Let $\Bbb Z_2$ be the ring of $2$-adic integers. I want to show $\Bbb Z_2^2/2\Bbb Z_2^2 \cong (\Bbb Z_2/2\Bbb Z_2)^2$.
My approach:
I think the Chinese Remainder Theorem doesn't help to factor the quotient ring $\Bbb Z_2^2/2\Bbb Z_2^2 $.
On the other hand, if I define the map $h:\Bbb Z_2^2=\Bbb Z_2 \times \Bbb Z_2 \to (\Bbb Z_2/2\Bbb Z_2)^2=\Bbb Z_2/2\Bbb Z_2 \times \Bbb Z_2/2\Bbb Z_2$ by $$ h(a,b)=(a+2 \Bbb Z_2, b+2\Bbb Z_2),$$ then it is a homomorphism.
I want to show that $\ker(h)=2\Bbb Z_2^2=2(\Bbb Z_2 \times\Bbb Z_2)$.
Now $2\Bbb Z_2^2=2(\Bbb Z_2 \times\Bbb Z_2)=\{2(u,v): (u,v) \in \Bbb Z_2 \times\Bbb Z_2 \}$. Then, \begin{align} h(2(u,v)) =(2u+2\Bbb Z_2, 2v+2\Bbb Z_2)&=(0+2\Bbb Z_2,0+2\Bbb Z_2) \\ &=\text{zero element in} \ \Bbb Z_2/2\Bbb Z_2 \times\Bbb Z_2/2\Bbb Z_2. \end{align} If the above two lines are correct then we get the result.
Please help