A consequence of the ideal theoretic version of Chinese Remainder Theorem Let $I_i$, $1\leq i\leq n$, be comaximal ideals in a commutative ring $R$ and $I=\cap _{1\leq i \leq n}I_i$. Prove that $(R/I)^\times$ is isomorphic to $(R/I_1)^\times \times \cdots \times (R/I_n)^\times$. ($(R/I)^\times$ means the invertible elements of the quotient ring.)
 A: Hints:


*

*If $R$ and $S$ are isomorphic rings, then $R^\times\cong S^\times$ (unit group is functorial).

*$(R\times S)^\times=R^\times\times S^\times$ (literally equals).
A: Could you "guess" what would be the natural map between $R/I$ and $R/I_1 \times R/I_2 \times \dots \times R/I_n $

$\eta :R/I \rightarrow R/I_1 \times R/I_2 \times \dots \times R/I_n $ with 
  $x+I\rightarrow (x+I_1,x+I_2,\dots,x+I_n)$

do you atleast see that $\eta$ is injective...
suppose $\eta(x+I)=0\in R/I_1 \times R/I_2 \times \dots \times R/I_n$
i.e., $(x+I_1,x+I_2,\dots,x+I_n)=(0+I_1,0+I_2,\dots,0+I_n)$ 
i.e., $x+I_1=0+I_1, x+I_2=0+I_2,\dots x+I_n=0+I_n$ i.e., $x\in I_i$ for all $1\leq i\leq n$
i.e.,  $x\in \cap_{1\leq i\leq n} I_i$ as we know that $\cap_{1\leq i\leq n} I_i=I$,we see that $x\in I$ 
i.e., $x+I=0+I$ i.e., $\eta$ is injective.
So, we now have a hope that it could "possibly be surjective" and thus ISOMORPHISM 
(In above step I assume you know how to prove $\eta$ is homomorphism)
let us try proving this first for $n=2$ and hope induction works neatly...
we now thus assume $I_1,I_2$ are comaximal and $I=I_1\cap I_2$.
As they are comaximal, we have $R=I_1+I_2$ i.e, for some $x\in I_1, y\in I_2$ we have $x+y=1$
let us see where does $x\in I_1$ and $y\in I_2$ goes under $\eta$ we have $\eta (x)=(x+I_1,x+I_2)$... 
As $x$ is already in $I_1$ we have $x+I_1=0+I_1$
as $x+y=1$ we have $x=1-y$ and thus $x+I_2=1-y+I_2$ 
we know that $y\in I_2$ thus $y+I_2=0+I_2$ i.e., $x+I_2=1-y+I_2=1+I_2$ 
So, we have $\eta(x)=(x+I_1,x+I_2)=(0+I_1,1+I_2)$ 
For similar reasons, we see that $\eta(y)=(y+I_1,y+I_2)=(1+I_1,0+I_2)$ 
Now, take an arbitrary element $(r_1+I_1,r_2+I_2) \in R/I_1\times R/I_2$
we ant to know where does $(r_1+I_1,r_2+I_2)$ comes from under $\eta$.
we have $(r_1+I_1,r_2+I_2)=(0+I_1,r_2+I_2)+(r_1+I_1,0+I_2)$
Now,  

$(0+I_1,r_2+I_2)=(r_2+I_1,r_2+I_2)(0+I_1,1+I_2)$
  $(r_1+I_1,0+I_2)=(r_1+I_1,r_1+I_2)(1+I_1,0+I_2)$

So, $(r_1+I_1,r_2+I_2)=(r_2+I_1,r_2+I_2)(0+I_1,1+I_2)+(r_1+I_1,r_1+I_2)(1+I_1,0+I_2)$
and we know that $\eta(x)=(x+I_1,x+I_2)=(0+I_1,1+I_2)$ and $\eta(y)=(y+I_1,y+I_2)=(1+I_1,0+I_2)$ 
So, we now have 
$(r_1+I_1,r_2+I_2)=(r_2+I_1,r_2+I_2)(0+I_1,1+I_2)+(r_1+I_1,r_1+I_2)(1+I_1,0+I_2)=\eta(r_2)\eta(x)+\eta(r_1)\eta(y)=\eta(r_2x+r_1y)$
So, $(r_1+I_1,r_2+I_2)$ comes from $r_2x+r_1y\in R$ (under $\eta$) .
Thus we see that $\eta : R/I \rightarrow R/I_1 \times R/I_2$ is surjective and hence ISOMORPHISM 
(Injectivity is proved already and homomorphism is left as easy to see condition)
i.e., we have $R/I\cong R/I_1\times R/I_2$
Now, for induction process, if you have $I_i : 1\leq i\leq n$ the, take $I_1=I_1$ and $I_2\cap I_3\cap \dots I_n= I_m$ then you will get $R/I\cong R/I_1\times R/I_m$ 
now repreat the same for $m$ at last you will end up with the case that
$R/I\cong R/I_1 \times R/I_2 \times \dots \times R/I_n $ and thus, we are done.
P.S : as Mr.Alex says, 

If $R\cong S$ then $R^*\cong S^*$ 
  $(R\times S)^*=R^* \times S^*$ 

and if we apply this to 
$R/I\cong R/I_1 \times R/I_2 \times \dots \times R/I_n $
we get $(R/I)^*\cong (R/I_1)^* \times (R/I_2)^* \times \dots \times (R/I_n)^* $ and
now it is complete :)
I prefer using $R^*$ for invertible elements of $R$ and please excuse me if that bothers you :)
