The Chinese Remainder Theorem for Rings.
Let $R$ be a ring and $I$ and $J$ be ideals in $R$ such that $I+J =
R$.(a) Show that for any $r$ and $s$ in $R$, the system of equations
\begin{align*}
x & \equiv r \pmod{I} \\
x & \equiv s \pmod{J}
\end{align*}
has a solution.(b) In addition, prove that any two solutions of the system are congruent
modulo $I \cap J$.(c) Let $I$ and $J$ be ideals in a ring $R$ such that $I + J = R$. Show
that there exists a ring isomorphism
$$
R/(I \cap J) \cong R/I \times R/J.
$$
Solution:
(a) Let's remind ourselves that $I + J = \{i + j : i \in I, j \in J\}$.
Because $I + J = R$, there are $i \in I, j\in J$ with $i + j = 1$.
The solution of the system is $rj + si$. We check both equations:
\begin{align*}
rj + si &\equiv rj \equiv ri + rj \equiv r(i + j) \equiv r \pmod{I} \\
rj + si &\equiv si \equiv si + sj \equiv s(i + j) \equiv s \pmod{J} \, .
\end{align*}
(b) Assume we have two different solutions $x$ and $x'$. Then
\begin{align*}
x &\equiv x' \pmod{I} \\
x &\equiv x' \pmod{J} \, ,
\end{align*}
or else one of them wouldn't even be a solution. So $x – x'$ is in $I$ and $J$, therefore $x – x' \in I \cap J$ and $x\equiv x' \pmod{I \cap J}$.
(c) The Cartesian product of two rings is a ring, so $R/I \times R/J$ is a ring.
We look at the map
\begin{align*}
\phi: R &\rightarrow R/I \times R/J \\
x &\mapsto (x + I, x + J) \, .
\end{align*}
»Componentwise« ring homomorphisms are ring homomorphisms, so $\phi$ is a ring homomorphism.
$\phi$ is surjective: by (a) for any $r\in R/I, s\in R/J$ there exists an $x \in R$ with $\phi(x) = (r, s)$.
The kernel of $\phi$ are the solutions of the system for $r = s = 0$. By (b) every other solution must be congruent to $0$ modulo $I \cap J$, so $\ker \phi = I \cap J$.
Then by the first isomorphism theorem for rings $$R/\ker(\phi) \cong \phi(R)$$ we obtain $$R/(I \cap J) \cong R/I \times R/J \, .$$
Could you please check, if my solution is correct? Thank you!
Best Answer
Note that 1 is not necessarily in $R$ for me, so your discussion for (a) is a little bit mistaken. Here is my thought for (a), you can consider it:
We have $I+J=R$, so there exist $i \in I,j \in J$ such that $i+j=s-r$. Put $p=r+i=s-j$, and we get $p$ is a solution of the system since $p\in r+I$ and $p\in s+J$.
Note: $x\in r+I$ is equivalent to $x\equiv r \pmod I$