# Showing an isomorphism between two quadratic rings

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

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

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.

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

This is the part that I am a bit confused on. The reverse direction is clear, since if $$a=b$$ they are just the same group so are of course isomorphic. It's the forward direction that I am stuck on.

• That's a command not a question. What work have you done on the problem? – Rob Arthan Dec 9 '18 at 1:04
• Hint: Suppose $f\colon R\to S$ is a ring homomorphism. Since $f(1)^2 = f(1)$, either $f(1)=1$ or $f(1)=0$. If $f(1)=0$, then... If $f(1)=1$, then let $f(\sqrt{a}) = r+s\sqrt{b}$ for some $r,s\in\mathbb{Z}$, and consider $(r+s\sqrt{b})^2$. – Arturo Magidin Dec 9 '18 at 2:32

Hint 1: as abelian groups, $$\mathbb{Z}[\sqrt{a}]\cong\mathbb{Z}\oplus\mathbb{Z}$$.
Hint 2: what integers have a square root in $$\mathbb{Z}[\sqrt{a}]$$?