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I was trying to understand the condition in proving the Chinese remainder theorem below

Let R be a ring and $I_1,\dots, I_m$ be ideals of R such that $I_j+\bigcap_{k\neq j}I_k=R$ for $0<j\leq m$. Then for $a_1,\dots,a_m\in R$, there always exist an $a\in R$ such that $a\equiv a_j$ (mod $I_j$).

The proof is quite simple and I knew how to exploit the condition to prove the claim based on the case when $m=2$, which basically means that $I_1, I_2$ are coprime ideals. But what I don't understand is why using the condition $I_j+\bigcap_{k\neq j}I_k=R$ in the statement instead of saying that any two ideals in $I_1,\dots, I_m$ are coprime. Can someone prove the claim using the latter or can you show that these two conditions are equivalent?

Arturo Magidin
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oscarmetal break
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1 Answers1

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The conditions are equivalent for unital rings (or for rings satisfying $R^2=R$).

Theorem. Let $R$ be a ring such that $R^2=R$, and let $I_1,\ldots,I_m$ be two-sided ideals. The following are equivalent:

  1. If $i\neq j$, then $I_i+I_j = R$ (the ideals are pairwise coprime);
  2. For each $j$, $I_j+\cap_{k\neq i}I_k = R$.

Proof. If 2 holds, then $$R = I_j+\cap_{k\neq i}I_k \subseteq I_j+ I_i \subseteq R,$$ so $I_j+I_i=R$.

Conversely, assume 1 holds. We proceed by induction on $m$. If $m=2$, the two statements are equal.

If $m=3$, and $I_1+I_2 = I_1+I_3 = R$, then $$\begin{align*} R=R^2 &= (I_1+I_2)(I_1+I_3)\\ &= I_1I_1 + I_1I_3 + I_2I_1+I_2I_3\\ &\subseteq I_1+I_2I_3\\ & \subseteq I_1+(I_2\cap I_3)\\ &\subseteq R, \end{align*}$$ so $I_1+(I_2\cap I_3)=R$. Similarly for $I_2+(I_1\cap I_3)$ and $I_3+(I_1\cap I_2)$.

Assume that the result holds for $k$ ideals and $m=k+1$. Then $$\begin{align*} R = R^2 &= (I_1+(I_2\cap\cdots\cap I_k))(I_1+I_{k+1})\\ &\subseteq I_1 + (I_2\cap\cdots \cdots \cap I_k)I_{k+1}\\ & \subseteq I_1+\cap_{j=2}^{k+1}I_j\\ &\subseteq R, \end{align*}$$ so we get the desired equality. $\Box$

You can weaken the condition $R^2=R$ to asking that $R=R^2+I_j$ for each $j$; this is what Hungerford does, for example. In the case of $m=3$, the argument proceeds as follows; as before $$R^2\subseteq I_1+(I_2\cap I_3).$$ Then we have $$R = I_1+R^2 \subseteq I_1+(I_1+(I_2\cap I_3)) = I_1+(I_2\cap I_3)\subseteq R,$$ proving the desired equality. The general case proceeds similarly.

This suggests that a problem may arise in a ring in which $R^2\neq R$ (and $R^2+I_j\neq R$ for at least one $j$). If $A$ is an abelian group and we define multiplication to be $ab=0$ for all $a,b\in A$, then ideals correspond to subgroups. We would be looking for three subgroups $B_1,B_2,B_3$ such that any two of them are comaximal, but the intersection of any two is contained in the third. An easy example is the Klein $4$-group $\mathbb{Z}_2\oplus\mathbb{Z}_2$, and take $B_1=\{(0,0),(1,0)\}$, $B_2=\{(0,0),(1,1)\}$, and $B_3=\{(0,0), (0,1)\}$. Then $B_1+B_2=B_1+B_3=B_2+B_3=\mathbb{Z}_2\oplus\mathbb{Z}_2$, but any two of them intersect trivially so $B_j+\cap_{i\neq j}B_i \neq \mathbb{Z}_2\oplus\mathbb{Z}_2$.

Arturo Magidin
  • 398,050
  • Thank you for the answer and the counter-example you attached for $R^2\neq R$. I wonder if it is often that $R^2\neq R$ appears in the context of abstract algebra since I wasn't warned about this when I was doing the problem. And this example that a ring is not being unital seems to not often be considered in the textbook, at least it wasn't until you posted your answer then I realized it is exitst. – oscarmetal break Nov 22 '23 at 18:25