# Implications between "for any $r_1, r_2\in R$, there exists an unique $r$ in $R$ such that $r-r_1\in J_1$ and $r-r_2\in J_2$" and "$J_1+J_2=R$"

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.

• 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. Feb 6, 2022 at 20:43

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 $$r-a\in J_1$$.