# Chinese Remainder Theorem induces a unit group isomorphism

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.

• If $A \cong B \times C$, then can you prove that $A^{\times} \cong B^{\times} \times C^{\times}$? – Alex Wertheim Feb 17 '16 at 7:51

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.