Showing when quadratic integer rings are isomorphic

[Mr Davis 97]

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?

 

[fresh_42]

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?

Always start with what you have. No assumptions, not jumping to conclusions, just a gathering of facts.

So let's assume we have an isomorphism $\varphi\, : \,R=\mathbb{Z}[\sqrt{a}] \longrightarrow \mathbb{Z}[\sqrt{b}] =S$.
I would start to find out, what $\varphi(\sqrt{a})$ can possibly be. Once we know that, we can consider $\varphi((x+y\sqrt{a})\cdot (u+v\sqrt{b}))$.

We already know that $\varphi(0)=0$ and $\varphi(1)=1$ because both rings are integral domains. We also know, that $(\sqrt{a})^2-a=0$.

I would start with this.

 

[Mr Davis 97]

Always start with what you have. No assumptions, not jumping to conclusions, just a gathering of facts.

So let's assume we have an isomorphism $\varphi\, : \,R=\mathbb{Z}[\sqrt{a}] \longrightarrow \mathbb{Z}[\sqrt{b}] =S$.
I would start to find out, what $\varphi(\sqrt{a})$ can possibly be. Once we know that, we can consider $\varphi((x+y\sqrt{a})\cdot (u+v\sqrt{b}))$.

We already know that $\varphi(0)=0$ and $\varphi(1)=1$ because both rings are integral domains. We also know, that $(\sqrt{a})^2-a=0$.

I would start with this.

Here is some of my reasoning.

If $\varphi (1) = 1$, then $\varphi (a) = a$, since $\varphi$ is a homomorphism, and $a$ is just an integer.

Now, suppose that $\varphi (\sqrt{a}) = r + s\sqrt{b}$ for some $r,s\in \mathbb{Z}$. Then, squaring both sides, we get $\varphi (\sqrt{a})^2 = r^2 + 2rs\sqrt{b} + bs^2$. Now we can use the homomorphism property to put the square inside, so that $\varphi (a) = r^2 + 2rs\sqrt{b} + bs^2$, and so $a= r^2 + 2rs\sqrt{b} + bs^2$. We have to somehow eliminate the $\sqrt{b}$ term on the RHS, since the LHS is just an integer, and we can do this in two ways: either $s=0$ or $r=0$. In the first case we get that $a = r^2$, and in the second case we get $a=bs^2$, but I'm not exactly sure what to do with either of these equations, or whether I am on the right track or not...

 

[fresh_42]

If $\varphi (1) = 1$, then $\varphi (a) = a$, since $\varphi$ is a homomorphism, and $a$ is just an integer.

O.k., let me see.

$\varphi(a)=\varphi(1\cdot a)=\varphi(1)\cdot \varphi(a) \Longrightarrow 0=\varphi(a)\cdot (1-\varphi(1)) \stackrel{(*)}{\Longrightarrow} 1=\varphi(1)$
$(*)$ since $0 \neq S$ is an integral domain and thus $\varphi(a)=\underbrace{\varphi(1)+\ldots +\varphi(1)}_{a\text{ times }}=\underbrace{1+\ldots +1}_{a\text{ times }}=a\,.$

Now, suppose that $\varphi (\sqrt{a}) = r + s\sqrt{b}$ for some $r,s\in \mathbb{Z}$. Then, squaring both sides, we get $\varphi (\sqrt{a})^2 = r^2 + 2rs\sqrt{b} + bs^2.$ Now we can use the homomorphism property to put the square inside, so that $\varphi (a) = r^2 + 2rs\sqrt{b} + bs^2,$ and so $a= r^2 + 2rs\sqrt{b} + bs^2.$ We have to somehow eliminate the $\sqrt{b}$ term on the RHS, since the LHS is just an integer, and we can do this in two ways: either $s=0$ or $r=0.$ In the first case we get that $a = r^2,$ and in the second case we get $a=bs^2,$ but I'm not exactly sure what to do with either of these equations, or whether I am on the right track or not...

From $s=0$ we get $\varphi(\sqrt{a})=r$ and so $\varphi^{-1}(r)=r=\sqrt{a}$ by the same argument as in the beginning, this time for $\varphi^{-1}$, but $\sqrt{a}\notin \mathbb{Z}$.
From $r=0$ we get - as you said - $a=bs^2$. However, $a$ is squarefree, so $s=\pm 1$.

 

[Mr Davis 97]

O.k., let me see.

$\varphi(a)=\varphi(1\cdot a)=\varphi(1)\cdot \varphi(a) \Longrightarrow 0=\varphi(a)\cdot (1-\varphi(1)) \stackrel{(*)}{\Longrightarrow} 1=\varphi(1)$
$(*)$ since $0 \neq S$ is an integral domain and thus $\varphi(a)=\underbrace{\varphi(1)+\ldots +\varphi(1)}_{a\text{ times }}=\underbrace{1+\ldots +1}_{a\text{ times }}=a\,.$From $s=0$ we get $\varphi(\sqrt{a})=r$ and so $\varphi^{-1}(r)=r=\sqrt{a}$ by the same argument as in the beginning, this time for $\varphi^{-1}$, but $\sqrt{a}\notin \mathbb{Z}$.
From $r=0$ we get - as you said - $a=bs^2$. However, $a$ is squarefree, so $s=\pm 1$.

So am I done, since what I wanted to show was that $a=b$? (And given that is there a way to eliminate the $s=-1$ case?)

 

[fresh_42]

So am I done, since what I wanted to show was that $a=b$? (And given that is there a way to eliminate the $s=-1$ case?)

