- #1

Mr Davis 97

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## Homework Statement

Let ##a,b## be squarefree integers and set ##R = \mathbb{Z}[\sqrt{a}]## and ##S = \mathbb{Z}[\sqrt{b}]##. Prove that

a) There is an isomorphism of abelian groups ##(R,+) \cong (S,+)##.

b) There is an isomorphism of rings ##R\cong S## if and only if ##a=b##.

## Homework Equations

## The Attempt at a Solution

a) Proof: Let ##\varphi : R \to \mathbb{Z} \times \mathbb{Z}## be a map of groups such that ##\varphi(x+y\sqrt{a}) = (x,y)##. This is a homomorphism since if ##x+y\sqrt{a}, w+z\sqrt{a} \in R##, then $$\varphi ((x+y\sqrt{a}) + (w+z\sqrt{a})) = \varphi((x+w) + (y+z)\sqrt{a}) = (x+w,y+z) = (x,y)+(w,z) = \varphi(x+y\sqrt{a}) + \varphi(w+z\sqrt{a}).$$

Now, let ##x+y\sqrt{a}\in \ker(\varphi)##. Then ##\varphi(x+y\sqrt{a}) = (x,y) = 0##, which is true if and only if ##x=0## and ##y=0##, i.e. ##x+y\sqrt{a} = 0##, and so the kernel is trivial, and the map is injective.

The map is certainly surjective since ##\varphi(x+y\sqrt{a}) = (x,y)##, and ##(x,y)## is a general element of ##\mathbb{Z} \times \mathbb{Z}##. Hence ##R \cong \mathbb{Z} \times \mathbb{Z}##.

Note that this proof did not depend on the value of ##a##, only that it was squarefree. Hence we also have that ##S \cong \mathbb{Z} \times \mathbb{Z}##, and so ##R \cong S## as groups under addition.

b) I think that the proof for a) is correct, but I am not sure where to start here. Clear if ##a=b##, then ##R\cong S##, so it is the other direction that I am confused by. How should I get started with this direction?