Proof of ideals and the Chinese remainder theorem

Question

Let $I$ and $J$ be ideals of a ring $R$. Prove that the pair of congruences $y\equiv r\,\mathrm{mod}\,I$ and $y\equiv s\,\mathrm{mod}\,J$ has a solution if and only if $r\equiv s\,\mathrm{mod}\,\, I+J$.

Answer 1

Hints:

For the forward direction, note that $y-r \in I$ and $y-s \in J$. Hence, $-(y-r)+(y-s) \in I+J$ and so...

For the reverse direction, suppose $r\equiv s \pmod{I+J}$. Then $r - s = i + j$ for some $i \in I$, $j \in J$, or $r - (s+j) = i$. Now take $y = s+j$. Then $y \equiv s \pmod J$, and $y \equiv r \pmod I$.