Showing when quadratic integer rings are isomorphic

• Mr Davis 97
In summary: 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##...?)Yes, you are right. You get ##b=s^2##, so you have your isomorphism ##\varphi : \mathbb{Z}[\sqrt{a}] \longrightarrow \mathbb{Z}[\sqrt{b}]##. In summary, The given conversation discusses the setting of ##a## and ##b## as squarefree integers and the sets ##R = \mathbb{Z}[\sqrt{a}]## and ##S = \mathbb{Z}[\sqrt{b}]##. The conversation then proceeds
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##.

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?

Mr Davis 97 said:
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##.

fresh_42 said:
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##.

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...

Mr Davis 97 said:
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\,.##

Mr Davis 97 said:
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##.

fresh_42 said:
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?)

Mr Davis 97 said:
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
fresh_42 said:
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##.

Mr Davis 97 said:
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.

Mr Davis 97

1. What is a quadratic integer ring?

A quadratic integer ring is a mathematical structure that is constructed using a quadratic polynomial and a set of integers. It is similar to a polynomial ring, but with the added restriction that the coefficients must be integers.

2. What does it mean for two quadratic integer rings to be isomorphic?

Two quadratic integer rings are isomorphic if they have the same structure, but their elements may be labeled differently. This means that there is a one-to-one correspondence between the elements of the two rings, and all mathematical operations will yield the same results.

3. How can you tell if two quadratic integer rings are isomorphic?

There are a few methods for determining if two quadratic integer rings are isomorphic. One way is to check if they have the same number of elements. Another method is to compare the characteristic polynomials of the rings, which should be identical if they are isomorphic.

4. What is the significance of showing when quadratic integer rings are isomorphic?

Showing when quadratic integer rings are isomorphic can help us understand the properties and relationships between different mathematical structures. It can also be useful in solving mathematical problems and making connections between seemingly unrelated concepts.

5. How is the isomorphism of quadratic integer rings related to the concept of symmetry?

The isomorphism of quadratic integer rings is related to symmetry because it shows that two structures that may appear different on the surface can actually have the same underlying structure. This is similar to how symmetrical objects may look different, but have the same structure when rotated or reflected.

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