in $\,\Bbb Z.\ \ $ $\endgroup$ – Bill Dubuque Feb 19 '17 at 14:23
  • $\begingroup$ I've never seen a proof of CRT using the equation you mentioned. $\endgroup$ – user26857 Feb 19 '17 at 16:42
  • $\begingroup$ Hi, do you mean $(I+K)\cap (J+K) = I\cap (J +K )$ instead of $(I+K)\cap (J+K) = I\cap J +K $? $\endgroup$ – hasManyStupidQuestions Jun 4 at 16:25
  • $\begingroup$ No I don't, and note that plugging in $I = 0$ in your version gives the equation $K\cap (J+K) = 0 \cap (J+K)$ which implies that $K = 0$ so it couldn't possibly be true for all $I,J,K$ in any ring. $\endgroup$ – Jef L Jun 4 at 18:50

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