I want to show that $ker(h)=2Bbb Z_2^2=2(Bbb Z_2 timesBbb Z_2)$

Let $Bbb Z_2$ be the ring of $2$-adic integers. I want to show $Bbb Z_2^2/2Bbb Z_2^2 cong (Bbb Z_2/2Bbb Z_2)^2$.

My approach:

I think the Chinese Remainder Theorem doesn’t help to factor the quotient ring $Bbb Z_2^2/2Bbb Z_2^2 $.

On the other hand, if I define the map $h:Bbb Z_2^2=Bbb Z_2 times Bbb Z_2 to (Bbb Z_2/2Bbb Z_2)^2=Bbb Z_2/2Bbb Z_2 times Bbb Z_2/2Bbb Z_2$ by $$ h(a,b)=(a+2 Bbb Z_2, b+2Bbb Z_2),$$ then it is a homomorphism.

I want to show that $ker(h)=2Bbb Z_2^2=2(Bbb Z_2 timesBbb Z_2)$.

Now $2Bbb Z_2^2=2(Bbb Z_2 timesBbb Z_2)={2(u,v): (u,v) in Bbb Z_2 timesBbb Z_2 }$. Then, begin{align} h(2(u,v)) =(2u+2Bbb Z_2, 2v+2Bbb Z_2)&=(0+2Bbb Z_2,0+2Bbb Z_2) \ &=text{zero element in} Bbb Z_2/2Bbb Z_2 timesBbb Z_2/2Bbb Z_2. end{align}
If the above two lines are correct then we get the result.

Please help