Exercise: 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 $$R/(I \cap J) \cong R/I \times R/J \, .$$ $\square$

Could you please check, if my solution is correct? Thank you!


Yes, your solution appears to be complete and correct.

  • $\begingroup$ Is there a solution where you don't assume a ring has multiplicative identity? $\endgroup$ – Yunus Syed Feb 28 '16 at 2:28
  • $\begingroup$ @YunusSyed I'm not aware of a version like that, no. $\endgroup$ – rschwieb Feb 28 '16 at 3:08
  • $\begingroup$ My textbook has the exact same problem but doesn't assume rings have unity. $\endgroup$ – Yunus Syed Feb 28 '16 at 3:09
  • $\begingroup$ @YunusSyed Which text? Perhaps you can adjoin a $1$, use this version, then descend to your ring without unity. $\endgroup$ – rschwieb Feb 28 '16 at 3:13
  • $\begingroup$ Judson "Abstract algebra theory and Applications" $\endgroup$ – Yunus Syed Feb 28 '16 at 3:14

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