1. ## Ring isomorphism

Hello I want to solve the following problem:

Let R = Z( $sqrt2$ ) = { a+b $sqrt2$|a,b belong to Z.

1. Prove that R is integral domain.

2. If I = ( $sqrt 2$) is the ideal generated by $sqrt(2)$, prove that R/I~=Z2 (~= means isomorphic).

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In 1. I showed that elements of R are commutative and without zero-divisors.

In 2. I understood I as all elements of R multiplied by $sqrt(2)$) . Am I correct?

How shall I prove isomorphism? I tried the following function from R to Z2: f(a+ $sqrt2$b+I)= a mod 2. But it is only hommomorphism in my opinion, not isomorphism.

Can you help me?

2. Originally Posted by andreas
Hello I want to solve the following problem:

Let R = Z( $sqrt2$ ) = { a+b $sqrt2$|a,b belong to Z.

2. If I = ( $sqrt 2$) is the ideal generated by $sqrt(2)$, prove that R/I~=Z2 (~= means isomorphic).

In 2. I understood I as all elements of R multiplied by $sqrt(2)$) . Am I correct?

How shall I prove isomorphism? I tried the following function from R to Z2: f(a+ $sqrt2$b+I)= a mod 2. But it is only hommomorphism in my opinion, not isomorphism.
The mapping $f:R\to\mathbb{Z}_2$ defined by $f(a+b\sqrt{2})=a\,({\rm mod}\,2)$ is an homomorphism, and it is obviously surjective. As a consequence, it suffices to show that its kernel is $I$ to conclude that $R/I\simeq \mathbb{Z}_2$.
Its kernel consists of numbers $x=a+b\sqrt{2}$ where $a,b\in\mathbb{Z}$ and $a=2a'$ is even. Can you go on?

3. Thank you! It was very helpful!! I can go on