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

Mathematics Asked by Masmath on December 20, 2020

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.

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