Let $R$ be a commutative ring and $J_1,J_2$ are two non-zero proper ideals of $R$. Consider the following two statements

$P:$ for any $r_1, r_2\in R$, there exists an unique $r$ in $R$ such that $r-r_1$ belongs to $J_1$ and $r-r_2$ belongs to $J_2$.

$Q:$ $J_1+J_2=R$.

Then which of the following options are correct?

$1$. Statement $P$ implies $Q$, but $Q$ does not implies $P$.

$2$. Statement $Q$ implies $P$, but $P$ does not implies $Q$.

$3$. Neither $P$ implies $Q$ nor $Q$ implies $P$.

$4$. Statement $P$ implies $Q$ and $Q$ implies $P$.

If I consider $R=\mathbb Z$ and ideals $2\mathbb Z$ and $3\mathbb Z$ then choose $r_1=2$ and $r_2=3$ so that the elements 6, 12,18 etc will work for $r$. So statement $Q$ does not imply $P$. I have no idea whether $P$ implies $Q$ or not. Please help me. Thank you.

  • 3
    $\begingroup$ Define a ring hom $\,h\, :\, R\to R/J_1\times R/J_2\,$ by $\,h(r) = (r+J_1,r+J_2).\,$ Then the existence of $\,r\,$ in statement $P$ is equivalent to $h$ is surjective (onto) which is equivalent to $J_1 + J_2 = (1),\,$ i.e. $J_1$ and $J_2$ are comaximal ("coprime"), and the uniqueness is equivalent to $\,h\,$ is injective ($1$-to-$1$) $ $ [$\!\!\iff\! J_1\cap J_2 = (0)\,$]. Combining we see that $P$ is equivalent to $R \,\cong\, R/J_1\times R/J_2$. This is CRT = Chinese Remainder Theorem in ideal form. $\endgroup$ Feb 6, 2022 at 20:43

1 Answer 1


Take some $a\in R$. To say that $a\in J_1+J_2$ amounts to saying that there exists some $r\in J_2$ such that $r-a\in J_1$. Can you prove that indeed this is equivalent?

Then try to find some $r_1,r_2\in R$ such that the conditions $r-r_1\in J_1$ and $r-r_2\in J_2$ are equivalent to $r\in J_2$ and $r-a\in J_1$.


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