The Chinese Remainder Theorem for rings states that if $R$ is a commutative ring with $I_1,\ldots, I_n$ ideals that are comaximal, i.e., $I_i+I_j=R$ if $i \neq j$, then the canonical map $\phi:R \rightarrow R/I_1 \times\ldots \times R/I_n$ induces a ring isomorphism:$$R/(I_1 \ldots I_n) \cong R/I_1 \times\ldots \times R/I_n$$ My question is, does the ring isomorphism also implies an isomorphism of multiplicative groups: $$(R/(I_1 \ldots I_n))^{\times} \cong (R/I_1)^{\times} \times\ldots \times (R/I_n)^{\times}$$ If so, is there an elegant way of showing this? That is, a method without going through the argument we prove the ring isomorphism using the first isomorphism theorem.

  • 1
    $\begingroup$ If $A \cong B \times C$, then can you prove that $A^{\times} \cong B^{\times} \times C^{\times}$? $\endgroup$ Feb 17, 2016 at 7:51

1 Answer 1


In general, if you know $R\cong S$, where $R$ and $S$ are rings, then by definition you also have $R^\times\cong S^\times$. Hence, to solve your question, all you need to do is show that for rings $R_1,\ldots,R_n$ we have$$(R_1\times\ldots\times R_n)^\times\cong R_1^\times\times\ldots\times R_n^\times.$$This, too, can be proved directly from the very basic definitions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.