That's easy to check, isn't it. Is $\varphi(x+y\sqrt{a})=x-y\sqrt{b}$ a ring isomorphism or not? My guess is, it is, so they are two possibilities, but I haven't done the math. But $\mathbb{Z}[\sqrt{a}]=\mathbb{Z}[-\sqrt{a}]$ is strong evidence.

 

[Mr Davis 97]

That's easy to check, isn't it. Is $\varphi(x+y\sqrt{a})=x-y\sqrt{b}$ a ring isomorphism or not? My guess is, it is, so they are two possibilities, but I haven't done the math. But $\mathbb{Z}[\sqrt{a}]=\mathbb{Z}[-\sqrt{a}]$ is strong evidence.

So would the following be a complete solution to part b)?

The reverse direction is easy, since if $a=b$ then by the reflexive property of isomorphisms, $\mathbb{Z}[\sqrt{a}] \cong \mathbb{Z}[\sqrt{b}]$, so we will focus on the other direction. Suppose that $\mathbb{Z}[\sqrt{a}] \cong \mathbb{Z}[\sqrt{b}]$. We want to show that $a=b$. Let $\phi : R \to S$ be an isomorphism. Since both $R$ and $S$ are integral domains, we know that $\phi (1) = 1$. Hence we then know that $\phi(a)=\underbrace{\phi(1)+\ldots +\phi(1)}_{a\text{ times }}=\underbrace{1+\ldots +1}_{a\text{ times }}=a\,$. Now, suppose that $\varphi (\sqrt{a}) = r + s\sqrt{b}$ for some $r,s\in \mathbb{Z}$. Then, squaring both sides, we get $\varphi (\sqrt{a})^2 = r^2 + 2rs\sqrt{b} + bs^2$. Now we can use the homomorphism property to put the square inside, so that $\varphi (a) = r^2 + 2rs\sqrt{b} + bs^2$, and so $a= r^2 + 2rs\sqrt{b} + bs^2$. We have to somehow eliminate the $\sqrt{b}$ term on the RHS, since the LHS is just an integer, and we can do this in two ways: either $s=0$ or $r=0$. In the first case we get that $a = r^2$, and in the second case we get $a=bs^2$. If $a=r^2$, then $\phi (\sqrt{a}) = r$, and so $\phi^{-1}(r) = \sqrt{a}$. But this is a contradiction since $\phi^{-1}$ is a homomorphism and so must map to another integer. So we must have that $a=bs^2$. But it was given that $a$ is squarefree, so it must be the case that $s =\pm 1$. Either way, we have that $a=b$.

 

[fresh_42]

So would the following be a complete solution to part b)?

The reverse direction is easy, since if $a=b$ then by the reflexive property of isomorphisms, $\mathbb{Z}[\sqrt{a}] \cong \mathbb{Z}[\sqrt{b}]$, so we will focus on the other direction. Suppose that $\mathbb{Z}[\sqrt{a}] \cong \mathbb{Z}[\sqrt{b}]$. We want to show that $a=b$. Let $\phi : R \to S$ be an isomorphism. Since both $R$ and $S$ are integral domains, we know that $\phi (1) = 1$. Hence we then know that $\phi(a)=\underbrace{\phi(1)+\ldots +\phi(1)}_{a\text{ times }}=\underbrace{1+\ldots +1}_{a\text{ times }}=a\,$. Now, suppose that $\varphi (\sqrt{a}) = r + s\sqrt{b}$ for some $r,s\in \mathbb{Z}$. Then, squaring both sides, we get $\varphi (\sqrt{a})^2 = r^2 + 2rs\sqrt{b} + bs^2$. Now we can use the homomorphism property to put the square inside, so that $\varphi (a) = r^2 + 2rs\sqrt{b} + bs^2$, and so $a= r^2 + 2rs\sqrt{b} + bs^2$.

The last three occurrences of $\varphi$ should probably be $\phi$.

We have to somehow eliminate the $\sqrt{b}$ term on the RHS, since the LHS is just an integer, and we can do this in two ways: either $s=0$ or $r=0$. In the first case we get that $a = r^2$, and in the second case we get $a=bs^2$. If $a=r^2$, then $\phi (\sqrt{a}) = r$, and so $\phi^{-1}(r) = \sqrt{a}.$

I wouldn't use this, as it looks like taking the square root, and I don't see at once that this is without trouble, resp. further argumentation. You can either use that $a$ is squarefree so it can't be equal to $r^2$, or directly that by $s=0$ we have $\phi(\sqrt{a})=r$ and thus $\phi^{-1}(r)=r=\sqrt{a}$ by the previous argument, that all integers are fix points of $\phi$ and accordingly of $\phi^{-1}$ which also leads to the squarefree-ness of $a$ in a way.

I don't immediately get your argument: $\phi(\sqrt{a})^2=r^2 \Longrightarrow \phi(\sqrt{a})=r$ why this has to be true. Theoretically there can be plenty of elements which squared resulted in $a$ or $\phi(a)$. And even if it was only $r$, there is still the $-r$ case to be mentioned. Saying $a=r^2 \in \mathbb{Z}$ can't be, since $a$ is squarefree is much easier.

But this is a contradiction since $\phi^{-1}$ is a homomorphism and so must map to another integer. So we must have that $a=bs^2$. But it was given that $a$ is squarefree, so it must be the case that $s =\pm 1$. Either way, we have that $a=b$.

So the only thing I didn't understand was "If $a=r^2$ then $\phi(\sqrt{a})=r.$"
If you had written "If $s=0$ then $\phi(\sqrt{a})=r$" everything would have been perfect. Except the "$\varphi$" typo of course.